# AIM : FREE EDUCATION TO ALL APUL 

## CHAPTER

21

## Vector Algebra

## Chapter Highlights

Scalars and vectors, Representation of Vectors, Types of vectors, Equal Vectors, Fixed Vectors, Free Vectors, Angle between Two Vectors, Addition (sum or resultant) of Two Vectors, Position Vector, Component of a Vector, Linear Combination, Linearly Dependent and Independent System of Vectors, Collinearity of Three Points, Coplanarity of Four Points, Some Results on linearly dependent and independent Vectors, Product Of Two Vectors, Scalar Product of Two Vectors, Some Useful identities, Work done by a Force, Vector Product of Two Vectors, Moment of a Force about a point, Scalar Triple Product.

## SCALARS AND VECTORS

Scalar Quantity A quantity which has only magnitude and no direction is called a scalar quantity or simply a scalar.

Examples of scalars are mass, temperature, volume, work and so on. To specify a scalar, two things are needed.

1. a unit in terms of which it is measured
2. a real number (+ve, -ve or zero)

Vector Quantity A quantity which has magnitude as well as direction is called a vector quantity or simply a vector.

Examples of vectors are displacement, velocity, acceleration, force and so on. To specify a vector, three things are needed.

1. a unit in terms of which it is measured
2. a real number (+ve, -ve or zero)
3. a particular direction

## REPRESENTATION OF VECTORS

The best way to represent a vector is with the help of a directed line segment. Suppose $A$ and $B$ are two points, then by the vector $\vec{A} B$. , we mean a quantity whose magnitude is the length $A B$ and whose direction is from $A$ to $B$.
$A$ and $B$ are called the end points of the vector $\vec{A} B$. In particular $A$ is called the initial point and $B$ is called the terminal point.

Sometimes a vector $\vec{A} B$. is expressed by a single letter a (which is always written in bold
type, to distinguish it from a scalar). Sometimes, however, we write the vector $a$ as $\vec{a}$ or $\vec{a}$.

## IMPORTANT POINTS

Every vector $(\overrightarrow{A B})$ has the following three characteristics:
(a) Length: The length of $\overrightarrow{A B}$ is denoted by $|\overrightarrow{A B}|$ or simply $A B$.
(b) Support: The line of unlimited length of which $A B$ is segment is called the support of vector $\overrightarrow{A B}$.
(c) Sense: The sense of $\overrightarrow{A B}$ is from $A$ to $B$ and that of $\overrightarrow{B A}$ will be from $B$ to $A$. The sense of directed line segment is from its initial point to the terminal point.

## Modulus (or Magnitude) of a Vector

The positive real number which is the measure of the length of the vector, is called the modulus, length, magnitude, absolute value or norm of the vector.

The modulus of a vector $a$ or $O A$ is usually denoted by $|a|$ or $|\overrightarrow{O A}|$ or by the corresponding letter ' $a$ ' (not in bold-faced type), i.e.,

$$
|\overrightarrow{O A}|=O A \text { and }|a|=a
$$

## Multiplication of a Vector by a Scalar

The product of a scalar $m$ and a vector $a$, is defined as a vector $m a$ or $a m$ whose magnitude is the product of the magnitudes of $m$ and $a$ and whose direction is that of $a$ or opposite to $a$ accordingly as $m$ is positive or negative.

---

# 21.2 ATM : FREE EDUCATION TO ALL APUL 

## TYPES OF VECTORS

Equal Vectors Two vectors $a$ and $b$ are equal when they have (1) the same magnitude and (2) the same direction. Symbolically such vectors are written as: $a=b$.
Unit Vectors A vector whose magnitude is unity is called a unit vector. The unit vector having the same direction as that of given vector $a$ is usually denoted by the symbol $\hat{a}$ (read as ' $a$ cap'), i.e.,

A vector $=$ Modulus of vector $\times$ Unit vector in its direction
or

$$
a=|a| \hat{a}
$$

Also, Unit vector in a direction

$$
\begin{aligned}
& =\frac{\text { vector in that direction }}{\text { modulus of vector }} \\
\hat{a} & =\frac{a}{|a|}
\end{aligned}
$$

## CAUTION

No units are to be attached with a unit vector, i.e., unit vector is dimensionless physical quantity

Zero or Null Vector A vector whose magnitude is zero, is called a zero vector. For such a vector, initial and terminal points are coincident so that its direction is indeterminate. A zero vector is denoted by the bold-faced symbol $O$ or $O$.
Collinear (or Parallel Vectors) The vectors which are parallel to the same straight line are called collinear vectors.

Vectors which are not parallel to the same line are called non-collinear vectors.
Like and Unlike Vectors Collinear vectors having the same direction are called like vectors and those having the opposite directions are called unlike vectors.
Remark: If two vectors $a$ and $b$ are collinear, then there exists a scalar $m$ such that $b=m a, m$ being positive or negative according as $a$ and $b$ are like or unlike vectors.

Conversely, if $b=m a$ be given, then $a$ and $b$ must be collinear (or parallel) vectors such that $|b|=|m||a|$.
Reciprocal Vector Let $|a|$ be the modulus of the given vector $a$. Then a vector whose direction is that of $a$ but modulus is $1 /|a|$ (reciprocal of the modulus of $a$ ) is called the reciprocal of $a$ and is written as $a^{-1}$. Thus,

$$
a^{-1}=\frac{1}{|a|} \hat{a}=\frac{|a|}{|a|^{2}} \hat{a}=\frac{a}{|a|^{2}}
$$

Coplanar and Non-coplanar Vectors Three or more vectors are said to be coplanar when they are parallel to the same plane. Otherwise they are said to be non-coplanar vectors.

Co-initial Vectors The vectors which have the same initial point are called co-initial vectors.
Negative of a Vector A vector having the same modulus as that of a given vector $a$ and the direction opposite to that of $a$, is called the negative of $a$ and is denoted by $-a$. Clearly, if $O A=a$, then

$$
A O=-a, \text { and therefore, } O A=-A O
$$

## EQUAL VECTORS

Two vectors $a$ and $b$ are said to be equal when they have equal magnitudes and same direction. Geometrically, if head of one vector coincides with, the head of other and so do the tails coincide then the vectors are said to be equal.

## CAUTION

If $a=b$, then $a=b$, always.
But if, $a=b$ doesn't always imply $a=b$

## FIXED VECTORS

Fixed vector is that vector whose initial point or tail is fixed. It is also called localised vector. For example, position vector and displacement vector are fixed vectors.

## FREE VECTORS

Free vector is that whose initial point or tail is not fixed. It is also known as non localised vector, For example, velocity vector of a particular moving particle along a straight line is a free vector.
![img-0.jpeg](img-0.jpeg)

Fig. 21.2

## ANGLE BETWEEN TWO VECTORS

The angle between two vectors $a$ and $b$ represented by $O A$ and $O B$, is defined as the angle $A O B$ which does not exceed $\pi$. This is also known as the inclination of given vectors $a$ and $b$. If the angle $A O B$ be $\theta$, then $0 \leq \theta \leq \pi$.

---

![img-1.jpeg](img-1.jpeg)

Fig 21.3
If $\theta=\frac{\pi}{2}$, then vectors are said to be perpendicular or orthogonal and if $\theta=0$ or $\pi$, then vectors are said to be parallel or coincident.

## IMPORTANT POINTS

Whenever finding angle between two vectors, make sure that either their heads coincide or their tails coincide.
![img-2.jpeg](img-2.jpeg)
i.e., if heads coincide or tails coincide then internal angle is the angle between two vectors (whether acute or obtuse) as in (1), (2), (3) and (4).

If heads coincide with tall then external angle is the angle between the two vectors as in (5) and (6).
![img-3.jpeg](img-3.jpeg)

## ADDITION (SUM OR RESULTANT) OF TWO VECTORS

![img-4.jpeg](img-4.jpeg)

Fig 21.6
Let $a, b$ be two vectors. Take any point $O$ and draw the vectors $O A=a$ and $A B=b$ such that the terminal point of the vector $a$ is the initial point of vector $b$. Join $O B$. Then the vector $O B$ is defined as the sum of $a$ and $b$ and is written as

$$
O B=O A+A B=a+b
$$

This method of addition of vectors is known as the triangle law of addition.

Completing the parallelogram $O A B C$. Since

$$
\begin{gathered}
A B=O C=b \\
O B=O A+A B=O A+O C
\end{gathered}
$$

That is, the sum of two co-initial vectors is the vector represented by the diagonal of the parallelogram formed with the component vectors as adjacent sides.

This method of addition of vectors is known as the parallelogram law of addition.
Remark From Equation (1), $-B O=O A+A B$ or $O A+A B$ $+B O=O$, showing that the sum of vectors determined by the sides of a triangle, taken in order, is zero.

## Properties of Vector Addition

1. Vector addition is commutative For any two vectors $a$ and $b$, we have

$$
a+b=b+a
$$

2. Vector addition is associative For any three vectors $a, b$ and $c$, we have

$$
(a+b)+c=a+(b+c)
$$

3. Existence of additive identity For every vector $a$, we have

$$
a+O=a=O+a
$$

where $O$ is the null vector.
4. Existence of additive inverse For a given vector $a$, there exists a vector $-a$ such that

$$
a+(-a)=(-a)+a=0
$$

The vector $-a$ is called the additive inverse of $a$.

## Properties of Multiplication of Vector by a Scalar

1. If $m=o$, then $m a=o$
2. If $m$ and $n$ be two scalars, then

$$
m(n a)=m n a=n(m a)
$$

3. If $m$ and $n$ be two scalars, then

$$
(m+n) a=m a+n a
$$

4. If $a, b$ are any two vectors and $m$ be any scalar, then

$$
m(a+b)=m a+m b
$$

## Subtraction (or Difference) of Two Vectors

Subtraction of vectors If $a$ and $b$ are two vectors, then their subtration $a-b$ is defined as $a-b=a+(-b)$ where $-b$ is the negative of $b$ having magnitude equal to that of $b$ and direction opposite to $b$.

---

Then $a-b=\left(a_{1}-b_{1}\right) i+\left(a_{2}-b_{2}\right) j+\left(a_{3}-b_{3}\right) k$
![img-5.jpeg](img-5.jpeg)

Fig 21.7

## Properties of Vector Subraction

(i) $a-b \neq b-\mathrm{a}$
(ii) $(a-b)-c \neq a-(b-c)$
(iii) Since any one side of a triangle is less than the sum and greater than the difference of the other two sides, so for any two vectors $a$ and $b$, we have
(A) $|a+b| \leq|a|+|b|$
(B) $|a+b| \geq|a|-|b|$
(C) $|\mathrm{a}-b| \leq|a|+|b|$
(D) $|a-b| \geq|a|-|b|$
![img-6.jpeg](img-6.jpeg)

Fig 21.8

## POSITION VECTOR

If a point $O$ is fixed as the origin in space (or plane) and $P$ is any point, the $O P$ is called the position vector of $P$ with respect to $O$.

If we say that $P$ is the point $r$, then we mean that the position vector of $P$ is $r$ with respect to some origin $O$.

1. $A B$ in terms of the position vectors of points $A$ and $B$ : If $a$ and $b$ are position vectors of points $A$ and $B$ respectively. Then $O A=a, \mathrm{OB}=b$
$\therefore A B=$ (Position vector of $B$ ) - (Position vector of $A$ )

$$
=O B-O A=b-a
$$

2. Position vector of a dividing point The position vectors of the points dividing the line $A B$ in the ratio $m: n$ internally or externally are $\frac{m b+n a}{m+n}$
or $\frac{m b-n a}{m-n}$

## TRICK(S) FOR PROBLEM SOLVING

- If $P$ is the mid-point of $A B$, then it divides $A B$ in the ratio $1: 1$. Thus position vector of $P$ is given by

$$
O P=\frac{a+b}{2}
$$

- Since $O P=r=\frac{m b+n a}{m+n}=\frac{m}{m+n} b+\frac{n}{m+n} a$
$=\lambda a+\mu b$
where $\lambda=\frac{n}{m+n}$ and $\mu=\frac{m}{m+n}$
Thus, p.v. of any point $P$ on $A B$ can always be taken as $r=\lambda a$ $+\mu b$, where $\lambda+\mu=1$.
- Since $O P=r=\frac{m b+n a}{m+n}$
$\Rightarrow \quad(m+n) r=m b+n a$
$\Rightarrow \quad n \cdot O A+m \cdot O B=(n+m) O P$,
where $P$ is a point on $A B$ dividing it in the ratio $m: n$.
In particular, if $P$ is the mid-point of $A B$, then

$$
O A+O B=2 O P
$$

That is, twice the position vector of the middle point is equal to the sum of the vectors of the ends.

## SOLVED EXAMPLES

1. In a regular hexagon $A B C D E F$,
$A B+A C+A D+A E+A F=k A D$, where $k$ is equal to
(A) 1
(B) 2
(C) 3
(D) none of these

## Solution (C)

$A B+A C+A D+A E+A F$
$=E D+A C+A D+A E+C D(\because A B=E D, A F=C D)$
![img-7.jpeg](img-7.jpeg)
$=(A C+C D)+(A E+E D)+A D$
$=A D+A D+A D=3 A D$
2. If $M$ and $N$ are the mid points of the diagonals $A C$ and $B D$ respectively of a quadrilateral $A B C D$, then $A B+A D+C B+C D=$
(A) 4 NM
(B) $4 M N$
(C) $2 M N$
(D) none of these

## Solution (B)

In $\triangle A B D, N$ is the mid-point of $B D$,
$\therefore \quad A B+A D=2 A N$

---

![img-8.jpeg](img-8.jpeg)

In $\triangle C B D, N$ is the mid point of $B D$,

$$
\therefore \quad C B+C D=2 C N
$$

Adding (1) and (2), we have

$$
A B+A D+C B+C D=2(A N+C N)
$$

In $\triangle A N C, M$ is the mid-point of $A C$

$$
\therefore \quad A N+C N=2 M N
$$

From (3), we get

$$
A B+A D+C B+C D=2(2 M N)=4 M N
$$

3. Five forces $A B, A C, A D, A E, A F$ act at the vertex $A$ of a regular hexagon $A B C D E F$. If $O$ is the centroid of the hexagon, then their resultant is a force given by
(A) $4 A O$
(B) $5 A O$
(C) $6 A O$
(D) none of these

Solution (C)
If $R$ is the resultant of given forces, then
![img-9.jpeg](img-9.jpeg)

$$
\begin{aligned}
R & =A B+A C+A D+A E+A F \\
& =E D+A C+A D+A E+C D \\
& =(A C+C D)+(A E+E D)+A D \\
& =A D+A D+A D=3 A D=6 A O
\end{aligned}
$$

4. $A B C D$ is parallelogram. If $L$ and $M$ are the middle points of $B C$ and $C D$, then $A L+A M=$
(A) $\frac{1}{2} A C$
(B) $\frac{3}{2} A C$
(C) $A C$
(D) none of these

Solution (B)

$$
\begin{aligned}
A L & =A B+B L=A B+\frac{1}{2} B C=A B+\frac{1}{2} A D \\
A M & =A D+D M=A D+\frac{1}{2} D C=A D+\frac{1}{2} A B
\end{aligned}
$$

![img-10.jpeg](img-10.jpeg)

Adding, $A L+A M=\frac{3}{2}(A B+A D)$

$$
=\frac{3}{2}(A B+B S)=\frac{3}{2} A C
$$

5. $A B C D$ is a quadrilateral and $E$ the point of intersection of the lines joining the middle points of opposite sides. If $O$ is any point, then the resultant of $O A, O B, O C$ and $O D$ is equal to
(A) $2 O E$
(B) $O E$
(C) $4 O E$
(D) none of these

Solution (C)
Let $P, Q, R, S$ be the mid-points of sides $B C, C D, D A$, $A B$ respectively of a quadrilateral $A B C D$. By geometry the figure formed by joining the mid-points $P, Q$, $R, S$ will be a parallelogram. Hence, its diagonals will bisect each other, say at $E$.
![img-11.jpeg](img-11.jpeg)

Now, $P$ is the mid-point of $B C$

$$
\therefore \quad O B+O C=2 O P
$$

And $R$ is the mid-point of $A D$

$$
\therefore \quad O A+O D=2 O R
$$

Adding (1) and (2), we have
$O A+O B+O C+O D=2(O P+O R)=2 \cdot 2 O E=4 O E$
( $\because E$ is mid-point of $P R, \therefore O P+O R=2 O E$ )
6. Two forces act at the vertex $A$ of a quadrilateral $A B C D$ represented by $A B, A D$ and two at $C$ represented by $C B$ and $C D$. If E and F are the middle points of $A C$ and $B D$ respectively, then their resultant is represented by
(A) $E F$
(B) $2 E F$
(C) $\frac{3}{2} E F$
(D) $4 E F$

Solution (D)
We have,
$A B+A D=2 A F$, where F is mid-point of $B D$

---

![img-12.jpeg](img-12.jpeg)

Also, $C B+C D=2 C F$;
$\therefore \quad A B+A D+C B+C D$
$=2(A F+C F)=-2(F A+F C)=-2[2 F E]$,
where $E$ is the mid-point of $A C=-4 F E=4 E F$.
7. Let $O A=i+3 j-2 k$ and $O B=3 i+j-2 k$. The vector $O C$ bisecting the angle $A O B$ where $C$ is a point on the line $A B$ is
(A) $2(i+j-k)$
(B) $4(i+j-k)$
(C) $i+j-k$
(D) none of these

## Solution (A)

Taking $O$ as origin, the position vectors of $A$ and $B$ are $a=i+3 j-2 k$ and $b=3 i+j-2 k$ respectively. We have $|a|=|b|=\sqrt{14}$
![img-13.jpeg](img-13.jpeg)

So, the bisector $O C$ of $\angle A O B$ meets $A B$ at its midpoint $C$.

$$
\therefore \quad O C=\frac{1}{2}(O A+O B)=2(i+j-k)
$$

8. The vector $c$, directed along the internal bisector of the angle between the vectors $a=7 i-4 j-4 k$ and $b=-2 i$ $-j+2 k$ with $|c|=5 \sqrt{6}$ is
(A) $\frac{5}{3}(5 i+5 j+2 k)$
(B) $\frac{5}{3}(i+7 j+2 k)$
(C) $\frac{5}{3}(-5 i+5 j+2 k)$
(D) $\frac{5}{3}(i-7 j+2 k)$

## Solution (D)

The required vector $\mathbf{c}$ is given by

$$
\begin{aligned}
c & =\lambda(a+b)=\lambda\left(\frac{a}{|a|}+\frac{b}{|b|}\right) \\
& =\lambda\left\{\frac{1}{9}(7 i-4 j-4 k)+\frac{1}{3}(-2 i-j+2 k)\right\}
\end{aligned}
$$

or, $\quad c=\frac{\lambda}{9}(\hat{i}-7 j+2 k)$
$\Rightarrow \quad|c|= \pm \frac{\lambda}{9} \sqrt{1+49+4}= \pm \frac{\lambda}{9} \sqrt{54}$.

But $|c|=5 \sqrt{6}$
$\therefore \quad \pm \frac{\lambda}{9} \sqrt{54}=5 \sqrt{6} \Rightarrow \lambda= \pm 15$.
Hence, $c= \pm \frac{15}{9}(i-7 j+2 k)= \pm \frac{5}{3}(i-7 j+2 k)$
9. If the points $P, Q, R, S$ have position vectors $p, q, r, s$ such that $p-q=2(s-r)$, then
(A) $P Q$ and $R S$ bisect each other
(B) $P Q$ and $P R$ bisect each other
(C) $P Q$ and $R S$ trisect each other
(D) $Q S$ and $P R$ trisect each other

## Solution (D)

We have, $p-q=2(s-r)$
$\Rightarrow \quad p+2 r=q+2 s \Rightarrow \frac{p+2 r}{1+2}=\frac{q+2 s}{1+2}$
$\therefore \quad$ Point dividing $P R$ in the ratio $2: 1$ is same as the point dividing $Q S$ in the ratio $2: 1$
Hence $Q S$ and $P R$ trisect each other.
10. If $a$ and $b$ are position vectors of $A$ and $B$ respectively, then the position vector of a point $C$ in $A B$ produced such that $A C=3 A B$ is
(A) $3 a-b$
(B) $3 b-a$
(C) $3 a-2 b$
(D) $3 b-2 a$

## Solution (D)

$A C=3 A B \quad \Rightarrow \quad c-a=3(b-a) \quad \Rightarrow \quad c=3 b-2 a$.
11. A vector $a$ has components $2 p$ and 1 w.r.t a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If w.r.t the new system, $a$ has components $p+1$ and 1 , then
(A) $p=0$
(B) $p=1$ or $p=-\frac{1}{3}$
(C) $p=-1$ or $p=\frac{1}{3}$
(D) $p=1$ or $p=-1$

## Solution (B)

Let $i, j$ be unit vectors along the co-ordinate axes
$\therefore \quad a=2 p i+1 \cdot j$
On rotation, let $b$ be the vector having components $p+1$ and 1.
$\therefore \quad b=(p+1) i+1 \cdot j$
where $i, j$ are unit vectors along the new co-ordinate axes.
But on rotation $|b|=|a| \quad \Rightarrow \quad|b|^{2}=|a|^{2}$
$\Rightarrow(p+1)^{2}+1=(2 p)^{2}+1 \quad \Rightarrow \quad 3 p^{2}-2 p-1=0$
$\Rightarrow(3 p+1)(p-1)=0 \quad \Rightarrow \quad p=1$ or $-\frac{1}{3}$

---

12. Let $A B C D E F$ be a regular hexagon. If $A D=x B C$ and $C F=y A B$, then $x y=$
(A) 4
(B) -4
(C) 2
(D) -2

## Solution (B)

Since $A B C D E F$ is a regular hexagon, from plane geometry, we have
![img-14.jpeg](img-14.jpeg)

$$
\begin{aligned}
& A D=2 B C \text { and } \mathrm{FC}=2 A B \\
& \therefore \quad A D=2 B C \text { and } F C=2 A B
\end{aligned}
$$

Given that $A D=x B C$.

$$
\begin{aligned}
& \therefore \quad 2 B C=x B C \text {, by (1) } \\
& \Rightarrow \quad x=2
\end{aligned}
$$

Again, given that $C F=y A B$ or $-F C=y A B$.

$$
\begin{aligned}
& \therefore \quad-2 A B=y A B \text {, using (2) } \\
& \Rightarrow \quad y=-2
\end{aligned}
$$

From (2) and (3), $x y=2(-2)=-4$.
13. A vector $a$ is collinear with vector $6 i-8 j-\frac{15}{2} k$ of magnitude 50 making an obtuse angle with $z$-axis is
(A) $24 i-32 j-30 k$
(B) $-24 i+32 j+30 k$
(C) $24 i+32 j-30 k$
(D) none of these

## Solution (A)

Let $b=6 i-8 j-\frac{15}{2} k$.
A unit vector along $b$ is $\pm \frac{2}{25}(6 i-8 j-\frac{15}{2} k)$.
$\therefore \quad a=$ a vector of length 50 along $b$

$$
= \pm 4\left(6 i-8 j-\frac{15}{2} k\right)
$$

Since $a$ makes obtuse angle with $z$-axis, so we must have

$$
a \cdot k<0
$$

Thus, $a=24 i-32 j-30 k$
14. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with lower base $A B C D$, upper base $A_{1} B_{1} C_{1} D_{1}$ and the lateral edges $A A_{1}, B B_{1}$, $C C_{1}$, and $D D_{1} ; M$ and $M_{1}$ are the centres of the faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ respectively. $O$ is a point on line $M M_{1}$, such that
$O A+O B+O C+O D=O M_{1}$, then $O M=\lambda . O M_{1}$ if $\lambda=$
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C) $\frac{1}{6}$
(D) $\frac{1}{8}$

## Solution (A)

$$
\begin{aligned}
O M_{1} & =O A+O B+O C+O D(\text { given }) \\
& =O M+M A+O M+M B+O M+M C \\
& +O M+M D \\
& =4 O M+(M A+M C)+(M B+M D) \\
& =4 O M(\because M A=-M C, M B=-M D) \\
\therefore \quad & O M=\frac{1}{4} O M_{1} \quad \therefore \quad \lambda=\frac{1}{4}
\end{aligned}
$$

15. In a trapezoid the vector $B C=\lambda A D$ and $p=\mu A D$ and $p=A C+B D$ is collinear with $A D$. Then
(A) $\mu=\lambda+1$
(B) $\lambda=\mu+1$
(C) $\lambda+\mu=1$
(D) $\mu=2+\lambda$

## Solution (A)

We have,
$p=A C+B D=A C+B C+C D=A C+\lambda A D+C D$
$=(A C+C D)+\lambda A D=A D+\lambda A D=(1+\lambda) A D$
Since $p=\mu A D \quad \therefore \quad \mu=1+\lambda$
16. $A B=3 i+j-k$ and $A C=i-j+3 k$. If the point $P$ on the line segment $B C$ is equidistant from $A B$ and $A C$, then $A P$ is
(A) $2 i-k$
(B) $i-2 k$
(C) $2 i+k$
(D) none of these

## Solution (C)

A point equidistant from $A B$ and $A C$ is on the bisector of the angle $B A C$.
A vector along the internal bisector of the angle $B A C$

$$
\begin{aligned}
& =\frac{A B}{|A B|}+\frac{A C}{|A C|} \\
& =\frac{3 i+j-k}{\sqrt{9+1+1}}+\frac{i-j+3 k}{\sqrt{1+1+9}}=\frac{1}{\sqrt{11}}(4 i+2 k) \\
\therefore \quad & A P=t(2 i+k) \\
\therefore \quad & B P=A P-A B=t(2 i+k)-(3 i+j-k) \\
& =(2 t-3) i-j+(t+1) k \\
& \text { Also } \quad B C=A C-A B=(i-j+3 k)-(3 i+j-k) \\
& =-2 i-2 j+4 k . \\
& \text { But } \quad B P=s B C . \\
& \therefore \quad(2 t-3) i-j+(t+1) k=s(-2 i-2 j+4 k) \\
& \therefore \quad 2 t-3=-2 s,-1=-2 s, t+1=4 s \\
& \therefore \quad s=\frac{1}{2} \text { and } t=1 \quad \therefore \quad A P=2 i+k .
\end{aligned}
$$

17. The position vectors $a, b, c$ and $d$ of four distinct points $A, B, C$ and $D$ lie on a plane are such that $|a-d|=$ $|b-d|=|c-d|$ then the point $D$ is the

---

(A) centroid of $\triangle A B C$
(B) orthocentre of $\triangle A B C$
(C) circumcentre of $\triangle A B C$
(D) none of these

## Solution (C)

We have, $a-d=-A D, b-d=-B D$ and $c-d=-C D$. According to the given condition $|A D|=|B D|=|C D|$. Thus, $D$ is the circumcentre of $\triangle A B C$.

## COMPONENT OF A VECTOR

The component of a vector $P Q$ on a line $l$ is $R S$, where $R$ and $S$ are the feet of perpendiculars from $P$ and $Q$ on the line $l$.

The vector component of $P Q$ on $l$ will be denoted by $R S$.

1. If $\theta$ is the angle between $P Q$ and $R S$, then the component of $P Q$ on $l=P Q \cos \theta=|P Q| \cos \theta$ and the vector component of $P Q$ on $l=P Q \cos \theta$.
2. If $r$ is the position vector of a point $P$, having coordinates $(x, y, z)$ then $r=x i+y j+z k$, where $i, j, k$ are unit vectors along $x, y$ and $z$ axes respectively.
3. $x i, y j, z k$ are the vector components of $r$ on $x, y$ and $z$ axes respectively.
4. If a point $P$ in space has coordinates $(x, y, z)$, then its $p . v$. $r$ is $x i+y j+z k$ and $|r|=|r|=\sqrt{x^{2}+y^{2}+z^{2}}$.

## SOLVED EXAMPLE

18. The triangle $A B C$ is defined by the vertices $A(1,-2$, 2), $B(1,4,0)$ and $C(-4,1,1)$. Let $M$ be the foot of the altitude drawn from the vertex $B$ to side $A C$. Then $B M$ $=$
(A) $(-20 / 7,-30 / 7,10 / 7)$
(B) $(-20,-30,10)$
(C) $(2,3,-1)$
(D) none of these

## Solution (A)

Since $M B$ is the component of $A B \perp$ to $A C$
![img-15.jpeg](img-15.jpeg)
$M B=A B-A M$
$=A B-\frac{(A B \cdot A C) A C}{(A C)^{2}}$
$=(6 j-2 k)-\frac{\{(6 j-2 k) \cdot(-5 i+3 j-k)\}(-5 i+3 j-k)}{(25+9+1)}$

$$
\begin{aligned}
& =(6 j-2 k)-\frac{20}{35}(-5 i+3 j-k)=\frac{10}{7}(2 i+3 j-k) \\
& \therefore \quad B M=-\frac{10}{7}(2 i+3 j-k)
\end{aligned}
$$

## LINEAR COMBINATION

A vector $r$ is said to be a linear combination of the given vectors $a, b, c \ldots$, and so on if there exist a system of scalars $x, y, z, \ldots$, and so on such that

$$
r=x a+y b+z c+\ldots
$$

## LINEARLY DEPENDENT AND INDEPENDENT SYSTEM OF VECTORS

The system of $n$ vectors $a_{1}, a_{2}, \ldots, a_{n}$ is said to be linearly dependent, if there exist scalars $x_{1}, x_{2}, \ldots, x_{n}$ not all zero such that

$$
x_{1} a_{1}+x_{2} a_{2}+\ldots+x_{n} a_{n}=0
$$

The same system of vectors is said to be linearly independent, if all scalers are zero, i.e. $x_{1}=x_{2}=\ldots=x_{n}=0$.

## REMARK

When the system of $n$ vectors $a_{1}, a_{2}, \ldots, a_{n}$ is linearly dependent (or independent), then $n$ vectors $a_{1}, a_{2}, \ldots, a_{n}$ are said to be linearly dependent (or independent).

## COLLINEARITY OF THREE POINTS

The necessary and sufficient condition for three points with position vectors $a, b$ and $c$ to be collinear is that there exist three scalars $x, y, z$, not all zero, such that

$$
x a+y b+z c=0, \text { where } x+y+z=0
$$

Test of Collinearity of Two Vectors To prove that two vectors $a$ and $b$ are collinear, find a scalar $m$ such that one of the vectors is $m$ times the other. In case no such scalar $m$ exists, then the two vectors will be non-collinear vectors.

## Test of Collinearity of Three Points

Method 1: To prove that three points $A, B, C$ are collinear, find the vectors $A B$ and $A C$ and show that there exists a scalar $m$ such that $A B=m A C$.

If no such scalar $m$ exists, then the points are not collinear.
Method 2: To prove that three points $A, B, C$ with position vectors $a, b, c$ respectively are collinear, find three scalars $x, y, z$ (not all zero) such that

---

$$
x a+y b+z c=0, \text { where } x+y+z=0
$$

If no such scalars $x, y, z$ exist, then the points are not collinear.

## SOLVED EXAMPLES

19. If the vectors $a$ and $b$ are non-collinear, then the value of $x$, for which, the vectors $c=(x-2) a+b$ and $d=(2 x$ +1 ) $a-b$ are collinear is
(A) $\frac{4}{3}$
(B) $\frac{2}{3}$
(C) $\frac{1}{3}$
(D) none of these

## Solution (C)

The vector $c$ is non-zero since the coefficient in $b$ is different from zero, and so the vectors $c$ and $d$ are collinear if for some number $y$ we have
$d=y c$ that is

$$
(2 x+1) a-b=y(x-2) a+y b
$$

or $(y x-2 y-2 x-1) a+(y+1) b=0$
Since $a, b$ are non-collinear, we must have

$$
y x-2 y-2 x-1=0 \text { and } y+1=0
$$

Solving these equations, we get $y=-1$ and $x=1 / 3$
20. With reference to a right handed system of mutually perpendicular unit vectors $i, j, k \alpha=3 i-j$ and $\beta=2 i+$ $j-3 k$
If $\beta=\beta_{1}+\beta_{2}$, where $\beta_{1}$ is parallel to $\alpha$ and $\beta_{2}$ is perpendicular to $\alpha$, then
(A) $\beta_{1}=\frac{3}{2} i+\frac{1}{2} j$
(B) $\beta_{1}=\frac{3}{2} i-\frac{1}{2} j$
(C) $\beta_{2}=\frac{1}{2} i+\frac{3}{2} j-3 k$
(D) $\beta_{2}=\frac{1}{2} i-\frac{3}{2} j-3 k$

## Solution (B, C)

Since $\beta_{1}$ is parallel to $\alpha$,
let $\beta_{1}=\lambda \alpha$, where $\lambda$ is a scalar.

$$
\begin{aligned}
& \beta=\beta_{1}+\beta_{2} \\
\therefore & \beta_{2}=\beta-\beta_{1} \\
& =2 i+j-3 k-\lambda \alpha=2 i+j-3 k-\lambda(3 i-j) \\
& =(2-3 \lambda) i+(1+\lambda) i-3 k
\end{aligned}
$$

Since $\beta_{2}$ is perpendicular to $\alpha$
$\therefore \beta_{2} \cdot a=0$
$\Rightarrow[(2-3 \lambda) i+(1+\lambda) i-3 i] \cdot(3 i-i)=0$
$\Rightarrow 3(2-3 \lambda)-(1+\lambda)=0$

$$
\begin{aligned}
& \Rightarrow 6-9 \lambda-1-\lambda=0 \Rightarrow 5-10 \lambda=0, \quad \therefore \quad \lambda=\frac{1}{2} \\
& \therefore \quad \beta_{1}=\lambda \alpha=\frac{1}{2}(3 i-i)=\frac{3}{2} i-\frac{1}{2} j \\
& \beta_{2}=\left(2-\frac{3}{2}\right) i+\left(1+\frac{1}{2}\right) j-3 k=\frac{1}{2} i+\frac{3}{2} j-3 k \\
& \therefore \quad 2 i+j-3 k=\lambda \alpha+\beta_{2} \\
& \text { Hence, } \quad \beta=\beta_{1}+\beta_{2}
\end{aligned}
$$

## COPLANARITY OF FOUR POINTS

The necessary and sufficient condition for four points with position vectors $a, b, c$ and $d$ to be coplanar is that there exist scalars $x, y, z$ and $w$, not all zero, such that

$$
x a+y b+z c+w d=0
$$

where $x+y+z+w=0$.
Test of Coplanarity of Three Vectors To prove that three vectors $a, b$ and $c$ are coplanar, express one of these vectors as the linear combination of the other two such as $c=x a+y b$.

Now, compare the coefficients from the two sides and find the values of $x$ and $y$. If real values of scalars $x$ and $y$ exist, then the vectors are coplanar otherwise non-coplanar.

## Test of Coplanarity of Four Points

Method 1: To prove that four points $A, B, C$ and $D$ are coplanar, find the vectors $A B, A C$ and $A D$ and show that these three vectors are coplanar.
Method 2: To prove that four points $A, B, C$ and $D$ with position vectors $a, b, c$ and $d$ respectively are coplanar, find four scalars $x, y, z, w$ (not all zero) such that

$$
x a+y b+z c+w d=0 \text { where } x+y+z+w=0
$$

If no such scalars $x, y, z, w$ exist, then the points are non-coplanar.

## SOME RESULTS ON LINEARLY DEPENDENT AND INDEPENDENT VECTORS

1. If $a, b, c$ are non-coplanar vectors, then these are linearly independent and conversely if $a, b, c$ are linearly independent, then they are non-coplanar.
2. If $a=a_{1} i+a_{2} j+a_{3} k, b=b_{1} i+b_{2} j+b_{3} k$ and $c=c_{1} i$ $+c_{2} j+c_{3} k$ are three linearly dependent vectors, then $\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0$
3. Let $a, b, c$ be three non-coplanar vectors. Then, vectors $x_{1} a+y_{1} b+z_{1} c, x_{2} a+y_{2} b+z_{2} c$ and $x_{3} a+y_{3} b+z_{3} c$ will be coplanar if

---

$$
\left|\begin{array}{lll}
x_{1} & x_{2} & x_{3} \\
y_{1} & y_{2} & y_{3} \\
z_{1} & z_{2} & z_{3}
\end{array}\right|=0
$$

4. Two non-zero, non-collinear vectors are linearly independent.
5. Any two collinear vectors are linearly dependent.
6. Any three non-coplanar vectors are linearly independent.
7. Any three coplanar vectors are linearly dependent.
8. Any four vectors in 3-dimensional space are linearly dependent.

## PRODUCT OF TWO VECTORS

Between two vectors, two distinct kinds of products are defined. One being a pure number is called the scalar product while the other being a vector quantity is called the vector product.

## SCALAR PRODUCT OF TWO VECTORS

The scalar product or dot product of two vectors $a$ and $b$ is defined as: $|a||b| \cos \theta$, where $\theta$ is the angle between them such that $0 \leq \theta \leq \pi$. It is denoted by placing a dot between the vectors $a$ and $b$. Thus,

$$
a \cdot b=|a||b| \cos \theta
$$

If either $a$ or $b$ is $O$, we define $a \cdot b=O$.
![img-16.jpeg](img-16.jpeg)

Fig 21.9

## Key Results on Scalar Product

1. Scalar product is commutative. For any two vectors $a$ and $b$ we have

$$
a \cdot b=b \cdot a
$$

2. If $m$ is any scalar and $a, b$ be any two vectors, then $(m a) \cdot b=m(a \cdot b)=a \cdot(m b)$
3. Scalar product is distributive with respect to vector addition, i.e., for any three vectors $a, b$ and $c$, we have

$$
a \cdot(b+c)=a \cdot b+a \cdot c
$$

4. Magnitude of a vector as a scalar product: For any vector $a$

$$
a \cdot a=|a|^{2}=a^{2}
$$

5. Scalar product of two perpendicular vectors is zero, i.e., if $a$ and $b$ are two perpendicular vectors, then $a \cdot b=0$.
However, if $a \cdot b=0 \Rightarrow$ Either $a=0$ or $b=0$ or $a \perp b$.
6. Scalar product of mutually orthogonal unit vectors $i$, $j, k$ :

$$
i \cdot i=1=j \cdot j=k \cdot k
$$

and $\quad i \cdot j=j \cdot k=k \cdot i=0$
7. Scalar porduct of two vectors in terms of components: If
$a=a_{1} i+a_{2} j+a_{3} k$ and $b=b_{1} i+b_{2} j+b_{3} k$, then $a \cdot b=$ $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}$
Thus, the scalar product of two vectors is equal to the sum of the products of their corresponding components.
8. Angle between two vectors in terms of the components of the given vectors.
If $\theta$ is the angle between two vectors
$a=a_{1} i+a_{2} j+a_{3} k$ and $b=b_{1} i+b_{2} j+b_{3} k$, then
$\cos \theta=\frac{a \cdot b}{|a||b|}=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}$

## IMPORTANT POINTS

If $\theta$ is acute, $a . b$ is positive and if $\theta$ is obtuse, $a . b$ is negative
9. Components of a vector $b$ along and perpendicular to vector $a$
Component of $b$ along $a=\left(\frac{a \cdot b}{|a|^{2}}\right) a$
Component of $b$ perpendicular to $a=b-\left(\frac{a \cdot b}{|a|^{2}}\right) a$
10. Any vector $r$ can be expressed as:

$$
r=(r \cdot i) i+(r \cdot j) j+(r \cdot k) k
$$

## SOME USEFUL IDENTITIES

Since scalar product satisfies commutative and distributive laws, we have

1. $(a+b)^{2}=a^{2}+b^{2}+2 a \cdot b$
2. $(a-b)^{2}=a^{2}+b^{2}-2 a \cdot b$
3. $(a+b) \cdot(a-b)=a^{2}-b^{2}$

## WORK DONE BY A FORCE

Work done by a force $F$ in displacing a particle from $A$ to $B$ is defined by

$$
W=F \cdot A B
$$

---

IMPORTANT POINTS

- $(\alpha+b+c)^{2}=|\alpha|^{2}+|b|^{2}+|c|^{2}+2 \mid(\alpha \cdot b+b \cdot c+c \cdot$ $\alpha)$
- $|\vec{i}+\vec{j}+\vec{k}|=\sqrt{3}$
- Cauchy - Schwarz Inequality $(\alpha \cdot b) \leq|\alpha|^{2}|b|^{2}$
- If a number of forces are acting on a particle, then the sum of the works done by the separate forces is equal to the work done by the resultant force


## SOLVED EXAMPLES

21. If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest value of $|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|$ is
(A) $2 \sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) $4 \sqrt{2}$

## Solution (A)

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.
Then, $\vec{a} \cdot \vec{b}=\cos \theta$
Now, $|\vec{a}+\vec{b}|=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=2+2 \cos \theta$

$$
=4 \cos ^{2} \frac{\theta}{2}
$$

$\Rightarrow|\vec{a}+\vec{b}|=2 \cos \frac{\theta}{2}$.
Similarly, $|\vec{a}-\vec{b}|=2 \sin \frac{\theta}{2}$
$\therefore|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=2\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right) \leq 2 \sqrt{2}$
22. If $a+b+c=0$ and $|a|=3,|b|=5,|c|=7$, then the angle between $a$ and $b$ is
(A) $\frac{\pi}{2}$
(B) $\frac{\pi}{3}$
(C) $\frac{\pi}{4}$
(D) $\frac{\pi}{6}$

## Solution (B)

Let $\theta$ be the angle between $a$ and $b$

$$
\begin{aligned}
& \therefore \quad a \cdot b=|a||b| \cos \theta \\
& \therefore \quad \cos \theta=\frac{a \cdot b}{|a||b|}=\frac{a \cdot b}{(3)(5)}=\frac{a \cdot b}{15}
\end{aligned}
$$

Now $a+b+c=0$

$$
\begin{aligned}
& \therefore \quad a+b=-c \\
& \Rightarrow|a+b|=|-c|=|c| \Rightarrow|a+b|^{2}=|c|^{2} \\
& \Rightarrow(a+b) \cdot(a+b)=(7)^{2} \Rightarrow|a|^{2}+2 a \cdot b+|b|^{2}=49 \\
& \Rightarrow(3)^{2}+2 a \cdot b+(5)^{2}=49
\end{aligned}
$$

$\Rightarrow 2 a \times b=49-9-25=15 \Rightarrow a \times b=\frac{15}{2}$
$\therefore$ From (1), we get
$\cos \theta=\frac{15}{2 \times 15}=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=60^{\circ}$
23. $\vec{a}$ and $\vec{c}$ are unit vectors and $|\vec{b}|=4$. The angle between $\vec{a}$ and $\vec{c}$ is $\cos ^{-1}\left(\frac{1}{4}\right)$. If $\vec{b}-2 \vec{c}=\lambda \vec{a}$, then $\lambda$ is equal to
(A) 3,4
(B) $-3,4$
(C) $3,-4$
(D) $\frac{1}{4}, \frac{3}{4}$

## Solution (C)

Given: $|\vec{a}|=1,|\vec{c}|=1$ and $|\vec{b}|=4$.
$\Rightarrow|\vec{a} \cdot \vec{c}|=1 \cdot 1 \cdot \frac{1}{4}=\frac{1}{4}$
Now, $\vec{b}-2 \vec{c}=\lambda \vec{a} \Rightarrow \vec{a} \cdot \vec{b}-2 \vec{a} \cdot \vec{c}=\lambda a^{2}$

$$
\begin{aligned}
& \Rightarrow \vec{a} \cdot \vec{b}-2 \cdot \frac{1}{4}=\lambda \\
& \Rightarrow \vec{a} \cdot \vec{b}=\lambda+\frac{1}{2}
\end{aligned}
$$

Again, $\vec{b}-2 \vec{c}=\lambda \vec{a} \Rightarrow \vec{b} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}=\lambda \vec{b} \cdot \vec{a}$

$$
\begin{aligned}
& \Rightarrow 16-2 \vec{b} \cdot \vec{c}=\lambda\left(\lambda+\frac{1}{2}\right) \\
& \Rightarrow \vec{b} \cdot \vec{c}=8-\frac{\lambda^{2}}{2}-\frac{\lambda}{4}
\end{aligned}
$$

Also, $\vec{b}-2 \vec{c}=\lambda \vec{a} \Rightarrow \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{c}=\lambda \vec{b} \cdot \vec{a}$

$$
\begin{aligned}
& \Rightarrow 8-\frac{\lambda^{2}}{2}-\frac{\lambda}{4}-2(1)=\lambda\left(\frac{1}{4}\right) \\
& \Rightarrow \lambda^{2}+\lambda-12=0
\end{aligned}
$$

$\therefore \lambda=-4,3$
24. The vectors $a=x i-3 j-k$ and $b=2 x i+x j-k$ include an acute angle and $b$ and positive $y$-axis include an obtuse angle. Then values of $x$ may be
(A) -2
(B) -3
(C) all $x<0$
(D) all $x>0$

## Solution (A, B, C)

According to the question

$$
a \cdot b>0 \text { and } b \cdot j<0 \Rightarrow 2 x^{2}-3 x+1>0 \text { and } x<0
$$

$\Rightarrow(2 x-1)(x-1)>0$ and $x<0$
i.e., $\left(x<\frac{1}{2}\right.$ or $\left.x>1\right)$ and $x<0 \Rightarrow x<0$
25. If the unit vectors $a$ and $b$ are inclined at angle $2 \theta(0 \leq$ $\theta \leq \pi)$ and $|a-b|<1$, then $\theta$ lies in the interval

---

(A) $\left[0, \frac{\pi}{6}\right)$
(B) $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$
(C) $\left[\frac{\pi}{2}, \frac{5 \pi}{6}\right]$
(D) none of these

Solution (A)
$a \cdot b=|a||b| \cos 2 \theta$
$\Rightarrow a \cdot b=(1)(1) \cos 2 \theta=\cos 2 \theta$.

$$
|a-b|<1
$$

$\Rightarrow a^{2}+b^{2}-2 a \cdot b<1 \Rightarrow 1+1-2 \cos \theta<1$
$\Rightarrow 2(1-\cos 2 \theta)<1 \Rightarrow 2\left(2 \sin ^{2} \theta\right)<1$
$\Rightarrow \sin ^{2} \theta<\frac{1}{4} \Rightarrow \theta$ lies in $\left[0, \frac{\pi}{6}\right)$
26. If $a, b, c$ are three vectors such that $|a|=3,|b|=4$, $|c|=5$ and $a, b, c$ are perpendicular to $b+c, c+a, a$ $+b$ respectively, then $|a+b+c|=$
(A) $6 \sqrt{2}$
(B) $4 \sqrt{2}$
(C) $3 \sqrt{2}$
(D) $5 \sqrt{2}$

Solution (D)
$\because a \perp(b+c)$
$\therefore \quad a \cdot(b+c)=0 \Rightarrow a \cdot b+a \cdot c=0$
Similarly $\quad b \perp(c+a) \Rightarrow b \cdot c+b \cdot a=0$
and $\quad c \perp(a+b)=0 \Rightarrow c \cdot a+c \cdot b=0$
Adding (1), (2), (3), we get

$$
2(a \cdot b+b \cdot c+c \cdot a)=0
$$

Now, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a \cdot b+b \cdot c+c \cdot a)$

$$
\begin{aligned}
& =|a|^{2}+|b|^{2}+|c|^{2}+0 \\
& =(3)^{2}+(4)^{2}+(c)^{2}=50
\end{aligned}
$$

Hence, $|a+b+c|=5 \sqrt{2}$
27. A parallelogram is constructed on the vector $a=3 p-q$ and $b=p+3 q$, given that $|p|=|q|=2$ and the angle between $p$ and $q$ is $\frac{\pi}{3}$. The length of a diagonal is
(A) $4 \sqrt{5}$
(B) $4 \sqrt{3}$
(C) $4 \sqrt{7}$
(D) none of these

## Solution (B)

The diagonals of the parallelogram are represented by the vectors.

$$
a+b=(3 p-q)+(p+3 q)=4 p+2 q
$$

and $\quad a-b=(3 p-q)-(p+3 q)=2 p-4 q$
Now, $|a+b|^{2}=|4 q+2 q|^{2}=16|p|^{2}+4|q|^{2}+16 p \cdot q$

$$
\begin{aligned}
& =16(2)^{2}+4(2)^{2}+16(2)(2) \cos \frac{\pi}{3} \\
& =64+16+12=112 \quad\left(\because \cos \frac{\pi}{3}=\frac{1}{2}\right)
\end{aligned}
$$

$\Rightarrow \quad|a+b|=\sqrt{112}=4 \sqrt{7}$
Similarly, $|a-b|=4 \sqrt{3}$
Hence the lengths of the diagonals are $4 \sqrt{3}$ and $4 \sqrt{3}$.
28. Let $a=2 i-j+k, b=i+2 j-k$ and $c=i+j-2 k$ be three vectors. A vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{\frac{2}{3}}$ is
(A) $2 i+3 j-3 k$
(B) $2 i+3 j+3 k$
(C) $-2 i-j+5 k$
(D) $2 i+j+5 k$

## Solution (A, C)

Any vector in the plane of $b$ and $c$ is

$$
\begin{aligned}
r & =b+\lambda c=(1+2 j+k)+\lambda(i+j-2 k) \\
& =(1+\lambda) i+(2+\lambda) j+(-1-2 \lambda) k
\end{aligned}
$$

Projection of $r$ on $a$ is $=\frac{r \cdot a}{|a|}$
$=\frac{2(1+\lambda)-(2+\lambda)-(1+2 \lambda)}{\sqrt{4+1+1}}=\frac{-\lambda-1}{\sqrt{6}}$
$\therefore \quad \frac{-\lambda-1}{\sqrt{6}}= \pm \sqrt{\frac{2}{3}} \Rightarrow-\lambda-1= \pm 2 ; \lambda=-3$ or 1
Hence, $r=-2 i-j+5 k$ or $r=2 i+3 j-3 k$
29. If the moduli of vectors $a, b, c$ are $3,4,5$ are respectively and $a$ and $b+c, b$ and $c+a, c$ and $a+b$ are mutually perpendicular then the modulus of $a+b+c$ is
(A) $\sqrt{12}$
(B) 12
(C) $5 \sqrt{2}$
(D) 50

## Solution (C)

According to the given condition, we have

$$
\begin{aligned}
& a \cdot(b+c)=0 \\
& b \cdot(c+a)=0 \\
& c \cdot(a+b)=0
\end{aligned}
$$

Now adding (1), (2) and (3), we get
$2(a \cdot b+b \cdot c+c \cdot a)=0 \quad(\because \quad a \cdot b=b \cdot a$ etc.)
Hence, $|a+b+c|^{2}$

$$
\begin{aligned}
& =a^{2}+b^{2}+c^{2}+2(a \cdot b+b \cdot c+c \cdot a) \\
& =3^{2}+4^{2}+5^{2}=9+16+25=50
\end{aligned}
$$

$\Rightarrow|a+b+c|=\sqrt{50}=5 \sqrt{2}$
30. If $p=i-2 j+3 k$ and $q=3 i+j+2 k$, than a vector $r$ which is linear combination of $p$ and $q$ and also perpendicular to $q$ is
(A) $i+5 j-4 k$
(B) $i-5 j+4 k$
(C) $\frac{-1}{2}(i+5 j-4 k)$
(D) none of these

---

Solution (C)
We have, $r=p+\lambda q$
$\Rightarrow \quad r \cdot q=p \cdot q+\lambda q \cdot q$
$\therefore \quad 0=7+14 \lambda$
$(\because \quad p \cdot q=3-2+6=7$ and $q \cdot q=9+1+4=14)$
$\Rightarrow \lambda=-\frac{7}{14}=-\frac{1}{2}, \quad \therefore \quad r=-\frac{1}{2}(i+5 j-4 k)$
31. A unit vector in $X Y$ plane that makes an angle of $45^{\circ}$ with the vector $i+j$ and an angle of $60^{\circ}$ with the vector $3 i-4 j$ is
(A) $i$
(B) $\frac{i+j}{\sqrt{2}}$
(C) $\frac{i-j}{\sqrt{2}}$
(D) none of these

## Solution (D)

Let the required unit vector in the $x-y$ plane be

$$
r=x i+y i, \quad \therefore \quad|r|=\sqrt{\left(x^{2}+y^{2}\right)}=1
$$

or $x^{2}+y^{2}=1$
Now the angle between $r$ and vector $3 i-4 j$ is $60^{\circ}$.

$$
\cos 60^{\circ}=\frac{(x i+y j) \cdot(3 i-4 j)}{|x i+y j||3 i-4 j|} \Rightarrow x+y=1
$$

and $3 x-4 y=\frac{5}{2}$
There exists no real values of $x$ and $y$, satisfying equations (1), (2) and (3).
32. If $A B C D E F$ is a regular hexagon, then $A B \cdot A F$ is equal to
(A) $\frac{1}{2} B C^{2}$
(B) $-\frac{1}{2} B C^{2}$
(C) $\frac{1}{2} A C^{2}$
(D) $\frac{-1}{2} A C^{2}$

## Solution (B)

Let $a$ be the length of each side of the hexagon $A B C D E F$ so that $A B=A F=B C=a$. Also from plane geometry, $\angle B A F=120^{\circ}$.
![img-17.jpeg](img-17.jpeg)

Hence, we have

$$
\begin{aligned}
A B \cdot A F & =|A B||A F| \cos 120^{\circ} \\
& =(\mathrm{AB})(\mathrm{AF})(-1 / 2) \\
& =-\left(a^{2} / 2\right)=-\frac{1}{2} B C^{2}
\end{aligned}
$$

33. Let $a, b, c$ be the three vectors such that $a \cdot(b+c)=b$. $(c+a)=0$ and $|a|=1,|b|=4,|c|=8$, then $|a+b$ $+c \mid=$
(A) 13
(B) 81
(C) 9
(D) 5

## Solution (C)

Adding $a \cdot(b+c)=b \cdot(c+a)=c \cdot(a+b)=0$, we get $2(a \cdot b+b \cdot c+c \cdot a)=0$
$\Rightarrow|a+b+c|^{2}=|a|^{2}+|b|^{2}+|c|^{2}+2(a \cdot b+b \cdot c+$ $c \cdot a)$

$$
=1+16+64=81
$$

Hence, $|a+b+c|=9$
34. If $e_{1}, e_{2}$ are two unit vectors and $\theta$ is the angle between them, then $\cos \frac{\theta}{2}$ is
(A) $\frac{1}{2}\left|e_{1}+e_{2}\right|$
(B) $\frac{1}{2}\left|e_{1}-e_{2}\right|$
(C) $\frac{\left|e_{1} e_{2}\right|}{2}$
(D) $\frac{\left|e_{1}+e_{2}\right|}{2\left|e_{1}\right|\left|e_{2}\right|}$

## Solution (A)

$\left(e_{1}+e_{2}\right)^{2}=e_{1} \cdot e_{1}+2 e_{1} \cdot e_{2}+e_{2} \cdot e_{2}$
$\Rightarrow\left|e_{1}+e_{2}\right|=|e|^{2}+2\left|e_{1}\right|\left|e_{2}\right| \cos \theta+\left|e_{2}\right|^{2}$
$\Rightarrow\left|e_{1}+e_{2}\right|^{2}=1+2 \cdot 1 \cdot 1 \cos \theta+1$

$$
\left(\because\left|e_{1}\right|=\left|e_{2}\right|=1\right)
$$

$\Rightarrow \quad\left|e_{1}+e_{2}\right|^{2}=2(1+\cos \theta)=2\left(2 \cos ^{2} \frac{\theta}{2}\right)$
$\Rightarrow \quad\left|e_{1}+e_{2}\right|^{2}=4 \cos ^{2} \frac{\theta}{2} \Rightarrow \cos \frac{\theta}{2}=\frac{1}{2}\left|e_{1}+e_{2}\right|$
35. The points $O, A, B, C, D$ are such that $O A=a, O B=b$, $O C=2 a+3 b$ and $O D=a-2 b$. If $|a|=3|b|$, then the angle between $B D$ and $A C$ is
(A) $\pi$
(B) $\frac{\pi}{2}$
(C) $\frac{\pi}{3}$
(D) none of these

## Solution (B)

$B D \cdot A C=|B D||A C| \cos \theta$ where $\theta$ is the angle between $B D$ and $A C$.
$\Rightarrow \quad(O D-O B) \cdot(O C-O A)$
$=|O D-O B||O C-O A| \cos \theta$
$\Rightarrow(a-2 b-b) \cdot(2 a+3 b-a)=|a-3 b||a+3 b| \cos \theta$
$\Rightarrow a^{2}-9 b^{2}=|a-3 b||a+3 b| \cos \theta$
$\Rightarrow 0=|a-3 b||a+3 b| \cos \theta(\because|a|=3|b|, \therefore a^{2}=$ $\left.9 b^{2}\right)$
$\Rightarrow \cos \theta=0, \quad \therefore \quad \theta=\frac{\pi}{2}$

---

36. $A, B, C, D$ are four points on a plane with position vectors $a, b, c, d$ respectively such that $(a-d) \cdot(b-c)$ $=(b-d) \cdot(c-a)=0$. For $\triangle A B C, D$ is the
(A) Incentre
(B) orthocentre
(C) centroid
(D) none of these

## Solution (B)

Since $(a-d) \cdot(b-c)=0 \quad \therefore \quad D A \cdot C B=0$
$\therefore A D \perp B C$
Since $(b-d) \cdot(c-a)=0, \quad \therefore \quad D B \cdot A C=0$
$\therefore B D \perp C A$
Then $D$ is the intersection of the altitudes through $A$ and $B$. Therefore, $D$ is the orthocentre of the triangle $A B C$.
37. Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, \vec{i}=\hat{i}+\hat{j}-\hat{k}$. A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{\sqrt{3}}$, then the vector is:
(A) $2 \hat{i}+\hat{j}+2 \hat{k}$
(B) $4 \hat{i}-\hat{j}+4 \hat{k}$
(C) $2 \hat{i}-\hat{j}+4 \hat{k}$
(D) none of these

Solution (A, B)
Let vector $\vec{r}$ be coplanar to $\vec{a}$ and $\vec{b}$

$$
\begin{aligned}
\therefore \quad \vec{r} & =\vec{a}+t \vec{b} \\
\Rightarrow \quad \vec{r} & =(\hat{i}+2 \hat{j}+\hat{k})+t(\hat{i}-\hat{j}+\hat{k}) \\
& =\hat{i}(1+t)+\hat{j}(2-t)+\hat{k}(1+t)
\end{aligned}
$$

Then projection of $\vec{r}$ on $\vec{c}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \quad|2-t|= \pm 1 \Rightarrow t=1$ or 3
When $t=1$, we have $\vec{r}=2 \hat{i}+\hat{j}+2 \hat{k}$
When $t=3$, we have $\vec{r}=4 \hat{i}-\hat{j}+4 \hat{k}$
38. The values of $x$ for which the angle between the vectors $a=x i-3 j-k$ and $b=2 x i+x j-k$ is acute and the angle between the vector $b$ and the $y$-axis lies between $\frac{\pi}{2}$ and $\pi$ are
(A) $1, \frac{1}{2}$
(B) $0<x<\frac{1}{2}$
(C) all $x<0$
(D) $x<0$ or $x>1$

Solution (C)
$a \cdot b>0$ and $b \cdot j<0$.
$\Rightarrow 2 x^{2}-3 x+1>0$ and $x<0$
$\Rightarrow(2 x-1)(x-1)>0$ and $x<0$
$\Rightarrow(x<1 / 2$, or $x>1)$ and $x<0$
Hence, $x<0$ is the required solution.
39. A unit vector in $x y$ plane that makes an angle of $45^{\circ}$ with the vector $i+j$ and an angle of $60^{\circ}$ with the vector $3 i-4 j$ is
(A) $i$
(B) $(i+j) / \sqrt{2}$
(C) $(i-j) / \sqrt{2}$
(D) none of these

## Solution (D)

Let the required unit vector in the $x-y$ plane be

$$
r=x i+y j, \quad \therefore \quad|r|=\sqrt{\left(x^{2}+y^{2}\right)}=1
$$

or

$$
x^{2}+y^{2}=1
$$

Since angle between $r$ and vector $i+j$ is $45^{\circ}$ and the angle between $r$ and vector $3 i-4 j$ is $60^{\circ}$.
$\therefore \quad \cos 45^{\circ}=\frac{(x i+y j) \cdot(i+j)}{|x i+y j||i+j|}$
and $\quad \cos 60^{\circ}=\frac{(x i+y j) \cdot(3 i-4 j)}{|x i+y j||3 i-4 j|}$
$\Rightarrow \quad x+y=1$
$3 x-4 y=5 / 2$
No real values of $x$ and $y$ exist satisfying equations (1), (2), and (3).

## VECTOR PRODUCT OF TWO VECTORS

The vector product or cross product of two vectors $a$ and $b$ is defined as
![img-18.jpeg](img-18.jpeg)

Fig. 21.10

$$
a \times b=|a||b| \sin \theta n
$$

1. $|a||b| \sin \theta$ is the modulus of $a \times b, \theta$ being the angle between the directions of $a$ and $b$ and $0 \leq \theta \leq \pi$;
2. direction of $a \times b$ is that of the unit vector $n$ which is perpendicular to both $a$ and $b$ such that $a, b$ and $n$ form a right handed system.

## REMEMBER

- By right handed system we mean that as the first vector $a$ is turned towards the second vector $b$ through an angle $\theta$, $n$ will point in the direction in which a right handed screw would advance if turned in a similar manner.

---

- If either $a$ or $b$ is $O$, we have $a \cdot b=O$.
- $|a \times b|=|a||b| \sin \theta$
- A unit vector perpendicular to the plane of two given vectors $a$ and $b$ is given as $n=\frac{a \times b}{|a \times b|}$.
- $a \times b$ is perpendicular to the plane of $a$ and $b$.


## Key Results on Vector Product

1. Vector product is not commutative. For any two vectors $a$ and $b$

$$
a \times b \neq b \times a
$$

In fact, $a \times b=-b \times a$
2. Vector product is associative with respect to a scalar. If $m$ and $n$ be any scalars and $a, b$ any vectors, then

$$
\begin{aligned}
m(a \times b) & =(m a) \times b=a \times(m b)=(a \times b) m \\
(m a) \times(n b) & =(n a) \times(m b)=(m n a) \times b \\
& =a \times(m n b)=m n(a \times b)
\end{aligned}
$$

3. Vector product is distributive with respect to addition. For any three vectors $a, b$ and $c$

$$
a \times(b+c)=a \times b+a \times c
$$

4. If two vectors $a$ and $b$ are parallel, then $a \times b=0$. In particular, $a \times a=0$.
5. Vector product of mutually orthogonal unit vectors $i$, $j, k$ :

$$
i \times i=j \times j=k \times k=O
$$

and $\quad i \times j=k=-j \times i$,

$$
j \times k=i=-k \times j, k \times i=j=-i \times k
$$

6. Vector product in terms of components. Let

$$
a=a_{1} i+a_{2} j+a_{3} k
$$

and $b=b_{1} i+b_{2} j+b_{3} k$, then

$$
\begin{gathered}
a \times b=\left(a_{1} b_{1}-a_{1} b_{2}\right) i+\left(a_{2} b_{1}-a_{1} b_{3}\right) j \\
+\left(a_{1} b_{2}-a_{2} b_{1}\right) k \\
=\left|\begin{array}{ccc}
i & j & k \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|
\end{gathered}
$$

7. Angle between two vectors: If $\theta$ is the angle between two vectors $a$ and $b$, then $\sin \theta=\frac{|a \times b|}{|a||b|}$.

## Geometrical Interpretation of Cross Product

1. The area of a parallelogram with adjacent sides $a$ and $b$ is given by $|a \times b|$.
2. The area of a triangle with adjacent sides $a$ and $b$ is given by $\frac{1}{2}|a \times b|$.
3. The area of a triangle $A B C$ is

$$
\frac{1}{2}|A B \times A C| \text { or } \frac{1}{2}|B C \times B A| \text { or } \frac{1}{2}|C B \times C A|
$$

4. The area of a parallelogram with diagonals $a$ and $b$ is given by $\frac{1}{2}|a \times b|$.
5. The area of a quadrilateral $A B C D$ is given by $\frac{1}{2}|A C \times B D|$, where $A C$ and $B D$ are its diagonals.
6. Vector area of a $\triangle A B C$, when $a, b, c$ are the position vectors of $A, B, C$ respectively is given by

$$
\Delta A B C=\frac{1}{2}(a \times b+b \times c+c \times a)
$$

## MOMENT OF A FORCE ABOUT A POINT

The vector moment or torque $M$ of a force $F$ acting at a point $A$ about the point $O$ is given by

$$
M=r \times F=O A \times \mathrm{F}
$$

where $r=O A$ is the position vector of the point $A$ with respect to the point $O$.
![img-19.jpeg](img-19.jpeg)

Fig 21.11

## NOTE

The algebraic sum of the moments of a system of forces about any point is equal to the moment of their resultant about the same point.

## TRICK(S) FOR PROBLEM SOLVING

Lagrange's Identity
If $a$ and $b$ are any two vectors, then

$$
|a \times b|^{2}+(a \cdot b)^{2}=|a|^{2}|b|^{2}
$$

## SOLVED EXAMPLES

40. If $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors and $\vec{r}$ be any arbitrary vector, then

---

$(\vec{a} \times \vec{b}) \times(\vec{r} \times \vec{c})+(\vec{b} \times \vec{c}) \times(\vec{r} \times \vec{a})+(\vec{c} \times \vec{a}) \times(\vec{r} \times \vec{b})$
is equal to
(A) 0
(B) $[\vec{a} \vec{b} \vec{c}] \vec{r}$
(C) $2[\vec{a} \vec{b} \vec{c}] \vec{r}$
(D) $3[\vec{a} \vec{b} \vec{c}] \vec{r}$

## Solution (C)

We have, $(\vec{a} \times \vec{b}) \times(\vec{r} \times \vec{c})$

$$
\begin{aligned}
& =((\vec{a} \times \vec{b}) \cdot \vec{c}) \vec{r}-((\vec{a} \times \vec{b}) \cdot \vec{r}) \vec{c} \\
& =[\vec{a} \vec{b} \vec{c}] \vec{r}-[\vec{a} \vec{b} \vec{r}] \vec{c}
\end{aligned}
$$

Similarly, $(\vec{b} \times \vec{c}) \times(\vec{r} \times \vec{a})=[\vec{b} \vec{c} \vec{a}] \vec{r}-[\vec{b} \vec{c} \vec{r}] \vec{c}$
and, $(\vec{c} \times \vec{a}) \times(\vec{r} \times \vec{b})=[\vec{c} \vec{a} \vec{b}] \vec{r}-[\vec{c} \vec{a} \vec{r}] \vec{b}$

$$
\begin{aligned}
\therefore \quad & (\vec{a} \times \vec{b}) \times(\vec{r} \times \vec{c})+(\vec{b} \times \vec{c}) \times(\vec{r} \times \vec{a})+(\vec{c} \times \vec{a}) \times(\vec{r} \times \vec{b}) \\
& =3[\vec{a} \vec{b} \vec{c}] \vec{r}-([\vec{b} \vec{c} \vec{r}] \vec{a}+[\vec{c} \vec{a} \vec{r}] \vec{b}+[\vec{a} \vec{b} \vec{r}] \vec{c}) \\
& =3[\vec{a} \vec{b} \vec{c}] \vec{r}-[\vec{a} \vec{b} \vec{c}] \vec{r} \\
& =2[\vec{a} \vec{b} \vec{c}] \vec{r}
\end{aligned}
$$

41. If $A=2 i+k, B=i+j+k$ and $C=4 i-3 j+7 k$, then a vector $R$ which satisfies $R \times B=C \times B$ and $R \cdot A=O$, is
(A) $-i-8 j+2 k$
(B) $i-8 j+2 k$
(C) $i+8 j+2 k$
(D) none of these

## Solution (A)

Let $\quad R=x i+y j+z k$

$$
\begin{aligned}
& \therefore \quad R \cdot A=0 \quad \Rightarrow \quad 2 x+z=0 \\
& R \times B=C \times B \Rightarrow\left|\begin{array}{ccc}
i & j & k \\
x & y & z \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
i & j & k \\
4 & -3 & 7 \\
1 & 1 & 1
\end{array}\right| \\
& \Rightarrow \quad(y-z) i+(z-x) j+(x-y) k=-10 i+3 j+7 k \\
& \Rightarrow \quad y-z=-10 \\
& z-x=3 \\
& \text { and } \quad x-y=7
\end{aligned}
$$

Solving (1) and (2), we get $x=-1, z=2$
$\therefore$ From (2), $y=-8$.
Hence $R=-i-8 j+2 k$
42. Let $A(0,0,0), B(1,1,1), C(3,2,1)$ and $D(3,1,2)$ be four points. The angle between the planes through the points $A, B, C$ and through the points $A, B, D$ is
(A) $\frac{\pi}{2}$
(B) $\frac{\pi}{6}$
(C) $\frac{\pi}{4}$
(D) $\frac{\pi}{3}$

## Solution (D)

Let $n_{1}$ and $n_{2}$ be the vectors normal to the planes $A B C$ and $A B D$ respectively.

$$
\begin{aligned}
& n_{1}=A B \times A C=-i+2 j-k \\
& n_{2}=A B \times A D=i+j-2 k
\end{aligned}
$$

Let $\theta$ be the acute angle between the planes, then $\theta$ is the acute angle between their normals $n_{1}$ and $n_{2}$

$$
\begin{aligned}
& \therefore \quad \cos \theta=\frac{|-1+2+2|}{\sqrt{6} \cdot \sqrt{6}}=\frac{3}{2}=\frac{1}{2}=\cos \frac{\pi}{3} \\
& \Rightarrow \quad=\frac{\pi}{3}
\end{aligned}
$$

43. Given $A=a i+b j+c k, B=d i+3 j+4 k$ and $C=3 i+$ $j-2 k$. If the vectors $A, B$ and $C$ form a triangle such that $A=B+C$, then
(A) $a=-8, b=-4, c=2, d=-11$
(B) $a=-8, b=4, c=-2, d=-11$
(C) $a=-8, b=4, c=2, d=-11$
(D) none of these

## Solution (C)

Here $A, B, C$ are the vectors which represent the sides of the triangle $A B C$ where

$$
\begin{aligned}
& A=a i+b j+c k \\
& B=d i+3 j+4 k \\
& C=3 i+j-2 k
\end{aligned}
$$

Given that, $A=B+C$
$\therefore \quad a i+b j+c k=(d+3) i+4 j+2 k$
$\Rightarrow a=d+3, b=4, c=2$.

$$
\begin{aligned}
& B \times C=\left|\begin{array}{ccc}
i & j & k \\
d & 3 & 4 \\
3 & 1 & -2
\end{array}\right| \\
& =-10 i+(2 d+12) j+(d-9) k
\end{aligned}
$$

$\therefore$ Area of the $\triangle A B C=\frac{1}{2}|B \times C|$

$$
\begin{aligned}
& =\frac{1}{2} \sqrt{\left[100+(2 d+12)^{2}+(d-9)^{2}\right]} \\
& =5 \sqrt{6}(\text { Given }) \\
\Rightarrow & \sqrt{\left(5 d^{2}+30 d+325\right)}=10 \sqrt{6} \\
\Rightarrow & 5 d^{2}+30 d+325=600 \Rightarrow 5 d^{2}+30 d-275=0 \\
\Rightarrow & d^{2}+6 d-55=0 \Rightarrow(d+11)(d-5)=0 \\
\Rightarrow & d=5 \text { or }-11
\end{aligned}
$$

When $d=5, a=8, b=4, c=2$
and when $d=-11, a=-8, b=4, c=2$.
44. If $x+y=a, x \times y=b$ and $x \cdot a=1$, then
(A) $x=\frac{a+a \times b}{a^{2}}$
(B) $y=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$
(C) $x=\frac{b+a \times b}{a^{2}}$
(D) $y=\frac{\left(b^{2}-1\right) b-a \times b}{a^{2}}$

---

Solution (A, B)
Given

$$
\begin{aligned}
& x+y=a \\
& \Rightarrow \quad y=a-x \\
& x \times y=b \\
& x \cdot a=1
\end{aligned}
$$

From (1) and (2), we get

$$
\begin{aligned}
& x \times(a-x)=b \\
\Rightarrow & x \times a-x \times x=b \Rightarrow x \times a=b \\
\Rightarrow & a \times(x \times a)=a \times b \Rightarrow(a \cdot a) x-(a \cdot x) a=a \times b \\
\Rightarrow & |a|^{2} x-1 \cdot a=a \times b \\
& \Rightarrow x=\frac{(a+a \times b)}{a^{2}}
\end{aligned}
$$

and $y=a-x=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$
45. If $a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c$ and $(c \cdot c) a=c$, while $b$ and $c$ are non-collinear, then
(A) $\alpha=\frac{\pi}{2}, \beta=-1$
(B) $\alpha=\frac{\pi}{2}, \beta=1$
(C) $\alpha=\frac{\pi}{3}, \beta=-1$
(D) $\alpha=\frac{\pi}{3}, \beta=-1$

## Solution (B)

We have, $a \times(b \times c)+(a \cdot b) b$

$$
=(4-2 \beta-\sin \alpha) b+\left(b^{2}-1\right) c
$$

and $\quad(c \cdot c) a=c$
where $b$ and $c$ are non-collinear vectors and $\alpha, \beta$ are scalars
From (2), $\quad(c \cdot c) a \cdot c=c \cdot c$
$\therefore \quad a \cdot c=1$
From (1), we get

$$
\begin{aligned}
& (a \cdot c) b-(a \cdot b) c+(a \cdot b) b \\
& =(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c \\
\text { or } & {[1+(a \cdot b)] b-(a \cdot b) c} \\
& =(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c \\
\Rightarrow \quad & 1+(a \cdot b)=4-2 \beta-\sin \alpha \\
\text { and } \quad & a \cdot b=-\left(b^{2}-1\right) \\
\therefore \quad & \sin \alpha=1+(1-\beta)^{2} \Rightarrow \beta=1, \sin \alpha=1
\end{aligned}
$$

i.e., $\alpha=\frac{\pi}{2}+2 n \pi, n \in I$.
46. If $u=a-b, v=a+b$ and $|a|=|b|=2$, then $|u \times v|$ $=$
(A) $2 \sqrt{\left[16-(a \cdot b)^{2}\right]}$
(B) $\sqrt{\left[16-(a \cdot b)^{2}\right]}$
(C) $2 \sqrt{\left[4-(a \cdot b)^{2}\right]}$
(D) $\sqrt{\left[4-(a \cdot b)^{2}\right]}$

Solution (A)

$$
\begin{array}{rlrl}
u \times v & =(a-b) \times(a+b)=2 a \times b \\
\therefore \quad & |u \times v|= & 2|a \times b|
\end{array}
$$

$$
\begin{aligned}
& =2 \sqrt{\left[a^{2} b^{2} \sin ^{2} \theta\right]}=2 \sqrt{\left(a^{2} b^{2}-a^{2} b^{2} \cos ^{2} \theta\right)} \\
& =2 \sqrt{\left[16-(a \cdot b)^{2}\right]} \quad(\because|a|=|b|=2)
\end{aligned}
$$

47. If $a \cdot i=4$, then $(a \cdot j) \times(2 j-3 k)=$
(A) 12
(B) 2
(C) 0
(D) -12

## Solution (D)

We have, $(a \cdot j) \times(2 j-3 k)$

$$
\begin{aligned}
& =a \cdot(j \times(2 j-3 k))=a \cdot(-3(j \times k) \\
& =-3(a \cdot i) \quad(\because \quad j \times k=i) \\
& =-3(4)=-12
\end{aligned}
$$

48. If $a \cdot b=\beta$ and $a \times b=c$, then $b$ is equal to
(A) $(\beta a-a \times c) / a^{2}$
(B) $(\beta a+a \times c) / a^{2}$
(C) $(\beta c-a \times c) / a^{2}$
(D) $(\beta c+a \times c) / a^{2}$

## Solution (A)

Here $a$ and $c=a \times b$ are non-collinear vectors.
$\therefore \quad$ Let $b=x a+y(a \times c)$
$\therefore \quad \beta=a \cdot b=a \cdot[x a+y(a \times c)]$
$=x|a|^{2}+y a \cdot(a \times c)=x a^{2} \Rightarrow x=\beta / a^{2}$
And $\quad c=a \times b=a \times[x a+y(a \times c)]$

$$
\begin{aligned}
& =x a \times a+y a \times(a \times c) \\
& =0+y(a \cdot c) a-y(a \cdot a) c \\
& =y[a \cdot(a \times b)] a-y a^{2} c=-y a^{2} c \\
\Rightarrow \quad & y=-1 / a^{2}
\end{aligned}
$$

$\therefore$ from (1), $b=(\beta a-a \times c) / a^{2}$
49. Let $O A=a, O B=10 a+2 b$ and $O C=b$ where $O, A, C$ are non-collinear. Let $p$ denote the area of the quadrilateral $O A B C$ and $q$ denote the area of the parallelogram with $O A$ and $O C$ as adjacent sides. Then $\frac{p}{q}$ is equal to
(A) 4
(B) 6
(C) $\frac{1}{2} \frac{|a-b|}{|a|}$
(D) none of these

## Solution (B)

Given, $|a \times b|=q$
and $\frac{1}{2}|b \times(10 a+2 b)|+\frac{1}{2}|a \times(10 a+2 b)|=p$
$\therefore \quad 5|b \times a|+|a \times b|=p$
$\Rightarrow \quad 6|a \times b|=p \quad \Rightarrow \quad 6 q=p \quad \Rightarrow \quad \frac{p}{q}=6$.
50. Two planes are perpendicular to one another. One of them contains $a$ and $b$ and the other contains vector $c$ and $d$, then $(a \times b) \cdot(c \times d)$ is equal to
(A) 1
(B) 0
(C) $[a b c] d$
(D) $[b c d] a$

---

# 21.18 A'TM ${ }^{21}$ : FREE EDUCATION TO ALL APUL 

## Solution (B)

$a \times b$ is a vector perpendicular to the plane of $a$ and $b$ and $c \times d$ is a vector perpendicular to the plane of $c$ and $d$.
Since, these planes are $\perp$ to one another
$\therefore(a \times b) \cdot(c \times d)=0$
51. If the vectors $c, a=x i+y j+z k$ and $b=j$ are such that $a, c$ and $b$ form a right handed system then $c$ is
(A) $z i-x k$
(B) 0
(C) $y j$
(D) $-z i-x k$

## Solution (A)

Since $a, c, b$ form a right-handed system, so

$$
c=\lambda(b \times a), \lambda>0
$$

i.e., $\quad c=\lambda i \times(x i+y j+z k)=\lambda(z i-x k)$

Taking $\lambda=1, c=z i-x k$
52. $A B C D$ is a quadrilateral with $A B=a, A D=b$ and $A C$ $=2 \boldsymbol{a}+3 b$. If its area is $\alpha$ times the area of the parallelogram with $A B, A D$ as adjacent sides, then $\alpha$ is equal to
(A) 5
(B) $\frac{5}{2}$
(C) 1
(D) $\frac{1}{2}$

## Solution (B)

Area of quadrilateral $A B C D$

$$
\begin{aligned}
& =\text { Area of } \triangle A B C+\text { Area of } \triangle A C D \\
& =\frac{1}{2}[a \times(2 a+3 b)]+\frac{1}{2}[(2 a+3 b) \times b] \\
& =\frac{1}{2}[3 a \times b+2 a \times b]=\frac{5}{2} a \times b \\
& =\frac{5}{2} \text { Area of } \|_{g m} A B C D . \quad \therefore \quad \alpha=\frac{5}{2}
\end{aligned}
$$

53. Consider a tetrahedron with faces $F_{1}, F_{2}, F_{3}, F_{4}$. Let $V_{1}, V_{2}, V_{3}, V_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to their faces in outward direction. Then $\left|V_{1}+V_{2}+V_{3}+V_{4}\right|$ equals
(A) 1
(B) 4
(C) 0
(D) none of these

## Solution (C)

We have,

$$
\begin{aligned}
& v_{1}=\frac{1}{2}(a \times b), v_{2}=\frac{1}{2}(b \times c) \\
& v_{3}=\frac{1}{3}(c \times a) \text { and } v_{4}=\frac{1}{2}[(c-a) \times(b-a)] \\
\therefore & v_{1}+v_{2}+v_{3}+v_{4}=0 \\
\therefore & \left|v_{1}+v_{2}+v_{3}+v_{4}\right|=0
\end{aligned}
$$

## SCALAR TRIPLE PRODUCT

Scalar triple product of three vectros If $a, b, c$ are three vectors, then their scalar triple product is defined as the dot product of two vectors $a$ and $b \times c$. It is generally denoted by $a \cdot(b \times c)$ or $[a b c]$.

## Properties of Scalar Triple Product

(i) If $a, b, c$ are cyclicallly permuted, the value of scalar triple product remains the same, i.e., $(a \times b) . c$ $=(b \times c) \cdot a=(c \times a) . b$ or $[a b c]=[b c a]=[c a b]$
(ii) The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude i.e., $[a b c]=-[b a c]=-[c b a]$ $=-[a c b]$
(iii) In scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of the vectors remains same i.e., $(a \times b) . c=a .(b \times c)$
(iv) The scalar triple product of three vectors is zero if any two of them are equal.
(v) For any three vectors $a, b, \mathrm{c}$ and scalar $\lambda,[\lambda a b c]=\lambda$ $[a b c]$
(vi) The scalar triple product of three vectors is zero if any two of them are parallel or collinear.
(vii) If $a, b, c, d$ are four vectors, then $[(a+b) c d]=[a c$ $d]+[b c d]$
(viii) The necessary and sufficient condition for three nonzero non-collinear vectors $a, b, c$ to be coplanar is that $[\mathrm{a} b \mathrm{c}]=0$.
(ix) Four points with position vectors $a, b, c$ and $d$ will be coplanar, if $[a b c]+[d c a]+[d a b]=[\mathrm{a} b \mathrm{c}]$.
(x) Volume of parallelopiped whose coterminous edges are $a, b, c$ is $[a b c]$ or $a \cdot(b \times c)$.

## Scalar Triple Product in Terms of Components

(i) If $a=a_{1} i+a_{2} j+a_{3} k, b=b_{1} i+b_{2} j+b_{3} k$ and $c=c_{1} i+$ $c_{2} j+c_{2} j+c_{3} k$ be three vectros
then, $[a b c]=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$
(ii) If $a=a_{1} l+a_{2} m+a_{3} n, b=b_{1} l+b_{2} m+b_{3} n$ and $c=c_{1} l$ $+c_{2} m+c_{3} n$, then

$$
[a b c]=\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|[l m n]
$$

---

(iii) For any three vectors $a, b$ and $c$
(A) $[a+b b+c c+a]=2[a b c]$
(B) $[a-b b-c c-a]=0$
(C) $[a \times b b \times c c \times a]=[a b c]^{2}$

Tetrahedron A tetrahendron is a three-dimensional figure formed by four triangles. $O A B C$ is a tetrahedron with $\triangle A B C$ as the base. $O A, O B, O C, A B, B C$ and $C A$ are known as edges of the tetrahedron. $O A, B C ; O B, C A$ and $O C, A B$ are known as the pairs of oppposite edges. A tetrahedron in which all edges are equal, is called a regular tetrahedron. Any two edges of regular tetrahedron are perpendicular to each other.
![img-20.jpeg](img-20.jpeg)

Fig. 21.12

## Volume of Tetrahedron

(i) The volume of a tetrahendron
$=\frac{1}{3}$ (area of the base) (corresponding altitude)
$=\frac{1}{6}[A B B C A D]$
(ii) If $a, b, c$ are position vectors of vertices $A, B$ and $C$ with respect to $O$, then volume of tetrahedron $O A B C=\frac{1}{6}[a b c]$
(iii) If $a, b, c, d$ are position vectors of vertices $A, B, C, D$ of a tetrahedron $A B C D$, then its volume

$$
=\frac{1}{6}[b-a c-a d-a]
$$

Reciprocal system of vectors Let $a, b, c$ be three non-coplanar vectors, and let $a^{\prime}=\frac{b \times c}{[a b c]}, b^{\prime}=\frac{c \times a}{[a b c]}, c^{\prime}=\frac{a \times b}{[a b c]}$. $a^{\prime}, b^{\prime}, c^{\prime}$ are said to form a reciprocal system of vectors for the vectors $a, b, c$.

If $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$ form a reciprocal system of vectors then
(i) $a, a^{\prime}=b b^{\prime}=c . c^{\prime}=1$
(ii) $a . b^{\prime}=a . c^{\prime}=0 ; b . c^{\prime}=b . a^{\prime}=0 ; c . a^{\prime}=c . b^{\prime}=0$
(iii) $\left[a^{\prime} b^{\prime} c^{\prime}\right]=\frac{1}{[a b c]}$
(iv) $a, b, c$ are non-coplanar iff so are $a^{\prime}, b^{\prime}, c^{\prime}$.

## SOLVED EXAMPLES

54. If $i, j, k$ are the unit vectors and mutually perpendicular, then $[i-j j-k k-i]=$
(A) 0
(B) 1
(C) -1
(D) none of these

## Solution (A)

$$
\begin{aligned}
& (i-j) \cdot[(j-k) \times(k-i)] \\
& \quad=(i-j) \cdot(j \times k-j \times i-k \times k+k \times i) \\
& \quad=(i-j) \cdot(i+k-0+j) \\
& \quad=i \cdot i+i \cdot k+i \cdot j-j \cdot i-j \cdot k-j \cdot j \\
& \quad=i^{2}+0+0-0-0-j^{2}=i^{2}-j^{2}=1-1=0
\end{aligned}
$$

55. If $e_{1}{ }^{\prime}, e_{2}{ }^{\prime}, e_{3}{ }^{\prime}$ are vectors reciprocal to the non-coplanar vectors $e_{1}, e_{2}, e_{3}$ then $\left[e_{1}{ }^{\prime}, e_{2}{ }^{\prime}, e_{3}{ }^{\prime}\right]\left[e_{1} e_{2} e_{3}\right]=$
(A) $\frac{-1}{2}$
(B) 1
(C) 0
(D) 4

## Solution (B)

Since $\left[e_{1}^{\prime} e_{2}^{\prime} e_{3}^{\prime}\right]=\frac{1}{\left[e_{1} e_{2} e_{3}\right]}$
$\therefore\left[e_{1}^{\prime} e_{2}^{\prime} e_{3}^{\prime}\right]\left[e_{1} e_{3} e_{3}\right]=1$
56. If $r=\lambda(a \times b)+\mu(b \times c)+v(c \times a)$ and $[a b c]=\frac{1}{8}$, then $\lambda+\mu+v$ is equal to
(A) $r \cdot(a+b+c)$
(B) $8 r \cdot(a+b+c)$
(C) $4 r \cdot(a+b+c)$
(D) none of these

## Solution (B)

Clearly $r \cdot c=\lambda[a b c]=\frac{1}{8} \lambda$

$$
\begin{gathered}
r \cdot a=\mu[a b c]=\frac{1}{8} \mu \\
r \times b=v[a b c]=\frac{1}{8} v \\
\therefore \quad r \cdot(a+b+c)=\frac{1}{8}(\lambda+\mu+v) \\
\therefore \quad \lambda+\mu+v=8 r \cdot(a+b+c)
\end{gathered}
$$

## Vector triple product

Let $a, b, c$ be any three vectors, then the vectors $a \times(b \times \mathrm{c})$ and $(a \times b) \times c$ are called vector triple product of $a, b, c$.

Thus, $a \times(b \times c)=(a . c) b-(a . b) c$

## Properties of Vector Triple Product

(i) The vector triple product $a \times(b \times c)$ is a linear combination of those two vectors which are within brackets.
(ii) The vector $r=a \times(b \times c)$ is perpendicular to $a$ and lies in the plane of $b$ and $c$.

---

(iii) The formula $a \times(b \times c)=(a . c) b-(a . b) c$ is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula. Thus, $(b \times c) \times a$ $=-\{a \times(b \times c)\}=-\{(a \cdot c) b-(a \cdot b) c\}=(a$. b) $c-(a \cdot c) b$
(iv) Vector triple product is a vector quantity.
(v) $a \times(b \times c) \neq(a \times b) \times c$

## Rotation of a Vector About an Axis

Let $a=\left(a_{1}, a_{2}, a_{3}\right)$. If system is rotated about
(i) $x$-axis through an angle $\alpha$, then the new components of $a$ are $\left(a_{1}, a_{2} \cos \alpha+a_{3} \sin \alpha,-a_{2} \sin \alpha+\right.$ $\left.a_{3} \cos \alpha\right)$.
(ii) $y$-axis through an angle $\alpha$, then the new components of $a$ are $\left(-a_{3} \sin \alpha+a_{1} \cos \alpha, a_{2}, a_{3} \cos \alpha+\right.$ $\left.a_{1} \sin \alpha\right)$.
(iii) $z$-axis through an angle $\alpha$, then the new components of $a$ are $\left(a_{1} \cos \alpha+a_{2} \sin \alpha,-a_{1} \sin \alpha+a_{2}\right.$ $\left.\cos \alpha, a_{3}\right)$.

## SOLVED EXAMPLES

57. Given three unit vectors $a, b, c$ no two of which are collinear satisfying $a \times(b \times c)=\frac{1}{2} b$. The angle between $a$ and $b$ is
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{2}$
(D) none of these

## Solution (C)

We have, $a \times(b \times c)=\frac{1}{2} b$
$\Rightarrow(a \cdot c) b-(a \cdot b) c=1 / 2 b$
$\Rightarrow(a \cdot c-1 / 2) b=(a \cdot b) c$
But since $b$ and $c$ are non-parallel, so the only possibility is $a \cdot c=1 / 2$ and $a \cdot b=0$ Hence the angle between $a$ and $b$ is $\pi / 2$.
58. If $a$ and $b$ are two unit vectors, then the vector $(a+b)$ $\times(a \times b)$ is parallel to the vector
(A) $a-b$
(B) $a+b$
(C) $2 a-b$
(D) $2 a+b$

## Solution (A)

We have, $(a+b) \times(a \times b)$
$=a \times(a \times b)+b \times(a \times b)$
$=(a \cdot b) a-(a \cdot a) b+(b \cdot b) a-(b \cdot a) b$
$=(a \cdot b)(a-b)+a-b\left(b \cdot b=b^{2}=1, a \cdot a=a^{2}=1\right.$
as $a, b$ are unit vectors)
$=(a \cdot b+1)(a-b)$
$=x(a-b)$ where $x=a \cdot b+1$ is a scalar.
$\therefore \quad$ The given vector is parallel to $a-b$.

# EXERCISES 

## Single Option Correct Type

1. $a$ and $b$ are mutually perpendicular unit vectors. If $r$ is a vector satisfying $r \cdot a=0, r \cdot b=1$ and $[r a b]=1$, then $r$ is
(A) $a \times b+b$
(B) $a+(a \times b)$
(C) $b+(a \times b)$
(D) $a \times b+a$
2. $a, b, c$ are three vectors of magnitude, $\sqrt{3}, 1,2$ such that $a \times(a \times c)+3 b=O$. If $\theta$ is the angle between $a$ and $c$, then $\cos ^{2} \theta$ is equal to
(A) $\frac{1}{4}$
(B) $\frac{3}{4}$
(C) $\frac{1}{2}$
(D) none of these
3. A vector $a$ has components $2 p$ and 1 w.r.t. a rectangular cartesian system. This system is rotated through a
certain angle about the origin in the counter-clockwise sense. If w.r.t. the new system, $a$ has components $p+1$ and 1 , then
(A) $p=0$
(B) $p=1$ or $p=-\frac{1}{3}$
(C) $p=-1$ or $p=\frac{1}{3}$
(D) $p=1$ or $p=-1$
4. If $a \cdot b=\beta$ and $a \times b=c$, then $\beta$ is equal to
(A) $(\beta a-a \times c) / a^{2}$
(B) $(\beta a+a \times c) / a^{2}$
(C) $(\beta c-a \times c) / a^{2}$
(D) $(\beta c+a \times c) / a^{2}$
5. Let $A B C D E F$ be a regular hexagon. If $A D=x B C$ and $C F=y A B$, then $x y=$
(A) 4
(B) -4
(C) 2
(D) -2

---

6. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with lower base $A B C D$, upper base $A_{1} B_{1} C_{1} D_{1}$ and the lateral edges $A A_{1}, B B_{1}$, $C C_{1}$ and $D D_{1} ; M$ and $M_{1}$ are the centres of the faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ respectively. $O$ is a point on line $M M_{1}$, such that
$O A+O B+O C+O D=O M_{1}$, then $O M=\lambda O M_{1}$ if $\lambda=$
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C) $\frac{1}{6}$
(D) $\frac{1}{8}$
7. The triangle $A B C$ is defined by the vertices $A(1,-2,2)$, $B(1,4,0)$ and $C(-4,1,1)$. Let $M$ be the foot of the altitude drawn from the vertex $B$ to side $A C$. Then, $B M=$
(A) $(-20 / 7,-30 / 7,10 / 7)$
(B) $(-20,-30,10)$
(C) $(2,3,-1)$
(D) none of these
8. If $A B=3 i+j-k$ and $A C=i-j+3 k$. If the point $P$ on the line segment $B C$ is equidistant from $A B$ and $A C$, then $A P$ is
(A) $2 i-k$
(B) $i-2 \mathrm{k}$
(C) $2 i+k$
(D) none of these
9. $A, B, C, D$ are four points on a plane with position vectors $a, b, c, d$, respectively, such that $(a-d)$. $(b-c)=(b-d) \cdot(c-a)=0$. For $\triangle A B C, D$ is the
(A) incentre
(B) orthocentre
(C) centroid
(D) none of these
10. If $a$ and $b$ are two unit vectors, then the vector $(a+b)$ $\times(a \times b)$ is parallel to the vector
(A) $a-b$
(B) $a+b$
(C) $2 a-b$
(D) $2 a+b$
11. If $r=\lambda(a \times b)+\mu(b \times c)+v(c \times a)$ and $[a b c]=\frac{1}{8}$, then $\lambda+\mu+v$ is equal to
(A) $r \cdot(a+b+c)$
(B) $8 r \cdot(a+b+c)$
(C) $4 r \cdot(a+b+c)$
(D) none of these
12. In a parallelogram $A B C D,|A B|=a,|A D|=b$ and $|A C$ $|=c$. Then, $D B \cdot A B$ has the value
(A) $\frac{3 a^{2}+b^{2}-c^{2}}{2}$
(B) $\frac{a^{2}+3 b^{2}-c^{2}}{2}$
(C) $\frac{a^{2}-b^{2}+3 c^{2}}{2}$
(D) $\frac{a^{2}+3 b^{2}+c^{2}}{2}$
13. Let $a=a_{1} i+a_{2} j+a_{3} k, b=b_{1} i+b_{2} j+b_{3} k$, and $c=c_{1} i+c_{2} j+c_{3} k$ be three non-zero vectors such that $c$ is a unit vector perpendicular to both vectors $a$ and $b$. If the angle between vectors $a$ and $b$ is $\pi / 6$, then
$\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right|^{2}$ is equal to
(A) 0
(B) 1
(C) $\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{3}^{2}+b_{3}^{2}\right)$
(D) $\frac{3}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\left(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}\right)$
(E) none of these
14. If $a, b, c$ are non-coplanar unit vectors such that $a \times(b \times c)=\frac{b+c}{\sqrt{2}}$, then the angle between $a$ and $b$ is
(A) $\frac{3 \pi}{4}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{2}$
(D) $\pi$
15. If the vectors $a$ and $b$ are perpendicular to each other, then a vector $v$ in terms of $a$ and $b$ satisfying the equations $v \cdot a=0, v \cdot b=1$ and $[v a b]=1$ is
(A) $\frac{1}{|b|^{2}} b+\frac{1}{|a \times b|^{2}} a \times b$
(B) $\frac{b}{|b|}+\frac{a \times b}{|a \times b|^{2}}$
(C) $\frac{b}{|b|^{2}}+\frac{a \times b}{|a \times b|}$
(D) none of these
16. Let the unit vectors $a$ and $b$ be perpendicular to each other and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c=x a+y b+z(a \times b)$, then
(A) $x=\cos \theta, y=\sin \theta, z=\cos 2 \theta$
(B) $x=\sin \theta, y=\cos \theta, z=-\cos 2 \theta$
(C) $x=y=\cos \theta, z^{2}=\cos 2 \theta$
(D) $x=y=\cos \theta, z^{2}=-\cos 2 \theta$
17. If $S$ is the circumcentre, $O$ is the orthocentre of $\triangle A B C$, then $S A+S B+S C=$
(A) $S O$
(B) $2 S O$
(C) $O S$
(D) $2 O S$
18. If $a, c, d$ are non-coplanar vectors and $d \cdot\{a \times[b \times$ $(c \times d)]\}$ is equal to
(A) $(b \cdot d)[a c d]$
(B) $(a \cdot d)[a c d]$
(C) $(c \cdot d)[a c d]$
(D) none of these
19. If $4 a+5 b+9 c=0$, then
$(a \times b) \times[(b \times c) \times(c \times a)]$ is equal to
(A) A vector perpendicular to the plane of $a, b$ and $c$
(B) A scalar quantity
(C) 0
(D) none of these

---

20. If $\sum_{i=1}^{n} a_{i}=0$ where $\left|a_{i}\right|=1 \nvdash i$, then the value of is $\sum_{1 \leq i<j \leq n} \sum a_{i} \cdot a_{j}$
(A) $-\frac{n}{2}$
(B) $-n$
(C) $\frac{n}{2}$
(D) $n$
21. Forces $P, Q$ act at $O$ and have a resultant $R$. If any transversal cuts their lines of action at $A, B, C$, respectively, then
(A) $\frac{\mathrm{P}}{O A}+\frac{\mathrm{Q}}{O B}+\frac{\mathrm{R}}{O C}=0$
(B) $\frac{\mathrm{P}}{O A}+\frac{\mathrm{Q}}{O B}+\frac{\mathrm{R}}{O C}=1$
(C) $\frac{\mathrm{P}}{O A}+\frac{\mathrm{Q}}{O B}-\frac{\mathrm{R}}{O C}=0$
(D) $\frac{\mathrm{P}}{O A}+\frac{\mathrm{Q}}{O B}-\frac{\mathrm{R}}{O C}=1$.
22. A vector $A$ has components $A_{1}, A_{2}, A_{3}$ in a right-handed rectangular cartesian coordinate system $O x, O y, O z$. The coordinate system is rotated about the z-axis through an angle $\frac{\pi}{2}$. The components of $A$ in the new coordinate system are
(A) $A_{1},-A_{2}, A_{3}$
(B) $A_{2}, A_{1}, A_{3}$
(C) $A_{1}, A_{2},-A_{3}$
(D) $A_{2},-A_{1}, A_{3}$.
23. In a $\triangle O A B, \mathrm{E}$ is the mid-point of $O B$ and $D$ is a point on $A B$ such that $A D: D B=2: 1$. If $O D$ and $A E$ intersect at $P$, then the ratio $O P: P D$ is
(A) $1: 2$
(B) $2: 1$
(C) $3: 2$
(D) $2: 3$.
24. If $a, b, c$ are three non-parallel unit vectors such that $a \times(b \times c)=\frac{1}{2} b$, then the angles which a makes with $b$ and $c$ are
(A) $90^{\circ}, 60^{\circ}$
(B) $45^{\circ}, 60^{\circ}$
(C) $30^{\circ}, 60^{\circ}$
(D) none of these
25. If $a=i+j-k, b=i-j+k$ and $c$ is a unit vector perpendicular to the vector $a$ and coplanar with $a$ and $b$, then a unit vector $d$ perpendicular to both $a$ and $c$ is
(A) $\frac{1}{\sqrt{6}}(2 i-j+k)$
(B) $\frac{1}{\sqrt{2}}(i+j)$
(C) $\frac{1}{\sqrt{2}}(j+k)$
(D) $\frac{1}{\sqrt{2}}(i+k)$
26. If the $p$ th, $q$ th and $r$ th terms of a G. P. are positive numbers $a, b$ and $c$, respectively, then the angle between
the vectors $i l_{a} a+j l_{a} b+k l_{a} c$ and $i(q-r)+j(r-p)+k$ $(p-q)$ is
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{6}$
(C) $\frac{\pi}{2}$
(D) none of these
27. A vector $a$ is collinear with vector $b=\left(6,-8,-7 \frac{1}{2}\right)$ and make an acute angle with the positive direction of $z$-axis. If $|a|=50$, then $a=$
(A) $(24,32,30)$
(B) $(24,-32,30)$
(C) $(-24,32,30)$
(D) none of these
28. The perpendicular distance of a corner of a unit cube form a diagonal not passing through it is
(A) $\sqrt{6}$
(B) $\frac{\sqrt{6}}{3}$
(C) $\frac{3}{\sqrt{6}}$
(D) none of these
29. The vectors $a, b$ and $c$ are equal in length and taken pairwise, they make equal angles. If $a=i+j, b=j+$ $k$, and $c$ makes an obtuse angle with the base vector $i$, then $c$ is equal to
(A) $i+k$
(B) $-i+4 j-k$
(C) $\frac{-1}{3} i+\frac{4}{3} j-\frac{1}{3} k$
(D) $\frac{1}{3} i+\frac{-4}{3} j+\frac{1}{3} k$.
30. If the four points $a, b, c, d$ are coplanar, then $[b c d]+[c a d]+[a b d]=$
(A) 0
(B) 1
(C) -1
(D) $[a b c]$
31. A tetrahedron has vertices at $O(0,0,0), A(1,2,1)$, $B(2,1,3)$ and $C(-1,1,2)$. Then, the angle between the faces $O A B$ and $A B C$ will be
(A) $\cos ^{-1}\left(\frac{19}{35}\right)$
(B) $\cos ^{-1}\left(\frac{71}{31}\right)$
(C) $30^{\circ}$
(D) $90^{\circ}$
32. If a quadrilateral $A B C D$ is such that $A B=b, A D=d$ and $A C=m b+p d(m+p \geq 1)$, then the area of the quadrilateral is $k(p+m)|b \times d|$, where $k$ is equal to
(A) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{2}$
(D) none of these
33. Let $a$ be a unit vector and $b$ be a non-zero vector not parallel to $a$. If two sides of the triangle are represented

---

by the vectors $\sqrt{3}(a \times b)$ and $b-(a \cdot b) a$, then the angles of the triangle are
(A) $30^{\circ}, 90^{\circ}, 60^{\circ}$
(B) $45^{\circ}, 45^{\circ}, 90^{\circ}$
(C) $60^{\circ}, 60^{\circ}, 60^{\circ}$
(D) none of these
34. Let $u$ and $v$ be unit vectors. If $w$ is a vector such that $w$ $+(w \times u)=v$, then $|(u \times v) \cdot w|$
(A) $\leq \frac{1}{3}$
(B) $\leq \frac{1}{2}$
(C) $>\frac{1}{3}$
(D) $\geq \frac{1}{2}$
35. If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector, then $(a \cdot b) b+(a \cdot c) c+\frac{a \cdot(b+c)}{|b+c|^{2}}(b \times c)=$
(A) $a$
(B) $b$
(C) $c$
(D) none of these
36. If the vector $-i+j-k$ bisects the angle between $3 i+4 j$ and vector $c$, then the unit vector along $c$ is
(A) $\frac{-11 i-10 j-2 k}{15}$
(B) $\frac{-11 i+10 j+2 k}{15}$
(C) $\frac{-11 i+10 j-2 k}{15}$
(D) none of these
37. If $a, b$ and $c$ are three unit vectors such that $a+b+c$ is also a unit vector and $\theta_{1}, \theta_{2}$ and $\theta_{3}$ are angles between the vectors $a, b ; b, c$ and $c, a$, respectively, then among $\theta_{1}, \theta_{2}$ and $\theta_{3}$.
(A) all are acute angles
(B) all are right angles
(C) at least one is obtuse angle
(D) none of these
38. If $x$ and $y$ are two non-collinear vectors and $A B C$ is a triangle with side lengths $a, b, c$ satisfying
$(20 a-15 b) x+(15 b-12 c) y+(12 c-20 a)(x \times y)=\overline{0}$, then $\triangle A B C$ is
(A) an acute-angled triangle
(B) an obtuse-angled triangle
(C) a right-angled triangle
(D) an isosceles triangle
39. If $\alpha(a \times b)+\beta(b \times c)+\gamma(c \times a)=0$, then
(A) a, b, c are coplanar if all of $\alpha, \beta, \gamma \neq 0$
(B) a, b, c are coplanar if any one of $\alpha, \beta, \gamma \neq 0$
(C) a, b, c are non-coplanar for any $\alpha, \beta, \gamma$
(D) none of these
40. If $p \times q=r$ and $p . q=c$, then $q=$
(A) $\frac{c p-p \times r}{|p|^{2}}$
(B) $\frac{c p+p \times r}{|p|^{2}}$
(C) $\frac{c r-p \times r}{|p|^{2}}$
(D) $\frac{c r+p \times r}{|p|^{2}}$
41. Let $a=i+j$ and $b=2 i-k$. The point of intersection of the lines $r \times a=b \times a$ and $r \times b=a \times b$ is
(A) $-i+j+k$
(B) $3 i-j+k$
(C) $3 i+j-k$
(D) $i-j-k$
42. The sides of a parallelogram are $2 i+4 j-5 k$ and $i+2 j$ $+3 k$. The unit vector parallel to one of the diagonals size is
(A) $\frac{1}{7}(3 i+6 j-2 k)$
(B) $\frac{1}{7}(3 i-6 j-2 k)$
(C) $\frac{1}{7}(-3 i+6 j-2 k)$
(D) $\frac{1}{7}(3 i+6 j+2 k)$

# More than One Option Correct Type 

43. If $x+y=a, x \times y=b$ and $x \cdot a=1$, then
(A) $x=\frac{a+a \times b}{a^{2}}$
(B) $y=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$
(C) $x=\frac{b+a \times b}{a^{2}}$
(D) $y=\frac{\left(b^{2}-1\right) b-a \times b}{a^{2}}$
44. If $a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c$ and $(c \cdot c) a=c$, while $b$ and $c$ are non-collinear, then
(A) $\alpha=\frac{\pi}{2}$
(B) $\alpha=\frac{\pi}{3}$
(C) $\beta=1$
(D) $\beta=-1$
45. If $x \times b=c \times b$ and $x \perp a$ then $x$ is equal to
(A) $\frac{b \times(a \times c)}{b \cdot c}$
(B) $\frac{(b \times c) \times a}{b \cdot a}$
(C) $\frac{a \times(c \times b)}{a \cdot b}$
(D) none of these
46. Let $b=4 i+3 j$ and $c$ be two vectors perpendicular to each other in $x y$-plane, then the vector in the same plane having projections 1 and 2 along $b$ and $c$ respectively is

---

(A) $2 i-j$
(B) $-2 i+j$
(C) $2 i+j$
(D) none of these
47. A vector of magnitude 2 along a bisector of the angle between the two vectors $2 i-2 j+k$ and $i+2 j-2 k$ is
(A) $\frac{2}{\sqrt{10}}(3 i-k)$
(B) $\frac{1}{\sqrt{26}}(i-4 j+3 k)$
(C) $\frac{2}{\sqrt{26}}(i-4 j+3 k)$
(D) none of these
48. The value of $\lambda$ such that $(x, y, z) \neq(0,0,0)$ and $(i+j+3 k) x+(3 i-3 j+k) y+(-4 i+5 j) z$ $=\lambda(x i+y j+z k)$ is
(A) 0
(B) 1
(C) -1
(D) none of these
49. A unit vector $a$ makes an angle $\frac{\pi}{4}$ with $i$ and $\frac{\pi}{3}$ with $j$. If the angle between $a$ and $k$ is $\theta$, where $\theta$ is acute, then
(A) $a=\frac{1}{2} i+\frac{1}{\sqrt{2}} j+\frac{1}{2} k$
(B) $a=\frac{1}{\sqrt{2}} i+\frac{1}{2} j+\frac{1}{2} k$
(C) $\theta=\frac{\pi}{3}$
(D) $\theta=\frac{\pi}{6}$
50. If the three vectors $a=(12,4,3), b=(8,-12,-9)$ and $c=(33,-4,-24)$ define a parallelopiped, then
(A) the lengths of the edges are $13,17,41$
(B) areas of the faces are $220,435,455$
(C) volume of parallelopiped is 3696
(D) all of these
51. A vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $a=\frac{1}{3}(i-2 j+2 k)$,
$b=\frac{1}{5}(-4 i-3 k)$ and $c=j$ is given by
(A) $5 i-j-5 k$
(B) $-5 i+j+5 k$
(C) $5 i+j+5 k$
(D) none of these
52. $A, B, C$ and $D$ are four points such that $A B=$ $m(2 i-6 j+2 k), B C=i-2 j$ and $C D=$ $n(-6 i+15 j-3 k)$. If $A B$ and $C D$ intersect at some point $E$, then
(A) $m \geq \frac{1}{2}$
(B) $n \geq \frac{1}{3}$
(C) area of $\triangle B C E=\frac{1}{2} \sqrt{6}$
(D) all of these
53. The position vectors of two points $A$ and $C$ are $9 i-j+$ $7 k$ and $7 i-2 j+7 k$, respectively. The point of intersection of vectors $A B=4 i-j+3 k$ and $C D=2 i-j+2 k$ is $P$. If vector $P Q$ is perpendicular to $A B$ and $C D$ and $P Q$ $=15$ units, the position vector of $Q$ is
(A) $6 i-9 j-9 k$
(B) $-4 i+11 j+11 k$
(C) $6 i+9 j-9 k$
(D) none of these
54. If $D A=a, A B=b$ and $C B=k a$, where $k>0$ and $X, Y$ are the mid-points of $D B$ and $A C$ respectively such that $|a|=17$ and $|X Y|=4$, then $k$ is equal to
(A) $\frac{8}{17}$
(B) $\frac{9}{17}$
(C) $\frac{25}{17}$
(D) 1
55. Let $a$ and $b$ be two non-collinear unit vectors. If $u=a-(a \cdot b) b$ and $v=a \times b$, then $|v|$ is
(A) $|u|$
(B) $|u|+|u \cdot a|$
(C) $|u|+|u \cdot b|$
(D) $|u|+u \cdot(a+b)$
56. A non-zero vector $a$ is parallel to the line of intersection of the plane determined by the vectors $i, i+j$ and the plane determined by the vectors $i-j, i+k$. The angle between $a$ and the vector $i-2 j+2 k$ is
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{4}$
(C) $\frac{3 \pi}{4}$
(D) none of these
57. A unit vector coplanar with $i+j+2 k$ and $i+2 j+k$ and perpendicular to $i+j+k$ is
(A) $\frac{j-k}{\sqrt{2}}$
(B) $\frac{-j+k}{\sqrt{2}}$
(C) $\frac{j+k}{\sqrt{2}}$
(D) $\frac{-(j+k)}{\sqrt{2}}$
58. The vectors $a, b, c$ are of same length and taken pairwise, they form equal angles. If $a=i+j$ and $b=j+k$, then $c=$
(A) $i+k$
(B) $j+k$
(C) $i+k$
(D) $-\frac{i}{3}+\frac{4}{3} j-\frac{k}{3}$

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59. The vector $a=-i+2 j+k$ is rotate $D$ through a right angle Passing through the $y$-axis in its way. The new vector is
(A) $\sqrt{3}(i+j-k)$
(B) $-\sqrt{3}(i+j-k)$
(C) $\sqrt{2}(i+j-k)$
(D) $-\sqrt{2}(i+j-k)$
60. $a$ and $c$ are unit vectors and $|b|=4$ with $a \times b=2 a \times c$. The angle between $a$ and $c$ is $\cos ^{-1}\left(\frac{1}{4}\right)$. Then, $b-2 c$ $=1 a$, if $\lambda$ is
(A) 3
(B) -3
(C) 4
(D) -4
61. The vector c directed along the bisectors of the angle between the vectors $a=7 i-4 j-4 k$ and $\vec{b}=-2 i-j+2 k$ if $|c|=3 \sqrt{6}$, is given by
(A) $i-7 j+2 k$
(B) $2 i+7 j-3 k$
(C) $-i+7 j-2 k$
(D) $4 i+7 j-4 k$

# Passage Based Questions 

## Passage 1

The vector differential operator DEL, written $\nabla$, is defined by $\nabla=\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k=i \frac{\partial}{\partial x}+j \frac{\partial}{\partial y}+k \frac{\partial}{\partial z}$, where $\frac{\partial}{\partial x}$ represents the derivative w.r.t. $x$ regarding $y$ and $z$ as constant. Similarly, $\frac{\partial}{\partial y}$ represents the derivative w.r.t. $y$ regarding $x$ and $z$ as constant and $\frac{\partial}{\partial z}$ represents the derivative w.r.t. $z$ regarding $x$ and $y$ as constant. The operator $\nabla$ is also known as nabla.

Let $\phi(x, y, z)$ be defined and differentiable at each point $(x, y, z)$ in a certain region of space. Then, the gradient of $\phi$, written $\nabla \phi$ or grad $\phi$, is defined by

$$
\nabla \phi=\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) \phi=\frac{\partial \phi}{\partial x} i+\frac{\partial \phi}{\partial y} j+\frac{\partial \phi}{\partial z} k
$$

Let $r$ be any vector such that $r=x i+y j+z k$
62. If $\phi=\ln |r|$ then $\nabla \phi=$
(A) $\frac{r}{r^{2}}$
(B) $\frac{r}{r^{3}}$
(C) $\frac{r}{r^{4}}$
(D) $\frac{r}{r}$
63. If $\phi=\frac{1}{r}$, then $\nabla \phi=$
(A) $\frac{r}{r}$
(B) $\frac{r}{r^{2}}$
(C) $\frac{r}{r^{3}}$
(D) $-\frac{r}{r^{3}}$
64. $\nabla r^{n}=$
(A) $n r n^{-1} r$
(B) $n r n-2 r$
(C) $n r n r$
(D) none of these

## Passage 2

Let $\nabla(x, y, z)=V_{1} i+V_{2} j+V_{3} k$ be defined and differentiable at each point $(x, y, z)$ in a certain region of space. Then, the divergence of $V$, written $\nabla \cdot V$ or $\operatorname{div} V$ is defined by

$$
\begin{aligned}
\nabla \cdot V & =\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) \cdot\left(V_{1} i+V_{2} j+V_{3} k\right) \\
& =\frac{\partial V_{1}}{\partial x}+\frac{\partial V_{2}}{\partial y}+\frac{\partial V_{3}}{\partial z}
\end{aligned}
$$

Here, $\nabla=\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k$ is the del operator.
Note the analogy with $A \cdot B=A_{1} B_{1}+A_{2} B_{2}+A_{3} B_{3}$. Also, note that $\vec{\nabla} \cdot V \neq V \cdot \nabla$.
65. If $\phi=2 x^{3} y^{2} z^{4}$, then $\nabla \cdot \nabla \phi=k\left(3 x y^{2} z^{4}+x^{3} z^{4}+6 x^{3} y^{2} z^{2}\right)$, where $k=$
(A) 2
(B) 3
(C) 4
(D) 6
66. If $r=x i+y j+z k$ then $\nabla^{2}\left(\frac{1}{r}\right)=$
(A) 1
(B) -1
(C) 0
(D) none of these
67. If $r=x i+y j+z k$, then $\nabla \cdot\left(\frac{r}{r^{3}}\right)=$
(A) 0
(B) 1
(C) -1
(D) none of these

## Passage 3

Let $V(x, y, z)=V_{1} i+V_{2} j+V_{3} k$ be defined and differentiable at each point $(x, y, z)$ in a certain region of space. Then, the curl or roation of $\vec{V}$, written $\nabla \times \vec{V}$, curl $\vec{V}$ or $\operatorname{rot} \vec{V}$, is defined by

---

$$
\begin{aligned}
\nabla \times V & =\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) \times\left(V_{1} i+V_{2} j+V_{3} k\right) \\
& =\left|\begin{array}{ccc}
i & j & k \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
V_{1} & V_{2} & V_{3}
\end{array}\right| \\
& =\left(\frac{\partial V_{3}}{\partial y}-\frac{\partial V_{2}}{\partial z_{2}}\right) i+\left(\frac{\partial V_{3}}{\partial z}-\frac{\partial V_{2}}{\partial x}\right) j+\left(\frac{\partial V_{2}}{\partial x}-\frac{\partial V_{1}}{\partial y}\right) k
\end{aligned}
$$

Note that in the expansion of the determinant the operators $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$ must precede $V_{1}, V_{2}, V_{3}$.

## Match the Column Type

70. 

## Column-I

I. The points $O, A, B, C, D$ are such that $O A=a, O B=b, O C=2 a+3 b$ and $O D$ $=a-2 b$. If $|a|=3|b|$, then the angle between $B D$ and $A C$ is
II. $a, b, c$ are three unit vectors such that $a \times(b \times c)=\frac{1}{2}(b+c)$. If the vectors
(B) $\frac{2 \pi}{3}$
$b$ and $c$ are non-parallel, then the angle between $a$ and $b$ is
III. Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times(c \times d)=0$. Let $P_{1}$ and $P_{2}$ be planes determined by the pairs of vectors $a, b$ and $c, d$ respectively, then the angle between $P_{1}$ and $P_{2}$ is
IV. If $a$ and $b$ are two vectors such that $a \cdot b<0$ and $|a \cdot b|=|a \times b|$, then the angle between vectors $a$ and $b$ is
(С) $\frac{3 \pi}{4}$

## Assertion-Reason Type

Instructions: In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options:
(A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A)
(B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A)
(C) Assertion(A) is True, Reason(R) is False
(D) Assertion(A) is False, Reason(R) is True
71. Assertion: $i, j, k$ are orthonormal unit vectors and $a$ is any vector. If $a \times r=j=\hat{j}$, then $a \cdot r$ is arbitrary scalar.
Reason: For any two arbitrary vectors $a$ and $r$, $|a \times r|^{2}+|a \cdot r|^{2}=|a|^{2}|r|^{2}$
72. Assertion: If $a, b, c$ are three non-coplanar, non-zero vectors, then
$(a \cdot a) b \times c+(a \cdot b) c \times a+(a \cdot c) a \times b=[b c a] a$
Reason: If the vectors $a, b, c$ are non-coplanar, then so are $b \times c, c \times a, a \times b$
73. Assertion: If

$$
\begin{aligned}
& i \times[(a-j) \times i]+j \times[(a-k) \times j]+k \\
& \times[(a-i) \times k]=0, \text { then } \\
& a=\frac{1}{2}(i+j+k)
\end{aligned}
$$

Reason: $(a \cdot i) i+(a \cdot j) j+(a \cdot k) k=a$ for any vector $a$.

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# Previous Year's Questions 

74. Given two vectors are $\hat{i}-\hat{j}$ and $\hat{i}+2 \hat{j}$ the unit vector coplanar with the two vectors and perpendicular to first is:
[2002]
(A) $\frac{1}{\sqrt{ }}(\hat{i} \quad \hat{j})$
(B) $\frac{1}{\sqrt{5}}(2 \hat{i}+\hat{j})$
(C) $\pm \frac{1}{\sqrt{2}}(\hat{i}+\hat{k})$
(D) none of these
75. The vector $\hat{i}+x \hat{j}+3 \hat{k}$ is rotated through an angle $\theta$ and doubled in magnitude, then it becomes $4 \hat{i}+(4 x-2) \hat{j}+2 \hat{k}$. The value of $x$ are:
[2002]
(A) $\left\{-\frac{2}{3}, 2\right\}$
(B) $\left\{\frac{1}{3}, 2\right\}$
(C) $\left\{\frac{2}{3}, 0\right\}$
(D) $\{2,7\}$
76. If the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ from the sides $B C, C A$ and $A B$ respectively of a triangle $A B C$, then:
[2002]
(A) $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{b}=0$
(B) $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$
(C) $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$
(D) $\vec{a} \times \vec{a}+\vec{a} \times \vec{c}+\vec{c} \times \vec{a}=0$
77. If the vectors $\vec{a}-x \hat{i}+y \hat{j}+z \hat{k}$ and such that $\vec{a}, \vec{c}$ and $\vec{b}$ form a right handed system, then $\vec{c}$ is:
[2002]
(A) $z \hat{i}-x \hat{k}$
(B) $\overrightarrow{0}$
(C) $y \hat{j}$
(D) $-z \hat{i}+x \hat{k}$
78. The centre of the circle given by $\vec{r} \cdot(\hat{i}+2 \hat{j}+2 \hat{k})=15$ and $|\vec{r} \cdot(\hat{j}+2 \hat{k})|=4$ is:
[2002]
(A) $(0,1,2)$
(B) $(1,3,4)$
(C) $(-1,3,4)$
(D) none of these
79. If $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$ and vectors $\left(1, a, a^{2}\right)\left(1, b, b^{2}\right)$ and $\left(1, c, c^{2}\right)$ are non-coplanar, then the product $a b c$ equals
[2003]
(A) 2
(B) -1
(C) 1
(D) 0
80. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\vec{a}+2 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3 \vec{c}$ is collinear with $\vec{a}$ ( $\lambda$ being some non-zero scalar) then $\vec{a}+2 \vec{b}+6 \vec{c}$ equals
[2004]
(A) $\lambda \vec{a}$
(B) $\lambda \vec{b}$
(C) $\lambda \vec{c}$
(D) 0
81 A particle is acted upon by constant forces $4 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}+\hat{j}-\hat{k}$ which displace it from a point $\hat{i}+2 \hat{j}+3 \hat{k}$ to the point $5 \hat{i}+4 \hat{j}+\hat{k}$. The work done in standard units by the forces is given by
[2004]
(A) 40
(B) 30
(C) 25
(D) 15
82. If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}$ and $(2 \lambda-1) \bar{c}$ are non-coplanar for
[2004]
(A) all values of $\lambda$
(B) all except one value of $\lambda$
(C) all except two values of $\lambda$
(D) no value or $\lambda$
83. Let $\bar{u}, \bar{v}, \bar{w}$ be such that $|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3$. If the projection $\bar{v}$ along $\bar{u}$ is equal to that of $\bar{w}$ along $\bar{u}$ and $\bar{v}, \bar{w}$ are perpendicular to each other then $|\bar{u}-\bar{v}+\bar{w}|$ equals
[2004]
(A) 2
(B) $\sqrt{7}$
(C) $\sqrt{14}$
(D) 14
84. Let $\bar{a}, \bar{b}$ and $\bar{c}$ be non-zero vectors such that $(\bar{a} \times \bar{b}) \times \bar{c}=-|\bar{b}| \bar{c} \mid \bar{a}$. If $\theta$ is the acute angle between the vectors $\bar{b}$ and $\bar{c}$ then $\sin \theta$ equals [2004]
(A) $\frac{1}{3}$
(B) $\frac{\sqrt{2}}{3}$
(C) $\frac{2}{3}$
(D) $\frac{2 \sqrt{2}}{3}$
85. If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$, then
[2005]
(A) $\overrightarrow{P A}+\overrightarrow{P B}=2 \overrightarrow{P C}$
(B) $\overrightarrow{P A}+\overrightarrow{P B}=\overrightarrow{P C}$
(C) $\overrightarrow{P A}+\overrightarrow{P B}+2 \overrightarrow{P C}=0$
(D) $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=0$
86. The distance between the line $\vec{r}=2 \hat{i}-2 \hat{j}+3 \hat{k}$ $+\lambda(\hat{i}+\hat{j}+4 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5$ is
[2005]
(A) $\frac{10}{9}$
(B) $\frac{10}{3 \sqrt{3}}$
(C) $\frac{3}{10}$
(D) $\frac{10}{3}$
87. For any vector $\vec{a}$, the value of $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}$ is equal to
[2005]

---

(A) $3 \vec{a}^{2}$
(B) $\vec{a}^{2}$
(C) $2 \vec{a}^{2}$
(D) $4 \vec{a}^{2}$
88. If non-zero numbers $a, b, c$ are in H.P., then the straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$ always passes through a fixed point. That point is
(A) $(-1,2)$
(B) $(-1,-2)$
(C) $(1,-2)$
(D) $\left(1,-\frac{1}{2}\right)$
89. Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a \vec{i}+a \vec{j}+c \vec{k}, \vec{i}+\vec{k}$ and $c \vec{i}+c \vec{j}+b \vec{k}$ lie in a plane, then $c$ is
(A) the Geometric Mean of $a$ and $b$
(B) the Arithmetic Mean of $a$ and $b$
(C) equal to zero
(D) the Harmonic Mean of $a$ and $b$
90. If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$, is a real number then $[\lambda(\vec{a}+\vec{b}) \lambda^{2} \vec{b} \lambda \vec{c}]=[\vec{a} \vec{b}+\vec{c} \vec{b}]$ for [2005]
(A) exactly one value of $\lambda$
(B) no value of $\lambda$
(C) exactly three values of $\lambda$
(D) exactly two values of $\lambda$
91. Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}$ $+(1+x-y) \hat{k}$. Then $[\vec{a}, \vec{b}, \vec{c}]$ depends on
(A) only $y$
(B) only $x$
(C) both $x$ and $y$
(D) neither $x$ nor $y$
92. If $(\vec{a} \times \vec{b}) \times \vec{c}=\vec{a} \times(\vec{b} \times \vec{c})$, where $\vec{a}, \vec{b}$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} \neq 0, \vec{b} \cdot \vec{c} \neq 0$, then $\vec{a}$ and $\vec{c}$ are
[A) inclined at an angle of $\frac{\pi}{3}$ between them
(B) inclined at an angle of $\frac{\pi}{6}$ between them
(C) perpendicular
(D) parallel
93. The values of $a$, for which the points $A, B, C$ with position vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $a \hat{i}-3 \hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $C=\frac{\pi}{2}$ are
(A) 2 and 1
(B) -2 and -1
(C) -2 and 1
(D) 2 and -1
94. If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2 \hat{u} \times 3 \hat{v}$ is a unit vector for [2007]
(A) exactly two values of $\theta$
(B) more than two values of $\theta$
(C) no value of $\theta$
(D) exactly one value of $\theta$
95. Let $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\bar{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$. If the vector $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$, then $x$ equals
(A) 0
(B) 1
(C) -4
(D) -2
96. The vector $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\hat{j}+\hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$ ?
(A) $\alpha=2, \beta=2$
(B) $\alpha=1, \beta=2$
(C) $\alpha=2, \beta=1$
(D) $\alpha=1, \beta=1$
97. The non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are related by $\vec{a}=8 \vec{b}$ and $\vec{c}=7 \vec{b}$. Then the angle between $\vec{a}$ and $\vec{c}$ is[2008]
(A) 0
(B) $\pi / 4$
(C) $\pi / 2$
(D) $\pi$
98. If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers, then the equality $[3 \vec{u} \quad p \vec{v} \quad p \vec{w}]-[p \vec{v} \quad \vec{w} \quad q \vec{u}]$ $-[2 \vec{w} \quad q \vec{v} q \vec{u}]=0$ holds for
(A) exactly one value of $(p, q)$
(B) exactly two values of $(p, q)$
(C) more than two but not all values of $(p, q)$
(D) all values of $(p, q)$
99. The projections of a vector on the three coordinate axis are $6,-3,2$ respectively. The direction cosines of the vector are
(A) $6,-3,2$
(B) $\frac{6}{5},-\frac{3}{5}, \frac{2}{5}$
(C) $\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$
(D) $-\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$
100. Let $\vec{a}=\vec{j}-\vec{k}$ and $\vec{c}=\vec{i}-\vec{j}-\vec{k}$. Then, the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $a \cdot \vec{b}=3$ is
(A) $2 \hat{i}-\hat{j}+2 \hat{k}$
(B) $\hat{i}-\hat{j}-2 \hat{k}$
(C) $\hat{i}+\hat{j}-2 \hat{k}$
(D) $-\hat{i}+\hat{j}-2 \hat{k}$
101. If the vectors $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}$ $+\mu \hat{k}$ are mutually orthogonal, then the tuple $(\lambda, \mu)=$
(A) $(2,-3)$
(B) $(-2,3)$
(C) $(3,-2)$
(D) $(-3,2)$

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102. If $a=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $b=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of the expression $(2 z-b) \cdot[(a \times b) \times(a+2 b)]$ is
[A) -3
(B) 5
(C) 3
(D) -5
103. The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying: $\vec{b} \times \vec{c}=\vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=0$. Then, the vector $\vec{d}$ is equal to
[A) $c+\left(\frac{a \cdot c}{a \cdot b}\right) b$
(B) $b+\left(\frac{b \cdot c}{a \cdot b}\right) c$
(C) $c-\left(\frac{a \cdot c}{a \cdot b}\right) b$
(D) $b-\left(\frac{b \cdot c}{a \cdot b}\right) c$
104. Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\vec{c}=\hat{a}+2 \vec{b}$ and $d=5 \hat{a}-4 \hat{b}$ are perpendicular to each other, then the angle between $\hat{a}$ and $\vec{b}$ is
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{2}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{4}$
105. Let $A B C D$ be a parallelogram such that $\overrightarrow{A B}=\overrightarrow{q}, \overrightarrow{A D}=\overrightarrow{p}$ and $\square B A D$ be an acute angle. If $\vec{r}$ is the vector which coincides with the altitude directed from the vertex $B$ to the side $A D$, then $\vec{r}$ is given by
(A) $\vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$
(B) $\vec{r}=-\vec{q}+\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$
(C) $\vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$
(D) $\vec{r}=-3 \vec{q}+\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$
106. If the vectors $\overrightarrow{A B}=3 \hat{i}+4 \hat{k}$ and $\overrightarrow{A C}=5 \hat{i}+2 \hat{j}+4 \hat{k}$ represent the sides of a triangle $A B C$, then the length of the median through $A$ is
(A) $\sqrt{72}$
(B) $\sqrt{33}$
(C) $\sqrt{45}$
(D) $\sqrt{18}$
107. If $[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=\lambda[\vec{a} \vec{b} \vec{c}]^{2}$, then the value of $\lambda$ is equal to
(A) 2
(B) 3
(C) 0
(D) 1
108. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$. If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c}$, then a value of $\sin \theta$
(A) $\frac{-\sqrt{2}}{3}$
(B) $\frac{2}{3}$
(C) $\frac{-2 \sqrt{3}}{3}$
(D) $\frac{2 \sqrt{2}}{3}$
109. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a} \times(\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})$. if $\vec{b}$ is not parallel to $\vec{c}$, then the angle between $\vec{a}$ and $\vec{b}$ is
(A) $\frac{5 \pi}{6}$
(B) $\frac{3 \pi}{4}$
(C) $\frac{\pi}{2}$
(D) $\frac{2 \pi}{3}$

# ANSWER KEYS 

## Single Option Correct Type

1. (A)
2. (B)
3. (B)
4. (A)
5. (B)
6. (A)
7. (A)
8. (C)
9. (B)
10. (A)
11. (B)
12. (A)
13. (C)
14. (A)
15. (A)
16. (D)
17. (A)
18. (A)
19. (C)
20. (A)
21. (C)
22. (D)
23. (C)
24. (A)
25. (C)
26. (C)
27. (C)
28. (B)
29. (C)
30. (D)
31. (A)
32. (C)
33. (A)
34. (B)
35. (A)
36. (A)
37. (C)
38. (C)
39. (B)
40. (A)

## More than One Option Correct Type

43. (A, B) 44. (A, C) 45. (B, C) 46. (A, B) 47. (A, C) 48. (A, C) 49. (B, D) 50. (A, B, C, D)
44. (A, B) 52. (A, B, C, D)
45. (A, B)
46. (B, C)
47. (A, B)
48. (A, D)
49. (C, D)
50. (A, D)

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# 21.30 <br> ATM $^{21}$ : FREE EDUCATION TO ALL APUL 

## Passage Based Questions

62. (A) 63. (D) 64. (B)
63. (C)
64. (C)
65. (A)
66. (B) 69. (C)

## Match the Column Type

70. $\mathrm{I} \leftrightarrow(\mathrm{D}), \mathrm{II} \leftrightarrow(\mathrm{B}), \mathrm{III} \leftrightarrow(\mathrm{A}), \mathrm{IV} \leftrightarrow(\mathrm{C})$

## Assertion-Reason Type

71. (A)
72. (A)
73. (A)

## Previous Year's Questions

74. (A)
75. (A)
76. (B)
77. (A)
78. (B)
79. (B)
80. (D)
81. (A)
82. (C)
83. (C)
84. (A)
85. (A)
86. (B)
87. (C)
88. (C)
89. (C)
90. (B)
91. (D)
92. (D)
93. (A)
94. (D)
95. (D)
96. (D)
97. (D)
98. (A)
99. (C)
100. (B)
101. (D)
102. (D)
103. (C)
104. (C) 105. (B)
106. (B)
107. (D)
108. (D)
109. (A)

## HINTS AND SOLUTIONS

## Single Option Correct Type

1. Let $r=x_{1} a+x_{2} b+x_{3}(a \times b)$

Then, $r \cdot a=x_{1}|a|^{2}, r \cdot b=x_{2}|b|^{2}$
and, $r \cdot(a \times b)=x_{3}|a \times b|^{2}$
$\Rightarrow x_{1}=0, x_{2}=1, x_{3}=1 \therefore \quad r=a \times b+b$
2. We have, $a \times(a \times c)+3 b=0$
$\Rightarrow(a \cdot c) a-(a \cdot a) c+3 b=0$
$\Rightarrow(2 \sqrt{3} \cos \theta) a-3 c+3 b=0$
$\Rightarrow(2 \cos \theta) a-\sqrt{3} c+\sqrt{3} b=0$
$\Rightarrow|(2 \cos \theta) a-\sqrt{3} c|^{2}=|-\sqrt{3} b|^{2}$
$\Rightarrow 4 \cos ^{2} \theta|a|^{2}+3|c|^{2}-4 \sqrt{3} \cos \theta(a \cdot c)=3|b|^{2}$
$\Rightarrow 12 \cos ^{2} \theta+12-4 \sqrt{3} \cos \theta \times \sqrt{3} \times 2 \cos \theta=3$
$\Rightarrow 12 \cos ^{2} \theta+9-24 \cos ^{2} \theta=0$
$\Rightarrow 12 \cos ^{2} \theta=9 \Rightarrow \cos ^{2} \theta=\frac{9}{12}=\frac{3}{4}$.
3. Let $i, j$ be unit vectors along the coordinate axes
$\therefore \quad a=2 p i+1 \cdot j$
On rotation, let $b$ be the vector having components $p+1$ and 1 .
$\therefore \quad b=(p+1) i+1 \cdot j$
where $\mathrm{i}, j$ are unit vectors along the new coordinate axes.
But on rotation $|b|=|a| \Rightarrow|b|^{2}=|a|^{2}$
$\Rightarrow(p+1)^{2}+1=(2 p)^{2}+1 \Rightarrow 3 p^{2}-2 p-1=0$
$\Rightarrow(3 p+1)(p-1)=0 \Rightarrow p=1$ or $-\frac{1}{3}$.
4. Here, $a$ and $c=a \times b$ are non-collinear vectors.
$\therefore \quad$ Let $b=x a+y(a \times c)$
$\therefore \quad \beta=a \cdot b=a \cdot[x a+y(a \times c)]$
$=x|a|^{2}+y a \cdot(a \times c)=x a^{2} \Rightarrow x=\beta / a^{2}$.
And, $c=a \times b=a \times[x a+y(a \times c)]$
$=x a \times a+y a \times(a \times c)$
$=0+y(a \cdot c) a-y(a \cdot a) c$
$=y\{a \cdot(a \times b)\} a-y a^{2} c=-y a^{2} c$
$\Rightarrow y=-1 / a^{2}$
$\therefore$ from (1), $b=(\beta a-a \times c) / a^{2}$.
5. Since $A B C D E F$ is a regular hexagon, from plane geometry, we have
![img-21.jpeg](img-21.jpeg)

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$A D=2 B C$ and $F C=2 A B$
$\therefore \quad A D=2 B C$ and $F C=2 A B$.
Given that $A D=x B C$.
$\therefore \quad 2 B C=x B C$, by (i)
$\Rightarrow \quad x=2$.
Again, given that $C F=y A B$ or $-F C=y A B$.
$\therefore \quad-2 A B=y A B$, using (ii)
$\Rightarrow \quad y=-2$.
From (ii) and (iii), $x y=2(-2)=-4$.
6. $O M_{1}=O A+O B+O C+O D \quad$ (given)
$=O M+M A+O M+M B+O M+M C+O M+M D$
$=4 O M+(M A+M C)+(M B+M D)$
$=4 O M$
$(\because M A=-M C, M B=-M D)$
$\therefore \quad O M=\frac{1}{4} O M_{1}, \quad \therefore \quad \lambda=\frac{1}{4}$.
7. Since $M B$ is the component of $A B \perp$ to $A C$
![img-22.jpeg](img-22.jpeg)
$M B=A B-A M$
$=A B-\frac{(A B \cdot A C) A C}{(A C)^{2}}$
$=(6 j-2 k)-\frac{\{(6 j-2 k) \cdot(-5 i+3 j-k)\}(-5 i+3 j-k)}{(25+9+1)}$
$=(6 j-2 k)-\frac{20}{35}(-5 i+3 j-k)=\frac{10}{7}(2 i+3 j-k)$.
$\therefore \quad B M=-\frac{10}{7}(2 i+3 j-k)$.
8. A point equidistant from $A B$ and $A C$ is on the bisector of the angle $B A C$.
A vector along the internal bisector of the angle $B A C$
$=\frac{A B}{|A B|}+\frac{A C}{|A C|}$
$=\frac{3 i+j-k}{\sqrt{9+1+1}}+\frac{i-j+3 k}{\sqrt{1+1+9}}=\frac{1}{\sqrt{11}}(4 i+2 k)$
$\therefore \quad A P=t(2 i+k)$
$\therefore \quad B P=A P-A B=t(2 i+k)-(3 i+j-k)$
$=(2 t-3) i-j+(t+1) k$
Also, $B C=A C-A B=(i-j+3 k)-(3 i+j-k)$
$=-2 i-2 j+4 k$.
But $B P=s B C$.
$\therefore \quad(2 t-3) i-j+(t+1) k=s(-2 i-2 j+4 k)$
$\therefore \quad 2 t-3=-2 s,-1=-2 s, t+1=4 s$
$\therefore \quad s=\frac{1}{2}$ and $t=1 \quad \therefore \quad A P=2 i+k$.
9. Since $(a-d) \cdot(b-c)=0 \therefore D A \cdot C B=0$
$\therefore \quad A D \perp B C$.
Since $(b-d) \cdot(c-a)=0, \therefore D B \cdot A C=0$
$\therefore \quad B D \perp C A$.
Then, $D$ is the intersection of the altitudes through $A$ and $B$. Therefore, $D$ is the orthocentre of the triangle $A B C$.
10. We have, $(a+b) \times(a \times b)$
$=a \times(a \times b)+b \times(a \times b)$
$=(a \cdot b) a-(a \cdot a) b+(b \cdot b) a-(b \cdot a) b$
$=(a \cdot b)(a-b)+a-b(b \cdot b=b^{2}=1, a \cdot a=a^{2}=1$
as $a, b$ are unit vectors)
$=(a \cdot b+1)(a-b)$
$=x(a-b)$, where $x=a \cdot b+1$ is a scalar.
$\therefore \quad$ The given vector is parallel to $a-b$.
11. Clearly, $r \cdot c=\lambda[a b c]=\frac{1}{8} \lambda$
$r \cdot a=\mu[a b c]=\frac{1}{8} \mu$
$r \cdot b=v[a b c]=\frac{1}{8} v$
$\therefore \quad r \cdot(a+b+c)=\frac{1}{8}(\lambda+\mu+v)$
$\therefore \quad \lambda+\mu+v=8 r \cdot(a+b+c)$.
12. $\because \quad D B=D A+A B$
or, $D A=D B-A B$
![img-23.jpeg](img-23.jpeg)
$\therefore \quad(D A)^{2}=(D B)^{2}+(A B)^{2}-2 D B \cdot A B$
In parallelogram $2\left(a^{2}+b^{2}\right)=c^{2}+D B^{2}$
$\therefore \quad(D B)^{2}=2 a^{2}+2 b^{2}-c^{2}$
$\therefore \quad$ From (1), $b^{2}=2 a^{2}+2 b^{2}-c^{2}+a^{2}-2 A B \cdot D B$
$\therefore \quad A B \cdot D B=\frac{3 a^{2}+b^{2}-c^{2}}{2}$.
13. We have,

$$
\begin{aligned}
& \left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{1} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|^{2}=[a b c]^{2}=[(a \times b) \cdot c]^{2} \\
& =[|a||b| \sin \pi / 6 c \cdot c]^{2} \\
& =|a|^{2}|b|^{2} \sin ^{2} \pi / 6 \cdot 1 \\
& =\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right) \times 1 / 4 \\
& =1 / 4\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)
\end{aligned}
$$

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14. $\frac{1}{\sqrt{2}}(b+c)=a \times(b \times c)=(a \cdot c) b-(a \cdot b) c$
$\Rightarrow\left(a \cdot c-\frac{1}{\sqrt{2}}\right) b-\left(a \cdot b+\frac{1}{\sqrt{2}}\right) c=0$
Since $a, b, c$ are non-coplanar so $a, b, c$ are linearly independent. Hence,
$a \cdot b=-\frac{1}{\sqrt{2}}$.
$\therefore \quad \cos \theta=\frac{a \cdot b}{|a||b|}=a \cdot b=-\frac{1}{\sqrt{2}}$
( $\theta$ is the angle between $a$ and $b$ )
$\Rightarrow \theta=3 \pi / 4$.
15. We have $a \perp b \Rightarrow a, b, a \times b$ are linearly independent
$\Rightarrow \quad v$ can be expressed uniquely in terms of $a, b$ and $a \times b$.
Let $v=x a+y b+z a \times b$
Given that, : $a \cdot b=0, v \cdot a=0, v \cdot b=1,[v, a, b]=1$
$\therefore \quad v \cdot a=x a^{2}$ or $x a^{2}=0$ or $x=0$
$v \cdot b=x v \cdot b+y b^{2}+z b \cdot(a \times b)$
or, $\quad y b^{2}=1$ or $y=\frac{1}{b^{2}}$
$v \cdot a \times b=x \cdot 0+y \cdot 0+z|a \times b|^{2}$
or, $z|a \times b|^{2}=1$
or, $\quad z=\frac{1}{|a \times b|^{2}}$
From (1), (2), (3) and (4), we get
$v=\frac{1}{|b|^{2}} b+\frac{1}{|a \times b|^{2}} a \times b$.
16. We have, $c=x a+y b+z(a \times b)$
$\Rightarrow \quad c \cdot a=x$ and $c \cdot b=y \Rightarrow x=y=\cos \theta$
Now, $c \cdot c=|c|^{2}$
$\Rightarrow[x a+y b+z(a \times b)] \cdot[x a+y b+z(a \times b)]=|c|^{2}$
$\Rightarrow 2 x^{2}+z^{2}|a \times b|^{2}=1$
$\Rightarrow 2 x^{2}+z^{2}\left[|a|^{2}|b|^{2}-(a \cdot b)^{2}\right]=1$
$\Rightarrow 2 x^{2}+z^{2}[1-0]=1 \quad[\because \quad a \perp b \therefore a \cdot b=0]$
$\Rightarrow 2 x^{2}+z^{2}=1 \Rightarrow z^{2}=1-2 x^{2}=1-2 \cos ^{2} \theta=-\cos 2 \theta$.
17. From geometry
$2 S D=A O$, where $D$ is the mid-point of $B C$.
$\therefore \quad S A+S B+S C$
$=S A+S D+D B+S D+D C$
$=S A+2 S D \quad[\because D B+D C=0]$
$=S A+A O=S O$
$\therefore \quad S A+S B+S C=S O$.
![img-24.jpeg](img-24.jpeg)
18. We have, $d \cdot\{a \times[b \times(c \times d)]\}$
$=d \cdot(a \times((b \cdot d) c-(b \cdot c) d))$
$=d \cdot((b \cdot d)(a \times c)-(b \cdot c)(a \times d))$
$=(b \cdot d) d \cdot(a \times c)-(b \cdot c) d \cdot(a \times d)$
$=(b \cdot d)[a c d] \quad[\because d \cdot(a \times d)=0]$.
19. $4 a+5 b+9 c=0 \Rightarrow$ Vectors $a, b$ and $c$ are collinear $\Rightarrow(b \times c) \times(c \times a)=\hat{0}$
20. $\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} a_{i}\right)=\sum_{i=1}^{n}\left|a_{i}\right|^{2}+2 \sum_{1 \leq i<j \leq n} \sum a_{i} \cdot a_{j}$
$\Rightarrow 0=n+2 \sum_{1 \leq i<j \leq n} \sum a_{i} \cdot a_{j} \Rightarrow \sum_{1 \leq i<j \leq n} \sum a_{i} \cdot a_{j}=-\frac{n}{2}$
21. Let $O$ be taken as the origin of reference. Let $a, b, c$ be the position vectors of $A, B, C$ respectively so that $O A=a, O B=b, O C=c$.
![img-25.jpeg](img-25.jpeg)

Then, the vector representing the force $P$ along $O A$
$=P$ (unit vector along $O A)=P \frac{O A}{O A}=\frac{P}{O A} a$.
Similarly, the vector representing the force $Q$ along $O B$ $=\frac{Q}{O B} b$ and the vector representing the force $R$ along $O C$
$=\frac{R}{O C} c$.
Since $R$ is the resultant of $P$ and $Q$, we have
$(\mathrm{P} / O A) a+(\mathrm{Q} / O B) b=(\mathrm{R} / O C) c$
or, $(P / O A) a+(Q / O B) b-(R / O C) a=0$.
Since, $a, b, c$ are the position vectors of three collinear points, therefore, we have
the algebraic sum of the coefficients of $a, b, c$ in (i) $=0$
i.e., $(P / O A)+(Q / O B)-(R / O C)=0$

Hence, $\frac{P}{O A}+\frac{Q}{O B}=\frac{R}{O C}$.
22. When rotated through $\frac{\pi}{2}$, the new $x$-axis is along old $y$-axis and new $y$-axis is along the old negative $x$-axis; $z$-axis remains same as before. Hence, the components of $A$ in the new system are $A_{2},-A_{1}, A_{3}$.
23. Let $a$ and $b$ be the position vectors of $A$ and $B$ respectively w.r.t. $O$ as the origin.

Then p.v. of $E=O E$
$=\frac{1}{2} O B=\frac{1}{2} b$
and, p.v. of $D=O D=\frac{1}{3}(2 b+a), \because \quad A D: D B=2: 1$

---

Now let $O D$ and $A E$ intersect at $P$ such that $O P / O D=\lambda$ and $A P / A E=\mu$.
$\therefore \quad O P=\lambda O D$ and $A P=\mu A E$.
or, $\quad O P=\lambda O D=\frac{1}{3}(2 b+a) \lambda$.
and, $A P=\mu A E=\mu(O E-O A)=\mu\left(\frac{1}{2} b-a\right)$.
In $\triangle O A P, O A+A P=O P$
$\Rightarrow a+\mu\left(\frac{1}{2} b-a\right)=\frac{1}{3}(2 b+a) \lambda$
$\Rightarrow\left(1-\mu-\frac{1}{3} \lambda\right) a+\left(\frac{1}{2} \mu-\frac{2}{3} \lambda\right) b=0$
$\Rightarrow 1-\mu-\frac{1}{3} \lambda=0$ and $\frac{1}{2} \mu-\frac{2}{3} \lambda=0$,
$\because \quad a$ and $b$ are non-collinear.
Solving these equations, we get $\mu=\frac{4}{5}, \lambda=\frac{3}{5}$.
Now, $O P=\lambda O D, P D=O D-O P=(1-\lambda) O D$
$\therefore \frac{O P}{P D}=\frac{\lambda}{1-\lambda}=\frac{3 / 5}{1-3 / 5}=\frac{3}{2}$
$\Rightarrow O P=\frac{3}{2} P D \Rightarrow O P=\frac{3}{2} P D \Rightarrow O P: P D=3: 2$.
24. We have, $a \times(b \times c)=\frac{1}{2} b$
$\therefore(a \cdot c) b-(a \cdot b) c=\frac{1}{2} b$
$\Rightarrow\left(a \cdot c-\frac{1}{2}\right) b-(a \cdot b) c=0$
Since $b$ and $c$ are non-parallel,
(Given)
$\therefore \quad$ (i) exists if coefficients of $b$ and $c$ vanish separately
i.e., $a \cdot c-\frac{1}{2}=0$ and $a \cdot b=0$.
$\Rightarrow a \cdot c=\frac{1}{2}$ and $a \cdot b=0$.
Let $\theta_{1}$ and $\theta_{2}$ be the angles which $a$ makes with $b$ and $c$, respectively. Also $a, b, c$ are unit vectors,
$\therefore \cos \theta_{2}=\frac{1}{2}$ and $\cos \theta_{1}=0$
i.e., $\theta_{2}=60^{\circ}$ and $\theta_{1}=90^{\circ}$

Hence, $\theta_{1}=90^{\circ}$ and $\theta_{2}=60^{\circ}$.
25. Let $c=x a+y b$
$=x(i+j-k)+y(i-j+k)$
$=(x+y) i+(x-y) j-(x-y) k$
Since $c$ is $\perp$ to $a$
$\therefore \quad x+y+x-y+x-y=0 \Rightarrow 3 x=y$
Since $|c|=1$,
$\therefore \quad(x+y)^{2}+(x-y)^{2}+(x-y)^{2}=1$
$\therefore \quad(4 x)^{2}+(-2 x)^{2}+(-2 x)^{2}=1$.
$\Rightarrow 24 x^{2}=1 . \quad \therefore \quad x=\frac{1}{2 \sqrt{6}}$
$\therefore \quad c=4 x i+2 x j+2 x k=\frac{2}{\sqrt{6}} i-\frac{1}{\sqrt{6}} j+\frac{1}{\sqrt{6}} k$
i.e., $c=\frac{1}{\sqrt{6}}(2 i-j-k)$

Let $d=d_{1} i+d_{2} j+d_{3} k$
Since $d$ is $\perp$ to $a$ and $c$
$\therefore \quad d \cdot a=0 \Rightarrow d_{1}+d_{2}-d_{3}=0$
and, $d \cdot c=0 \Rightarrow 2 d_{1}-d_{2}+d_{3}=0$

$$
\Rightarrow 3 d_{1}=0, \therefore d_{1}=0, \therefore d_{2}=d_{3}
$$

Since $|d|=1$,
$\therefore \quad d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=1$
$\Rightarrow 0+d_{2}^{2}+d_{2}^{2}=1 \Rightarrow 2 d_{2}^{2}=1$
$\Rightarrow d_{2}^{2}=\frac{1}{2}$, or $d_{2}=\frac{1}{\sqrt{2}}, \therefore \quad d_{2}=d_{3}=\frac{1}{\sqrt{2}}$.
Hence, $d=\frac{1}{\sqrt{2}}(j+k)$.
26. Let $x$ be the first term and $y$ the common ratio of G.P.

Then, $a=p$ th term $=x y^{-p}-{ }^{1}$

$$
\begin{aligned}
& b=q \text { th term }=x y^{-q}-{ }^{1} \\
& c=r \text { th term }=x y^{-r}-{ }^{1}
\end{aligned}
$$

Now, $\left(i l_{a} a+j l_{a} b+k l_{a} c\right) \cdot[i(q-r)+j(r-p)+k(p-q)]$
$=(q-r) l_{a} a+(r-p) l_{a} b+(p-q) l_{a} c$
$=(q-r) l_{a}\left(x y^{p}-{ }^{1}\right)+(r-p) l_{a}\left(x y^{q}-{ }^{1}\right)+(p-q) l_{a}\left(x y^{r}-{ }^{1}\right)$
$=(q-r)\left\{l_{a} x-(p-1) l_{a} y\right\}+\ldots$
$=\{(q-r)+(r-p)+(p-q)\} l_{a} x+\{(q-r)(p-1)$
$+(r-p)(q-1)+(p-q)(r-1)\} l_{a} y=0$.
Hence, the vectors are perpendicular.
27. Since $a$ is collinear with $b=6 i-8 j-(15 / 2) k$.
$\therefore \quad$ there exists a scalar $t$, s.t.
$a=t b=6 t i-8 t j-(15 / 2) t k$.
Now, if a makes an angle $\theta$ with positve $z$-axis, then
$\cos \theta=\frac{a \cdot k}{|a||k|}=\frac{-(15 / 2) t}{|a|}$.
$\because \theta<90^{\circ} \quad \therefore \quad \cos \theta=\frac{-15 t}{2|a|}>0 \Rightarrow t<0$.
Now, $|a|=\sqrt{\left[(36+64+(225 / 4)) t^{2}\right]}=50$
$\Rightarrow t^{2}=16 \Rightarrow t=-4$,
[Rejecting $t=4, \because t<0$ ]
Hence, $a=-24 i+32 j+30 k=(-24,32,30)$.
28. Let the unit vectors $i, j$ and $k$ be denoted by the coterminous edges $O A, O B$ and $O C$, respectively of the unit cube. Let $C N$ be the perpendicular drawn from $C$ on the diagonal $O E$ of the cube which does not pass through $C$. Here, $O E=i+j+$ $k$. Let $e$ be the unit vector along $O E$. Then,

---

![img-26.jpeg](img-26.jpeg)

$$
\begin{aligned}
e & =\frac{i+j+k}{|i+j+k|}=\frac{i+j+k}{\sqrt{(1+1+1)}} \\
& =\frac{1}{\sqrt{3}}(i+j+k)
\end{aligned}
$$

$\therefore \quad O N=$ projection of $O C$ on $O E$
$=k \cdot e=k \cdot \frac{1}{\sqrt{3}}(i+j+k)=\frac{1}{\sqrt{3}}$.
$\therefore \quad \mathrm{CN}^{2}=O \mathrm{C}^{2}-O N^{2} \quad$ [In right triangle $\triangle O C N$ ]
$=1^{2}-\left(\frac{1}{\sqrt{3}}\right)^{2}=1-\frac{1}{3}=\frac{2}{3}$.
$\therefore \quad C N=\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$
29. The length of $a$ is $|a|=\sqrt{1^{2}+1^{2}}=\sqrt{2}$. Similarly, $|b|=\sqrt{2}$. Since the three vectors have equal length, $\therefore|c|=\sqrt{2}$. Let $c=c_{1} i+c_{2} j+c_{3} k$. Then, since $c$ makes an obtuse angle with $i$, we must have $c \cdot i=c_{1}<0$. We are also given that the angles between the vectors are equal, i.e.,
$\cos ^{-1} \frac{a \cdot b}{|a||b|}=\cos ^{-1} \frac{a \cdot c}{|a||c|}=\cos ^{-1} \frac{b \cdot c}{|b||c|}$
Now, $a \cdot b=1, a \cdot c=c_{1}+c_{2}$ and $b \cdot c=c_{2}+c_{3}$, so

$$
\begin{aligned}
\cos ^{-1} \frac{a \cdot b}{|a||b|} & =\cos ^{-2} \frac{1}{2}=\cos ^{-1} \frac{c_{1}+c_{2}}{2} \\
& =\cos ^{-1} \frac{c_{2}+c_{3}}{2}
\end{aligned}
$$

This gives the equations $c_{1}+c_{2}=1$ and $c_{2}+c_{3}=1$, from which we get $c_{2}=c_{1}$ and $c_{2}=1-c_{3}$. Putting these values in $c_{1}^{2}+c_{2}^{2}+c_{3}^{2}=2$, we get
$c_{1}^{2}+\left(1-c_{1}\right)^{2}+c_{1}^{2}=2$
$\Rightarrow \quad 3 c_{1}^{2}-2 c_{1}-1=0$
$\Rightarrow \quad\left(c_{1}-1\right)\left(3 c_{1}+1\right)=0 \quad \Rightarrow \quad c_{1}=1,-\frac{1}{2}$.
Since $c_{1}$ must be less than zero, we rule out the solution $c_{1}$ $=1$, giving $c_{1}=-1 / 3=c_{2}$ and $c_{2}=1-c_{1}=\frac{4}{3}$. Hence, the required vector is
$c=-\frac{1}{3} i+\frac{4}{3} j-\frac{1}{3} k$.
30. Let the given points be $A, B, C, D$, respectively. Now, if the points $A, B, C, D$ are coplanar, then $A B, A C, A D$ are coplanar.
$\therefore \quad A B \cdot(A C \times A D)=0$
$\Rightarrow \quad(b-a) \cdot\{(c-a) \times(d-a)\}=0$.
$\Rightarrow \quad(b-a) \cdot\{c \times d-c \times a-a \times d\}=0 \quad \because a \times a=0$
$\Rightarrow \quad b \cdot(c \times d)-b \cdot(c \times a)-b \cdot(a \times d)-a \cdot(c \times d)$
$+a \cdot(c \times a)+a \cdot(a \times d)=0$
$\Rightarrow \quad[b c d]-[b c a]-[b a d]-[a c d]+0+0=0$
$\Rightarrow \quad[b c d]-[a b c]+[a b d]+[c a d]=0$
$\Rightarrow \quad[b c d]+[c a d]+[a b d]=[a b c]$.
31. Vector $\perp$ to face $O A B$
$=O A \times O B=\left|\begin{array}{ccc}i & j & k \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right|=5 i-j-3 k$
Vector $\perp$ to the face $A B C$
$=A B \times A C=\left|\begin{array}{ccc}i & j & k \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right|=i-5 j-3 k$
Since the angle between the faces $=$ angle between their normals
$\therefore \quad \cos \theta=\left|\frac{5+5+9}{\sqrt{35} \sqrt{35}}\right|=\frac{19}{35}$
$\therefore \quad \theta=\cos ^{-1}\left(\frac{19}{35}\right)$.
32. We have,
$B C=B A+A C=-b+m b+p d$
$=(m-1) b+p d$
and, $C D=C A+A D=-m b-p d+d$
$=-m b+(1-p) d$
Area of the triangle $A B D=\frac{1}{2}|b \times d|$.
![img-27.jpeg](img-27.jpeg)

Area of the triangle $B C D$

$$
\begin{aligned}
& =\frac{1}{2}|(m-1) b+p d) \times(-m b+(1-p) d)| \\
& =\frac{1}{2}|-p m d \times b+(1-p)(m-1) b \times d| \\
& =\frac{1}{2}|m-1+p||b \times d|=\frac{1}{2}(p+m-1)|b \times d|
\end{aligned}
$$

---

Hence, the area of the quadrilateral $A B C D$
$=\frac{1}{2}(p+m)|b \times d| . \quad \therefore \quad k=\frac{1}{2}$.
33. Let $A B C$ be a triangle in which the given vectors are represented by the sides $A B$ and $A C$.
i.e., $A B=\sqrt{3}(a \times b)$
and, $A C=b-(a \cdot b) a$
![img-28.jpeg](img-28.jpeg)

$$
\begin{aligned}
\therefore \quad A B \cdot A C & =\sqrt{3(a \times b) \cdot[b-(a \cdot b) a]} \\
& =\sqrt{3[(a \times b) \cdot b-(a \cdot b)(a \times b) \cdot a]} \\
& =\sqrt{3[0-0]}=0
\end{aligned}
$$

Therefore, $\angle B A C=90^{\circ}$

$$
\begin{aligned}
A B^{2} & =[\sqrt{3}(a \times b)]^{2}=3(a \times b)^{2} \\
A C^{2} & =[b-(a \cdot b) a]^{2} \\
& =(b)^{2}+(a \cdot b)^{2} a^{2}-2(b \cdot a)(a \cdot b) \\
& =(b)^{2}+(a \cdot b)^{2}-2(a \cdot b)^{2} \\
& =(b)^{2}-(a \cdot b)^{2} \\
& =(b)^{2}-|a|^{2}|b|^{2} \cos ^{2} \theta \\
& =(b)^{2}\left[1-|a|^{2} \cos ^{2} \theta\right]=(b)^{2}\left(1-\cos ^{2} \theta\right) \\
& =(b)^{2} \sin ^{2} \theta=|a|^{2}|b|^{2} \sin ^{2} \theta=(a \times b)^{2}
\end{aligned}
$$

Dividing (i) by (ii), we get

$$
\begin{aligned}
& \frac{A B^{2}}{A C^{2}}=\frac{3(a \times b)^{2}}{(a \times b)^{2}} \\
\Rightarrow & A B^{2}=3 \cdot A C^{2} \Rightarrow A B=\sqrt{3} A C . \\
\tan C & =\frac{A B}{A C}=\frac{\sqrt{3} A C}{A C}=\sqrt{3}, \therefore \angle C=60^{\circ}
\end{aligned}
$$

So $\therefore \quad \angle A=180-90^{\circ}-60^{\circ}=30^{\circ}$.
Hence, angles of the triangle are $30^{\circ}, 90^{\circ}$ and $60^{\circ}$.
34. Given equation is $w+(w \times u)=v$

Taking cross-product with $u$, we get
$u \times[w+(w \times u)]=u \times v$
$\Rightarrow \quad u \times w+u \times(w \times u)=u \times v$
$\Rightarrow \quad u \times w+(u \cdot u) w-(u \cdot w) u=u \times v$
$\Rightarrow \quad u \times w+w-(u \cdot w) u=u \times v$
Taking scalar product of (1) with $u$ we get
$u \cdot w+u \cdot(w \times u)=u \cdot v$
$\Rightarrow \quad u \cdot w=u \cdot v \quad[\because \quad u \cdot(w \times u)=0]$
Taking scalar product of (1) with $v$ we get
$v \cdot w+v \cdot(w \times u)=v \cdot v$
$\Rightarrow \quad v \cdot w+[v w u]=1$
$\Rightarrow \quad(u \times v) \cdot w=1-v \cdot w$
Taking scalar product of (2) with $w$ we get
$(u \times w) \cdot w+w \cdot w-(u \cdot w)(u \cdot w)=(u \times v) \cdot w$
$\Rightarrow \quad 0+|w|^{2}-(u \cdot w)^{2}=(u \times v) \cdot w$
$\Rightarrow \quad(u \times v) \cdot w=|w|^{2}-(u \cdot w)^{2}$.

Taking scalar product of (1) with $w$ we get
$w \cdot w+(w \times u) \cdot w=v \cdot w$
$\Rightarrow|w|^{2}=v \cdot w$
$\Rightarrow|w|^{2}=1-(u \times v) \cdot w \quad$ [using (4)]
From (5) we get
$(u \times v) \cdot w=|w|^{2}-(u \cdot w)^{2}$
$=1-(u \times v) \cdot w-(u \cdot w)^{2}$
$\Rightarrow \quad 2(u \times v) \cdot w=1-(u \cdot v)^{2} \quad$ [Using (3)]
Thus, $|(u \times v) \cdot w|=\frac{1}{2}\left|1-(u \cdot v)^{2}\right|$

$$
\leq \frac{1}{2}[\therefore(u \cdot v)^{2} \geq 0]
$$

35. Let $I$ be a unit vector in the direction of $b, J$ in the direction of $c$. Note that $b=I$ and $c=J$.
We have $b \times c=|b||c| \sin \alpha k=\sin \alpha k$,
where $k$ is a unit vector perpendicular to $b$ and $c$.
$\Rightarrow \quad|b \times c|=\sin \alpha \quad \Rightarrow \quad k=\frac{b \times c}{|b \times c|}$
Any vector $a$ can be written as a linear combination of $I, J$ and $k$. Let
$a=a_{1} i+a_{2} J+a_{3} k$
Now, $a \cdot b=a \cdot I=a_{1}, a \cdot c=a \cdot J=a_{2}$
and, $a \cdot \frac{b \times c}{|b \times c|}=a \cdot k=a_{3}$
Thus, $(a \cdot b) b+(a \cdot c) c+\frac{a \cdot(b \times c)}{|b \times c|^{2}}(b \times c)$
$=a_{1} b+a_{2} c+a_{3} \frac{b \times c}{|b \times c|}$
$=a_{1} I+a_{2} J+a_{3} k=a$.
36. Let $c=x i+y j+z k$, where $x^{2}+y^{2}+z^{2}=1$

Unit vector along $3 i+4 j=\frac{1}{5}(3 i+4 j)$
$\therefore \quad$ equation of the bisector of these two is
$r=t\left[(x i+y j+z k)+\left(\frac{3 i+4 j}{5}\right)\right]$
But the bisector is $-i+j-k$.
$\therefore \quad t\left(x+\frac{3}{5}\right) i+t\left(y+\frac{4}{5}\right) j+t z=-i+j-k$
$\Rightarrow \quad x=-\frac{1}{t}-\frac{3}{5}=-\frac{3 t+5}{5 t}$,
$y=\frac{1}{t}-\frac{4}{5}=\frac{5-4 t}{5 t}$ and $z=-\frac{1}{t}$
But $x^{2}+y^{2}+z^{2}=1$
$\therefore \quad \frac{3 t+5}{5 t} \quad+\left(\frac{5-4 t}{5 t}\right)+\frac{1}{t^{2}}=1$

---

# 21.36 <br> ATM $^{21}$ : FREE EDUCATION TO ALL APUL 

$\Rightarrow 9 t^{2}+30 t+25+25+16 t^{2}-40 t+25=25 t^{2}$
$\Rightarrow 75-10 t=0, \therefore t=\frac{75}{10}=\frac{15}{2}$
Hence, $x=-\frac{\frac{45}{2}+5}{\frac{75}{2}}=\frac{-55}{75}=-\frac{11}{15}$
$y=\frac{5-30}{\frac{75}{2}}=\frac{-50}{75}=\frac{-10}{15}$
and, $z=-\frac{1}{\frac{15}{2}}=-\frac{2}{15}$
$\therefore$ unit vector along $c$

$$
=\frac{-11 i}{15}-\frac{10 j}{15}-\frac{2 k}{15} \text { or }-\frac{11 i+10 j+2 k}{15}
$$

37. Since $|a+b+c|=1 \Rightarrow(a+b+c) \cdot(a+b+c)=1$
$\Rightarrow 1+1+1+2(a \cdot b+b \cdot c+c \cdot a)=1$
$\Rightarrow a \cdot b+b \cdot c+c \cdot a=-1$
$\Rightarrow \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=-1$
So, at least one of $\cos \theta_{1}, \cos \theta_{2}$ and $\cos \theta_{3}$ must be negative.
38. Since $x$ and $y$ are linearly independent,
$20 a-15 b=15 b-12 c=12 c-20 a=0 \Rightarrow \frac{a}{3}=\frac{b}{4}=\frac{c}{5}$
$\Rightarrow c^{2}=a^{2}+b^{2} \Rightarrow \Delta A B C$ is right-angled.
39. We have,
$\alpha(a \times b)+\beta(b \times c)+\gamma(c \times a)=0$
Taking dot product with $c$, we have
$\alpha[a b c]+\beta[b c c]+\gamma[c a c]=0$
i.e., $\alpha[a b c]+0+0=0$
i.e., $\alpha[a b c]=0$

Similarly, taking dot product with b and c , we have $\gamma[a b c]=0, \beta[a b c]=0$

Now, even if one of $\alpha, \beta, \gamma \neq 0$, then we have $[a b c]=$ $0 \Rightarrow a, b, c$ are coplanar.
40. Since $p \times q=r$
$\therefore p \times(p \times q)=p \times r$
$\Rightarrow(p \cdot q) p-(p \cdot p) q=p \times r$
$\Rightarrow c p-(p)^{2} q=p \times r$
$\Rightarrow(p)^{2} q=c p-p \times r$
$\Rightarrow q=\frac{c p-p \times r}{|p|^{2}}$.
41. $r \times a=b \times a \Rightarrow(r-b) \times a=\overrightarrow{0}$
$\Rightarrow r-b$ is paralllel to $a$.
$\therefore r-b=\lambda a$ i.e. $r=b+\lambda a$
Similarly, $r \times b=a \times b$ can be written as

$$
r=a+\mu b
$$

For point of intersection of the two lines (1) and (2),
we get $b+\lambda a=a+\mu b \Rightarrow \lambda=\mu=1$
Hence, the required point of intersection is given by
$r=a+b=i+j+2 i-k=3 i+j-k$.
42. Let $a=2 i+4 j-5 k, b=i+2 j+3 k$
$\therefore$ diagonals of the parallelogram are
$p=a+b$ and $q=b-a$
i.e., $p=3 i+6 j-2 k$
and, $q=-i-2 j+8 k$
$\therefore$ unit vectors along the diagonals are

$$
\frac{3 i+6 j-2 k}{\sqrt{9+36+4}} \text { and } \frac{-i-2 j+8 k}{\sqrt{1+4+64}}
$$

i.e., $\frac{3 i+6 j-2 k}{7}$ and $\frac{-i-2 j+8 k}{\sqrt{69}}$
$\therefore$ unit vector parallel to one of the diagonals is

$$
\frac{1}{7}(3 i+6 j-2 k)
$$

## More than One Option Correct Type

43. Given $x+y=a$
$\Rightarrow y=a-x$
(1)
$x \times y=b$
$x \cdot a=1$
From (1) and (2), we get

$$
\begin{aligned}
& x \times(a-x)=b \\
\Rightarrow & x \times a-x \times x=b \Rightarrow x \times a=b \\
\Rightarrow & a \times(x \times a)=a \times b \Rightarrow(a \cdot a) x-(a \cdot x) a=a \times b \\
\Rightarrow & |a|^{2} x-1 \cdot a=a \times b \\
\Rightarrow & x=\frac{(a+a \times b)}{a^{2}}
\end{aligned}
$$

and, $y=a-x=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$.
44. We have, $a \times(b \times c)+(a \cdot b) b$
$=(4-2 \beta-\sin \alpha) b+(\beta^{2}-1) c$
and, $(c \cdot c) a=c$
where $b$ and $c$ are non-collinear vectors and $\alpha, \beta$ are scalars
From (2), $(c \cdot c) a \cdot c=c \cdot c$
$\therefore a \cdot c=1$
From (1), we get
$(a \cdot c) b-(a \cdot b) c+(a \cdot b) b$
$=(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c$
or, $\{1+(a \cdot b)\} b-(a \cdot b) c$
$=(4-2 \beta-\sin \alpha) b+(\beta^{2}-1) c$
$\Rightarrow 1+(a \cdot b)=4-2 \beta-\sin \alpha$

---

and, $a \cdot b=-(\beta f-1)$
$\therefore \quad \sin \alpha=1+(1-\beta)^{2} \Rightarrow \beta=1, \sin \alpha=1$
i.e., $\alpha=\frac{\pi}{2}+2 n \pi, n \in I$.
45. We have, $x \times b=c \times b$
$\Rightarrow a \times(x \times b)=a \times(c \times b)$
$\Rightarrow(a \cdot b) x-(a \cdot x) b=a \times(c \times b)$
$\Rightarrow x=\frac{a \times(c \times b)}{a \cdot b} \quad(\because a \perp x)$.
Also, $a \times(c \times b)=-(c \times b) \times a=(b \times c) \times a$
$\therefore x=\frac{(b \times c) \times a}{b \cdot a}$.
46. Let $c=c_{1} i+c_{2} j$. Since $c \perp b \therefore b \cdot c=0$
$\Rightarrow(4 i+3 j) \cdot\left(c_{1} i+c_{2} j\right)=0 \Rightarrow 4 c_{1}+3 c_{2}=0$
$\Rightarrow c_{2}=-\frac{4}{3} c_{1}$. Thus, $c=\frac{c_{1}}{3}[3 i-4 j]$.
Now, the projection of $a= \pm\left(a_{1} i+a_{2} j\right)$ on $b$ is 1 and on $c$ is 2
$\therefore 1=\left|\frac{4 a_{1}+3 a_{2}}{5}\right|$ and $2=\left|\frac{3 a_{1}-4 a_{2}}{5}\right|$
Solving, we get $a_{1}=2$ and $a_{2}=-1 \therefore a= \pm(2 i-j)$.
47. Let $a=2 i-2 j+k$ and $b=i+2 j-2 k$
$\therefore|a|=3$ and $|b|=3$.
$\therefore \quad$ a vector along bisectors
$=\frac{a}{|a|} \pm \frac{b}{|b|}=\frac{2 i-2 j+k}{3} \pm \frac{i+2 j-2 k}{3}$
$=i-\frac{1}{3} k, \frac{1}{3} i-\frac{4}{3} j+k$.
$\therefore \quad$ The required vector
$=2 \cdot \frac{i-\frac{1}{3} k}{\sqrt{1^{2}+\left(-\frac{1}{3}\right)^{2}}}, 2 \cdot \frac{\frac{1}{3} i-\frac{4}{3} j+k}{\left(\frac{1}{3}\right)^{2}+\left(-\frac{4}{3}\right)^{2}+1^{2}}$
$=\frac{2}{\sqrt{10}}(3 i-k), \frac{2}{\sqrt{26}}(i-4 j+3 k)$.
48. We have,
$(i+j+3 k) x+(3 i-3 j+k) y+(-4 i+5 j) z$
$=\lambda(x i+y j+x k)$
$\Rightarrow(x+3 y-4 z-\lambda x) i+(x-3 y+5 z-\lambda y) j$

$$
+(3 x+y+0 z-\lambda z) k=0
$$

Above is a relation of the form

$$
l i+m j+n k=0
$$

where $i, j$ and $k$ are non-coplanar and hence each of the coefficients $l, m, n$ is zero.
$\therefore \quad(1-\lambda) x+3 y-4 z=0$

$$
x-(3+\lambda) y+5 z=0
$$

$$
3 x+y-\lambda z=0
$$

Eliminating $x, y, z$ we get

$$
\begin{aligned}
& \left|\begin{array}{ccc}
1-\lambda & 3 & -4 \\
1 & -3-\lambda & 5 \\
3 & 1 & -\lambda
\end{array}\right|=0 \\
& \Rightarrow(1-\lambda)\left(3 \lambda+\lambda^{2}-5\right)-3(-\lambda-15)-4(1+9+3 \lambda)=0 \\
& \Rightarrow-\lambda^{3}+\lambda^{2}(1-3)+\lambda(5+3+3-12)+(-5+45-40)=0 \\
& \Rightarrow-\lambda^{3}-2 \lambda^{2}-\lambda=0 \text { or } \lambda\left(\lambda^{2}+2 \lambda+1\right)=0 \\
& \therefore \quad \lambda(\lambda+1)^{2}=0 \text { which gives } \lambda=0,-1,-1
\end{aligned}
$$

49. We have, $a=12 i+4 j+3 k$
$b=8 i-12 j-9 k$
$c=33 i-4 j-24 k$
$\therefore|a|=\sqrt{(12)^{2}+4^{2}+3^{2}}=\sqrt{144+16+9}$

$$
\begin{aligned}
& =\sqrt{169}=13 \\
|b| & =\sqrt{(8)^{2}+(-12)^{2}+(-9)^{2}} \\
& =\sqrt{64+144+81}=\sqrt{289}=17 \\
|c| & =\sqrt{(33)^{2}+(-4)^{2}+(-24)^{2}} \\
& =\sqrt{1689+16+576}=\sqrt{1681}=41
\end{aligned}
$$

![img-29.jpeg](img-29.jpeg)

Now, area $O A B D=$ area $C F G E=|a \times b|$.

$$
\begin{aligned}
& a \times b=\left|\begin{array}{ccc}
i & j & k \\
12 & 4 & 3 \\
8 & -12 & -9
\end{array}\right| \\
& =i(-36+36)-j(-108-24)+k(-144-32) \\
& =0 . i+132 j-176 k \\
& \therefore \quad|a \times b|=\sqrt{0+(132)^{2}+(-176)^{2}}=\sqrt{(484000)} \\
& =220
\end{aligned}
$$

In the same way, area $F A B G=$ area $C O D E=|a \times c|$
$|a \times c|=\sqrt{(189225)}$ or $|a \times c|=435$
area $F A O C=$ area $G B D E=|b \times c|$
$\therefore|b \times c| \sqrt{(207025)}=455$
Volume of parallelopiped $=[a b c]=\left|\begin{array}{ccc}12 & 1 & 3 \\ 8 & -12 & -9 \\ 33 & -4 & -24\end{array}\right|$

$$
\begin{aligned}
& =(4 \times 3)\left|\begin{array}{ccc}
12 & 1 & 1 \\
8 & -3 & -3 \\
33 & -1 & -8
\end{array}\right| \\
& =12\left|\begin{array}{ccc}
12 & 1 & 1 \\
8 & -3 & -3 \\
33 & -1 & -8
\end{array}\right|
\end{aligned}
$$

---

# 21.38 <br> ATM $^{21}$ : FREE EDUCATION TO ALL APUL 

$=12[12(24-3)-1(-64+99)+1(-8+99)]$
$=12(12 \times 21-1 \times 35+91)$
$=12(252+56)=12 \times 308=3696$.
$\therefore \quad$ Lengths of the edges are $13,17,41$.
Areas of the faces are 220, 435, 455 and volume 3696.
50. Let the required vector be $\alpha=d_{1} i+d_{2} j+d_{3} k$ where $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=51$ (given)
Since, each of the given vectors $a, b, c$ is a unit vector
$\therefore \quad \cos \theta=\frac{d \cdot a}{|d||a|}=\frac{d \cdot b}{|d||b|}=\frac{d \cdot c}{|d||c|}$
$\Rightarrow \quad d \cdot a=d \cdot b=d \cdot c$ as $|d|=\sqrt{51}$
cancels out and $|a|=|b|=|c|=1$
$\Rightarrow \frac{1}{3}\left(d_{1}-2 d_{2}+2 d_{3}\right)=\frac{1}{5}\left(-4 d_{1}+0 d_{2}-3 d_{3}\right)$

$$
=d_{2}
$$

$\therefore \quad d_{1}-5 d_{2}+2 d_{3}=0$ from 1 st and 3 rd
and $4 d_{1}+5 d_{2}+3 d_{3}=0$ from 2 nd and 3 rd. Solving the above two equations, we get
$\frac{d_{1}}{-15-10}=\frac{d_{2}}{8-3}=\frac{d_{3}}{5+20}$
$\Rightarrow \frac{d_{1}}{5}=\frac{d_{2}}{-1}=\frac{d_{3}}{-5}=\lambda$ (say)
Putting for $d_{1}, d_{2}$ and $d_{3}$ in (1), we get
$(25+1+25) \lambda^{2}=51, \therefore \quad \lambda= \pm 1$.
Hence, the required vectors are $\pm(5 i-j-5 k)$.
51. Since they are unit vectors, we have $|a|=|b|=1=|c|$. It is also given that the angle $\theta$ between $a$ and $c$ equals that between $b$ and $c$, i.e.,
$\frac{a \cdot c}{|a||c|}=a \cdot c=\cos \theta=\frac{b \cdot c}{|b||c|}=b \cdot c$.
Since $a \cdot b=0$, we get from the given value of $c$,
$a \cdot c=\alpha a \cdot a+\beta a \cdot b+\gamma a \cdot(a \times b)=\alpha$,
i.e., $\alpha=\cos \theta$, and similarly, $b \cdot c=\cos \theta=\beta$.

That is, $\alpha=\beta=\cos \theta$, so that answer (a) is correct. Next, we have
$1=c \cdot c=2 \alpha^{2}+\gamma^{2}|a \times b|^{2}$
$=2 \alpha^{2}+\gamma^{2}\left[|a|^{2}|b|^{2}-(a \cdot b)^{2}\right]=2 \alpha^{2}+\gamma^{2}$
$\Rightarrow \gamma^{2}=1-2 \alpha^{2}=1-2 \cos ^{2} \theta=-\cos 2 \theta$
$\Rightarrow \alpha^{2}=\beta^{2}=\frac{1-\gamma^{2}}{2}=\frac{1+\cos 2 \theta}{2}$
52. Let $E B=p A B, C E=q C D$
![img-30.jpeg](img-30.jpeg)
then, $0<p, q \leq 1$

Since $E B+B C+C E=0$
$\therefore \quad p m(2 i-6 j+2 k)+(i-2 j)+q n(-6 i+15 j-3 k)=0$
$\Rightarrow(2 p m+1-6 q n) i+(-6 p m-2+15 q n) j$
$+(2 p m-6 q n) k=0$
$\Rightarrow 2 p m-6 q n+1=0,-6 p m-2+15 q n=0$,
$2 p m-6 q n=0$.
Solving these, we get
$p=1 /(2 m)$ and $q=1 /(3 n)$.
$\therefore \quad 0<1 /(2 m) \leq 1$ and $0<1 /(3 n) \leq 1$
$\Rightarrow \quad m \geq 1 / 2$ and $n \geq 1 / 3$.
The area of $\triangle B C E=\frac{1}{2}|E B \times B C|=\frac{1}{2} \sqrt{6}$.
53. Given $\mathrm{A}=(9 i-j+7 k)$

$$
C=(7 i-2 j+7 k)
$$

$$
A B=4 i-j+3 k
$$

and, $C D=2 i-j+2 k$
![img-31.jpeg](img-31.jpeg)

Let the position vector of $P$, the point of intersection of $A B$ and $C D$ be $r$.
Then, $A P=s A B$
$\Rightarrow \quad r-(9 i-j+7 k)=s(4 i-j+3 k)$
and, $C P=t C D$
$\Rightarrow \quad r-(7 i-2 j+7 k)=t(2 i-j+2 k)$
$\therefore \quad(4 s+9) i-(s+1) j+(3 s+7) k$
$=(2 t+7) i-(t+2) j+(2 t+7) k$
$\therefore \quad$ Equating the coefficients and solving, we get
$s=-2, t=-3, \quad \therefore \quad r=i+j+k$
$\therefore \quad$ Position vector of $P=i+j+k$

$$
\begin{aligned}
& A B \times C D=\left|\begin{array}{ccc}
i & j & k \\
4 & -1 & 3 \\
2 & -1 & 2
\end{array}\right|=I-2 j-2 k . \\
& \therefore \quad P Q\|(i-2 j-2 k) . \text { As }|P Q|=15 \\
& \therefore \quad P Q= \pm 15 \frac{(i-2 j-2 k)}{3}= \pm 5(i-2 j-2 k) \\
& \Rightarrow \quad \text { p.v. of } Q-\text { p.v. of } P= \pm 5(i-2 j-2 k) \\
& \Rightarrow \quad \text { p.v. of } Q=(i+j+k) \pm 5(i-2 j-2 k) \\
&=(6 i-9 j-9 k) \text { or }(-4 i+11 j+11 k)
\end{aligned}
$$

54. We have, $D A=a, A B=b$ and $C B=k a$

Since X, Y are mid-points of DB and AC.
$\therefore \quad O X=\frac{O B+O D}{2}, O Y=\frac{O A+O C}{2}$
$\therefore \quad X Y=O Y-O X=\frac{O A+O C-O B-O D}{2}$

$$
=\frac{D A+B C}{2}=\frac{a-k a}{2}
$$

$\therefore \quad X Y=\frac{(1-k) a}{2}$
$\therefore \quad|X Y|= \pm\left(\frac{1-k}{2}\right)|a|=4$ (given).

---

$\Rightarrow\left(\frac{1-k}{2}\right) 17=4$ (taking positive sign)
$\Rightarrow 1-k=\frac{8}{17} \Rightarrow k=1-\frac{8}{17}=\frac{9}{17}$
and taking negative sign, $-\left(\frac{1-k}{2}\right) 17=4$
$\Rightarrow k-1=\frac{8}{17} \Rightarrow k=1+\frac{8}{17}=\frac{25}{17}$.
Hence, $k=\frac{9}{17}$ or $\frac{25}{17}$.
55. Let $\theta$ be the angle between unit vectors $a$ and $b$. As $a$ and $b$ are non-collinear, $\theta \neq 0$ and $\theta \neq \pi$.
We have $a \cdot b=|a||b| \cos \theta=\cos \theta$.
Now, $\quad|u|^{2}=|a-(a \cdot b) b|^{2}=|a-\cos \theta b|^{2}$

$$
\begin{aligned}
& =|a|^{2}+\cos ^{2} \theta|b|^{2}-2 \cos \theta(a \cdot b) \\
& =1+\cos ^{2} \theta-2 \cos ^{2} \theta=1-\cos ^{2} \theta \\
& =\sin ^{2} \theta
\end{aligned}
$$

Also, $\quad|v|^{2}=|a \times b|^{2}=|a|^{2}|b|^{2} \sin ^{2} \theta$
$\therefore \quad|v|^{2}=|u|^{2} \Rightarrow|v|=|u|$.
Also, $u \cdot a=(a-(a \cdot b) b) \cdot a=a \cdot a-(a \cdot b)(b \cdot a)$

$$
=|a|^{2}-\cos ^{2} \theta=1-\cos ^{2} \theta=\sin ^{2} \theta
$$

$\therefore \quad|u|+|u \cdot a|=\sin \theta+\sin ^{2} \theta \neq|u|$.
Next, $u \cdot b=a \cdot b-(a \cdot b)(b \cdot b)=a \cdot b-(a \cdot b)$ $=0$
$\therefore \quad|u|+|u \cdot b|=|u|+0=|u|=|v|$.
Also, $\quad u \cdot(a+b)=u \cdot a+u \cdot b=u \cdot a$
$\Rightarrow \quad|u|+u \cdot(a+b)=|u|+u \cdot a \neq|v|$.
56. Equation of the plane containing $i$ and $i+j$ is $[r-i i i+j]=0$
or, $\quad(r-i) \cdot[i \times(i+j)]=0$
or, $\{(x-1) i+y j+z k\} \cdot k=0$
or, $\quad z=0$
Equation of the plane containing $i-j$ and $i+k$ is
$[r-(i-j) i-j i+k]=0$
or, $\quad[(\bar{r}-i+j) \cdot[(i-j) \times(i+k)]=0$
or, $\quad[(x-1) i+(y+1) j+z k] \cdot(-i-j+k)=0$
or, $\quad-(x-1)-(y+1)+z=0$
or, $\quad x+y-z=0$.
Let $a=a_{1} j+a_{2} j+a_{3} k$.
Since $a$ is parallel to (1) and (2)
$a_{1}=0$ and $a_{1}+a_{2}-a_{3}=0$
$\Rightarrow a_{1}=-a_{2}, a_{3}=0$.
Thus, a vector in the direction of $a$ is $u=i-j$.
If $\theta$ is the angle between $a$ and $i-2 j+2 k$, then
$\cos \theta= \pm \frac{(1)(1)+(-1)(-2)}{\sqrt{1+1} \sqrt{1+4+4}}= \pm \frac{3}{(\sqrt{2})(3)}$
$\Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$.
57. Any vector coplanar with vectors $i+j+2 k$ and $i+2 j+k$ is $x(i+j+2 k)+y(i+2 j+k)$
i.e., $(x+y) i+(x+2 y) j+(2 x+y) k$.

If this is perpendicular to $i+j+k$, then
$(x+y)+(x+2 y)+(2 x+y)=0$
$\Rightarrow 4 x+4 y=0 \Rightarrow y=-x$.
Substituting $y=-x$ in (1), the vector coplanar with the given vectors is $x(-j+k)$.
So, the required unit vector $=\frac{x(-j+k)}{\sqrt{x^{2}\left(1^{2}+1^{2}\right)}}$

$$
= \pm \frac{(-j+k)}{\sqrt{2}}
$$

58. Let $c=x i+y j+z k$.

Given, $|c|=|a|=|b| \quad$ [equal magnitude]
$\Rightarrow x^{2}+y^{2}+z^{2}=2$
Also, $\frac{a \cdot b}{|a||b|}=\frac{b \cdot c}{|b||c|}=\frac{c \cdot a}{|c||a|}$
[equally inclined]
$\Rightarrow \frac{0+1+0}{\sqrt{2} \sqrt{2}}=\frac{y+z}{\sqrt{2} \sqrt{2}}=\frac{x+y}{\sqrt{2} \sqrt{2}}$
$\Rightarrow 1=y+z=x+y$
Solving equations (1) and (2), we have
$(1-y)^{2}+y^{2}+(1-y)^{2}=2$
i.e., $3 y^{2}-4 y=0$
$\Rightarrow y=0,4 / 3$ and corresponding $x=1,-1 / 3$ and $z=1,-1 / 3$.
Hence, $c=(1,0,1)$ or $\left(\frac{-1}{3}, \frac{4}{3}, \frac{-1}{3}\right)$.
59. We have $a=-i+2 j+1 k$

Let the new vector be $b=x i+y j+z k$
We have, $|a|=|b|$
i.e., $x^{2}+y^{2}+z^{2}=1^{2}+2^{2}+1^{2}=6$
and, $a \cdot b=0$
i.e., $-x+2 y+z=0$

Also $a, b$ and $y$-axis lie in one plane
$\Rightarrow a, b, j$ are coplanar
$\Rightarrow[a b j]=0$
i.e., $\left|\begin{array}{ccc}-1 & 2 & 1 \\ x & y & z \\ 0 & 1 & 0\end{array}\right|=0$
i.e., $z+x=0$

From equations (2) and (3), we have
$z=-x, y=x$
Putting in equation (1), we get $x= \pm \sqrt{2}$
Hence, $b= \pm \sqrt{2}(i+j-k)$.

---

60. Given $|a|=1,|c|=1,|b|=4$
$a \times b=2 a \times c$ and angle between $a$ and $c$ is $\cos ^{-1}\left(\frac{1}{4}\right)$
Now, $a \cdot c=|a||c| \cos \left[\cos ^{-1} \frac{1}{4}\right]=(1)(1)\left(\frac{1}{4}\right)=\frac{1}{4}$
$\therefore a \cdot c=\frac{1}{4}$
Again, $b-2 c=\lambda a$
$\Rightarrow a \cdot b-2 a \cdot c=\lambda a \cdot a$
$\Rightarrow a \cdot b-\frac{1}{2}=1 \quad\left[\because a \cdot a=a^{2}=a^{2}=1\right]$
$\Rightarrow a \cdot b=\lambda+\frac{1}{2}$
Again, from (2)
$b \cdot b-2 b \cdot c=\lambda a \cdot b$
$\Rightarrow 16-2 b \cdot c=\lambda\left(\lambda+\frac{1}{2}\right)$
(3)
$\Rightarrow b \cdot c=8-\frac{\lambda^{2}}{2}-\frac{\lambda}{4}$

## Passage Based Questions

62. $r=x i+y i+z k$. then $|r|=\sqrt{x^{2}+y^{2}+z^{2}}$
and, $\phi=\ln |r|=\frac{1}{2} \ln \left(x^{2}+y^{2}+z^{2}\right)$.
$\nabla \phi=\frac{1}{2} \nabla \ln \left(x^{2}+y^{2}+z^{2}\right)$
$=\frac{1}{2}\left\{i \frac{\partial}{\partial x} \ln \left(x^{2}+y^{2}+z^{2}\right)+j \frac{\partial}{\partial y} \ln \left(x^{2}+y^{2}+z^{2}\right)+\right.$
$\left.k \frac{\partial}{\partial z} \ln \left(x^{2}+y^{2}+z^{2}\right)\right\}$
$=\frac{1}{2}\left\{i \frac{2 x}{x^{2}+y^{2}+z^{2}}+j \frac{2 y}{x^{2}+y^{2}+z^{2}}+k \frac{2 z}{x^{2}+y^{2}+z^{2}}\right\}$
$=\frac{x i+y j+z k}{x^{2}+y^{2}+z^{2}}=\frac{r}{r^{2}}$.
63. $\nabla \phi=\nabla\left(\frac{1}{r}\right)=\nabla\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)$
$=\nabla\left\{\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\right\}$
$=i \frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}+j \frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}+$
$k \frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}$
$=i\left\{-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} 2 x\right\}+j\left\{-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} 2 y\right\}+$
$k\left\{-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} 2 z\right\}$
Also, from (2)
$b \cdot c-2 c \cdot c=\lambda \vec{a} \cdot \vec{c}$
$\therefore 8-\frac{\lambda^{2}}{2}-\frac{\lambda}{4}-2(1)=\lambda \cdot\left(\frac{1}{4}\right)$
$\Rightarrow 32-2 \lambda^{2}-\lambda-8=\lambda$
$\Rightarrow 2 \lambda^{2}+2 \lambda-24=0 \Rightarrow \lambda^{2}+\lambda-12=0$
$\Rightarrow(\lambda+4)(\lambda-3)=0, \therefore \lambda=3,-4$.
64. Let $a_{0}$ and $b_{0}$ be unit vectors along $a$ and $b$ respectively, then, $a_{0}=(1 / 9)(7 i-4 i-4 k)$
and, $b_{0}=\frac{1}{3}(-2 i-j+2 k)$
The required vector $c=\lambda\left(a_{0}+b_{0}\right)$ (where $\lambda$ is a scalar)
$=\lambda(1 / 9) i-(7 / 9) j+(2 / 9) k$
$\Rightarrow|c|^{2}=\lambda^{2} \times\left(\frac{1+49+4}{81}\right)$
$\Rightarrow(3 \sqrt{6})^{2}=(54 / 81) \lambda^{2} \Rightarrow \lambda= \pm 9$.
$\therefore c= \pm(i-7 j+2 k)$.
$=\frac{-x i-y i-z k}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}=-\frac{r}{r^{2}}$.
65. $\nabla r^{n}=\nabla\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)^{n}=\nabla\left(x^{2}+y^{2}+z^{2}\right)^{n / 2}$
$=i \frac{\partial}{\partial x}\left\{\left(x^{2}+y^{2}+z^{2}\right)^{n / 2}\right\}+j \frac{\partial}{\partial y}\left\{\left(x^{2}+y^{2}+z^{2}\right)^{n / 2}\right\}+$
$k \frac{\partial}{\partial z}\left\{\left(x^{2}+y^{2}+z^{2}\right)^{n / 2}\right\}$
$=i\left\{\frac{n}{2}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{n}{2}-1} 2 x\right\}+j\left\{\frac{n}{2}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{n}{2}-1} 2 y\right\}+$
$k\left\{\frac{n}{2}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{n}{2}-1} 2 z\right\}$
$=n\left(x^{2}+y^{2}+z^{2}\right)^{\frac{n}{2}-1}(x i+y j+z k)$
$=n\left(r^{2}\right)^{\frac{n}{2}-1} r=n r^{n-2} r$
66. $\nabla \phi=i \frac{\partial}{\partial x}\left(2 x^{2} y^{2} z^{4}\right)+j \frac{\partial}{\partial y}\left(2 x^{2} y^{2} z^{4}\right)+k \frac{\partial}{\partial z}\left(2 x^{2} y^{2} z^{4}\right)$
$=6 x^{2} y^{2} z^{4} i+4 x^{3} y z^{4} j+8 x^{3} y^{2} z^{2} k$
Then,

$$
\begin{aligned}
\nabla \cdot \nabla \phi=\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) & \cdot\left(6 x^{2} y^{2} z^{4} i+4 x^{3} y z^{4} j\right. \\
& \left.+8 x^{3} y^{2} z^{2} k\right)
\end{aligned}
$$

---

$$
\begin{aligned}
& =\frac{\partial}{\partial x}\left(6 x^{2} y^{2} z^{4}\right)+\frac{\partial}{\partial y}\left(4 x^{3} y z^{4}\right)+\frac{\partial}{\partial z}\left(8 x^{3} y^{2} z^{3}\right) \\
& =12 x y^{2} z^{4}+4 x^{2} z^{4}+24 x^{3} y^{2} z^{2} \\
& \therefore \quad k=4 . \\
& \text { 66. } \nabla^{2}\left(\frac{1}{r}\right)=\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right) \\
& \frac{\partial}{\partial x}\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)=\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} \\
& =-x\left(x^{2}+y^{2}+z^{2}\right)^{-3 / 2} \\
& \frac{\partial^{2}}{\partial x^{2}}\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)=\frac{\partial}{\partial x}\left[-x\left(x^{2}+y^{2}+z^{2}\right)^{-3 / 2}\right] \\
& =3 x^{2}\left(x^{2}+y^{2}+z^{2}\right)-^{3 / 2}-\left(x^{2}+y^{2}+z^{2}\right)-^{3 / 2} \\
& =\frac{2 x^{2}-y^{2}-x^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}
\end{aligned}
$$

Similarly,

$$
\begin{aligned}
& \frac{\partial^{2}}{\partial y^{2}}\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)=\frac{2 y^{2}-z^{2}-x^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \\
& \text { and, } \frac{\partial^{2}}{\partial z^{2}}\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)=\frac{2 z^{2}-x^{2}-y^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}
\end{aligned}
$$

Then, by addition,

$$
\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)\left(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)=0
$$

67. $\nabla \cdot\left(\frac{r}{r^{3}}\right)=\nabla \cdot\left(r^{-3} r\right)$

$$
=\left(\nabla r^{-3}\right) \cdot r+\left(r^{-3}\right) \nabla \cdot r[\because \nabla \cdot(\phi A)=(\nabla \phi) \cdot A+\phi(\nabla \cdot A)] \\
=-3 r^{-3} r \cdot r+3 r^{-3}=0[\because \nabla \cdot r=3]
\end{aligned}
$$

68. $\nabla \times(\nabla \phi)=\nabla \times\left(\frac{\partial \phi}{\partial x} i+\frac{\partial \phi}{\partial y} j+\frac{\partial \phi}{\partial z} k\right)$

$$
\begin{aligned}
& =\left[\begin{array}{ccc}
i & j & k \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}
\end{array}\right] \\
& =\left[\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial z}\right)-\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial y}\right)\right] i+\left[\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial x}\right)-\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial z}\right)\right] j+ \\
& {\left[\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial y}\right)-\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right)\right] k} \\
& =\left(\frac{\partial^{2} \phi}{\partial y \partial z}-\frac{\partial^{2} \phi}{\partial z \partial y}\right) i+\left(\frac{\partial^{2} \phi}{\partial z \partial x}-\frac{\partial^{2} \phi}{\partial x \partial z}\right) j+\left(\frac{\partial^{2} \phi}{\partial x \partial y}-\frac{\partial^{2} \phi}{\partial y \partial x}\right) k=0
\end{aligned}
$$

69. $\nabla \cdot(\nabla \times A)=\nabla \cdot\left|\begin{array}{ccc}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_{1} & A_{2} & A_{1}\end{array}\right|$

$$
\begin{aligned}
& =\nabla \cdot\left[\left(\frac{\partial A_{1}}{\partial y}-\frac{\partial A_{1}}{\partial z}\right) i+\left(\frac{\partial A_{1}}{\partial z}-\frac{\partial A_{1}}{\partial x}\right) j+\left(\frac{\partial A_{2}}{\partial x}-\frac{\partial A_{1}}{\partial y}\right) k\right] \\
& =\frac{\partial}{\partial x}\left(\frac{\partial A_{2}}{\partial y}-\frac{\partial A_{2}}{\partial z}\right)+\frac{\partial}{\partial y}\left(\frac{\partial A_{1}}{\partial z}-\frac{\partial A_{1}}{\partial x}\right)+\frac{\partial}{\partial z}\left(\frac{\partial A_{2}}{\partial x}-\frac{\partial A_{1}}{\partial y}\right) \\
& =\frac{\partial^{2} A_{2}}{\partial x \partial y}-\frac{\partial^{2} A_{2}}{\partial x \partial z}+\frac{\partial^{2} A_{1}}{\partial y \partial z}-\frac{\partial^{2} A_{1}}{\partial y \partial x}+\frac{\partial^{2} A_{2}}{\partial z \partial x}-\frac{\partial^{2} A_{1}}{\partial z \partial y}=0
\end{aligned}
$$

# Match the Column Type 

70 I. $B D \cdot A C=|B D||A C| \cos \theta$, where $\theta$ is the angle between $B D$ and $A C$.
$\Rightarrow \quad(O D-O B) \cdot(O C-O A)$
$=|O D-O B||O C-O A| \cos \theta$
$\Rightarrow \quad(a-2 b-b) \cdot(2 a+3 b-a)=|a-3 b||a+3 b| \cos \theta$
$\Rightarrow \quad a^{2}-9 b^{2}=|a-3 b||a+3 b| \cos \theta$
$\Rightarrow \quad 0=|a-3 b||a+3 b| \cos \theta[\because|a|=3|b|, \quad \therefore a^{2}=9 b^{2}]$
$\Rightarrow \cos \theta=0, \therefore \theta=\frac{\pi}{2}$.
II. Let angles between $a$ and $b$ and between $a$ and $c$ be $\alpha$ and $\beta$, respectively.

We have, $a \times(b \times c)=\frac{1}{2}(b+c)$
or, $\quad(a \cdot c) b-(a \cdot b) c=\frac{1}{2} b+\frac{1}{2} c$
or, $(a \cdot c-1 / 2) b-(a \cdot b+1 / 2) c=0$
or, $\quad a \cdot c=\frac{1}{2}, a \cdot b=-\frac{1}{2}$ (As $b$ and $c$ are non-parallel)
or, $\quad \cos \beta=\frac{1}{2}, \cos \alpha=-\frac{1}{2}[a, b, c$ are unit vectors $]$
or, $\beta=\frac{\pi}{3}, \alpha=\frac{2 \pi}{3}$.
Hence, angle between $a$ and $b$ is $\frac{2 \pi}{3}$.
III. A normal to the plane $P_{1}$ is given by $N_{1}=a \times b$ and a normal to the plane $P_{2}$ is given by $N_{2}=c \times d$.
Since $(a \times b) \times(c \times d)=0$, we get $N_{1} \times N_{2}=0$.
If $\theta$ is an angle between $P_{1}$ and $P_{2}$, we get
$\left|N_{1}\right| \times\left|N_{2}\right| \sin \theta=0$ or $\sin \theta=0 \Rightarrow \theta=0$.

---

IV. We have, $|a \cdot b|=|a \times b|$
$\Rightarrow|a||b|| \cos \theta|=|a||b|| \sin \theta|$
(where ' $\theta$ ' is the angle between $a$ and $b$ )

$$
\begin{aligned}
& \Rightarrow|\cos \theta|=|\sin \theta| \Rightarrow \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}(\text { as } 0 \leq \theta \leq \pi) \\
& \text { But } a \cdot b<0, \quad \therefore \quad \theta=\frac{3 \pi}{4}
\end{aligned}
$$

# Assertion-Reason Type 

71. We have, $|a \times r|^{2}+|a \cdot r|^{2}=|a|^{2}|r|^{2}$
$\Rightarrow \quad|J|^{2}+(a \cdot r)^{2}=|a|^{2}|r|^{2}$
$\Rightarrow \quad(a \cdot r)= \pm \sqrt{|a|^{2}|r|^{2}-1}$
This shows that $a \cdot r$ depends on $|r|$ for given $a$.
Hence, $a \cdot r$ is an arbitrary constant.
72. As $a, b, c$ are non-coplanar, $b \times c, c \times a, a \times b$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let $a=\lambda b \times c+\mu c \times a+v a \times b$
$\therefore \quad a \cdot a=\lambda[b c a], a \cdot b=\mu[c a b], a \cdot c=v[a b c]$
$\therefore a=\frac{(a \cdot a) b \times c}{[b c a]}+\frac{(a \cdot b) c \times a}{[c a b]}+\frac{(a \cdot c) a \times b}{[a b c]}$
$\therefore \quad(a \cdot a) b \times c+(a \cdot b) c \times a+(a \cdot c) a \times b=[b c a] a$.
73. We have,

$$
\begin{aligned}
i \times[(a-j) \times i] & =(i \cdot i)(a-j)-[i \cdot(a-j)] i \\
& =a-j-(i \cdot a) i \\
j \times[(a-k) \times j] & =(j \cdot j)(a-k)-[(j \cdot(a-k)] j \\
& =a-k-(j \cdot a) j \\
k \times[(a-i) \times k] & =(k \cdot k)(a-i)-[(k \cdot(a-i)] k \\
& =a-i-(k \cdot a) k
\end{aligned}
$$

According to the given condition, we have
$i \times[(a-j) \times i]+j \times[(a-k) \times j]+k \times[(a-i) \times k]=0$
i.e., $3 a-(i+j+k)-[(a \cdot i) i+(a \cdot j) j+(a \cdot k) k]=0$
i.e., $3 a-a=i+j+k$
i.e., $a=\frac{1}{2}(i+j+k)$.
75. The vector, $\hat{i}+x \hat{j}+z \hat{k}$ if it is doubled in magnitude becomes
$4 \hat{i}+(4 x-2) \hat{j}+2 \hat{k}$
$\therefore \quad 2|\hat{i}+x \hat{j}+3 \hat{k}|=|4 \hat{i}+(4 x-2) \hat{j}+2 \hat{k}|$
$\Rightarrow \quad 2 \sqrt{1+x^{2}+9}=\sqrt{16+(4 x-2)^{2}+4}$
$\Rightarrow 40+4 x^{2}=20+(4 x-2)^{2}$
$\Rightarrow 3 x^{2}-4 x-4=0$
$\Rightarrow 3 x^{2}-6 x+2 x-4=0$
$\Rightarrow 3 x(x-2)+2(x-2)=0$
$\Rightarrow \quad(x-2)(3 x+2)=0$
$\Rightarrow x=2,-\frac{2}{3}$
76. Key Idea: If the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ represent the sides of a triangle, then $\vec{a}+\vec{b}+\vec{c}=0$
$\because \vec{a}+\vec{b}+\vec{c}=0$
$\Rightarrow \vec{a}+\vec{b}=-\vec{c}$
$\Rightarrow(\vec{a}+\vec{b}) \times \vec{c}=-\vec{c} \times \vec{c}$
$\Rightarrow \vec{a} \times \vec{c}+\vec{b} \times \vec{c}=0$
$\Rightarrow \vec{b} \times \vec{c}=\vec{c} \times \vec{a}$
Similarly, $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}$
Hence $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$.
77. Given that $\vec{a}=x i+y \hat{j}+z \hat{k}$ and $\vec{b}=\hat{j}$ are such that $\vec{a}, \vec{c}$ and $\vec{b}$ form a right handed system.
$\therefore \quad \vec{c}=\vec{b} \times \vec{a}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ x & y & z\end{array}\right|$
$=\hat{i} z-x \hat{k}$
78. The equation of a line through the center represented by vector $\hat{j}+2 \hat{k}$ and normal to the given plane is
$\vec{r}=\hat{j}+2 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$
This meets the plane at a point for which we must have
$[(\hat{j}+2 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})] \cdot(\hat{i}+2 \hat{j}+2 \hat{k})=15$
$\Rightarrow 6+9 \lambda=15$

---

$\Rightarrow \lambda=1$
On putting $\lambda=1$ in Eq. (i), we get
$\vec{r}=\vec{i}+3 \vec{j}+4 \vec{k}$
$\therefore \quad$ Centre of the circle is $(1,3,4)$.
79. $\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|+\left|\begin{array}{llll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=0$
$\Rightarrow(1+a b c)\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right|=0$
$\Rightarrow a b c=-1$.
Hence, (8) is the correct answer
80. $(\vec{a}+2 \vec{b})=t_{1} \vec{c}$

And $\vec{b}+3 \vec{c}=t_{2} \vec{a}$
$(1)-2 \times(2) \Rightarrow \vec{a}\left(1+2 t_{2}\right)+\vec{c}\left(-t_{1}-6\right)=0 \Rightarrow 1+2 t_{2}=0$

$$
\Rightarrow t_{2}=-1 / 2 \& t_{1}=-6
$$

(since $\vec{a}$ and $\vec{c}$ are non-collinear)
Putting the value of $t_{1}$ and $t_{2}$ in (1) and (2), we get $\vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$.
81. Work done by the forces $\vec{F}_{1}$ and $\vec{F}_{2}$ is $\left(\vec{F}_{1}+\vec{F}_{2}\right) \cdot \vec{d}$ where $\vec{d}$ is the displacement.
Now, according to question
$\vec{F}_{1}+\vec{F}_{2}=(4 \vec{i}+\vec{j}-3 \vec{k})+(3 \vec{i}+\vec{j}-\vec{k})=7 \vec{i}+2 \vec{j}-4 \vec{k}$ and
$\vec{d}=(5 \vec{i}+4 \vec{j}+\vec{k})-(\vec{i}+2 \vec{j}+3 \vec{k})=4 \vec{i}+2 \vec{j}-2 \vec{k}$. Hence
$\left(\vec{F}_{1}+\vec{F}_{2}\right) \cdot \vec{d}$ is 40 .
82. Condition for given three vectors to be coplanar is $\left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2 \lambda-1\end{array}\right|=0 \Rightarrow \lambda=0,1 / 2$.
Hence given vectors will be non coplanar for all real values of $\lambda$. except for 0 and $1 / 2$.
83. Projection of $\vec{v}$ along $\vec{u}$ and $\vec{w}$ along $\vec{u}$ is $\frac{\vec{v} \cdot \vec{u}}{|\vec{u}|}$ and $\frac{\vec{w} \cdot \vec{u}}{|\vec{u}|}$ respectively
Now, according to question

$$
\begin{aligned}
& \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|}=\frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} \Rightarrow \vec{v} \cdot \vec{u}=\vec{w} \cdot \vec{u} \cdot \text { and } \vec{v} \cdot \vec{w}=0 \\
& |\vec{u}-\vec{v}+\vec{w}|^{2}=|\vec{u}|^{2}+|\vec{v}|^{2}+|\vec{w}|^{2}-2 \vec{u} \cdot \vec{w}=14 \\
& \Rightarrow|\vec{u}-\vec{v}+\vec{w}|=\sqrt{14}
\end{aligned}
$$

84. $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} \Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$
85. By triangle law,
$\overrightarrow{P A}+\overrightarrow{A C}+\overrightarrow{C P}=0$
$\overrightarrow{P B}+\overrightarrow{B C}+\overrightarrow{C P}=0$
Adding, we get
$\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{A C}+\overrightarrow{B C}+2 \overrightarrow{C P}=0$
Since $\overrightarrow{A C}=-\overrightarrow{B C}$
$\& \overrightarrow{C P}=-\overrightarrow{P C}$
$\Rightarrow \overrightarrow{P A}+\overrightarrow{P B}-2 \overrightarrow{P C}=0$
![img-32.jpeg](img-32.jpeg)
86. Distance between the line with equation $\vec{r}=2 \vec{i}-2 \vec{j}+3 \vec{k}$ $+\lambda(\vec{i}-\vec{j}+4 \vec{k})$ and the plane $\vec{r} \cdot(\vec{i}+5 \vec{j}+\vec{k})=5$ equivalently $x+5 y+z=5$ is equal to the perpendicular distance of point $(2,-2,3)$ from the plane
i.e., $\left|\frac{2-10+3-5}{\sqrt{1+5^{2}+1}}\right|=\frac{10}{3 \sqrt{3}}$.
87. Let $\vec{a}=x \vec{i}+y \vec{j}+z \vec{k}$, then
$\vec{a} \times \vec{i}=z \vec{j}-y \vec{k}$
$\Rightarrow(\vec{a} \times \vec{i})^{2}=y^{2}+z^{2}$
Similarly $(\vec{a} \times \vec{j})^{2}=x^{2}+z^{2}$
and $(\vec{a} \times \vec{k})^{2}=x^{2}+y^{2} \Rightarrow(\vec{a} \times \vec{i})^{2}=y^{2}+z^{2}$
Similarly $(\vec{a} \times \vec{k})^{2}=x^{2}+y^{2}$
$\Rightarrow(\vec{a} \times \vec{i})^{2}+(\vec{a}+\vec{j})^{2}+(\vec{a} \times \vec{k})^{2}=2\left(x^{2}+y^{2}+z^{2}\right)=2 \vec{a}^{2}$.
88. Given that a, b, c are in H.P.
$\Rightarrow \frac{2}{b}-\frac{1}{a}-\frac{1}{c}=0$
$\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$
$\Rightarrow \frac{x}{-1}=\frac{y}{2}=\frac{1}{-1}$
$\therefore \quad x=1, y=-2$
89. Vectors $a \vec{i}+a \vec{j}+c \vec{k}, \vec{i}+\vec{k}$ and $c \vec{i}+c \vec{j}+b \vec{k}$ are coplanar Therefore, $\left|\begin{array}{llll}a & a & c \\ 1 & 0 & 1 \\ c & c & b\end{array}\right|=0 \Rightarrow c^{2}=a b$
$\therefore a, b, c$ are in G.P.
90. $[\lambda(\vec{a}+\vec{b}) \lambda^{2} \vec{b} \lambda \vec{c}]=[\vec{a} \vec{b}+\vec{c} \vec{b}]$
$\Rightarrow\left|\begin{array}{ccc}\lambda & \lambda & 0 \\ 0 & \lambda^{2} & 0 \\ 0 & 0 & \lambda\end{array}\right|=\left|\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right|$
$\Rightarrow \lambda^{4}=-1$
Hence no real value of $\lambda$.

---

91. Given $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$.
$[\vec{a} \vec{b} \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})$

$$
\begin{aligned}
\Rightarrow \quad \vec{b} \times \vec{c}= & \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|=\hat{i}\left(1+x-x-y-x^{2}\right) \\
& -\hat{j}\left(x+x^{2}-x y-y+x y\right)+\hat{k}\left(x^{2}-y\right) \\
& \hat{a} \cdot(\vec{b} \times \vec{c})=1
\end{aligned}
$$

Which does not depend on $x$ and $y$.
92. Given condition $(\vec{a} \times \vec{b}) \times \vec{c}=\vec{a} \times(\vec{b} \times \vec{c}), \vec{a} \cdot \vec{b} \neq 0, \vec{b} \cdot \vec{c} \neq 0$

$$
\begin{aligned}
& \Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} \\
& \Rightarrow(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{b} \cdot \vec{c}) \vec{a} \\
& \Rightarrow \vec{a} \| \vec{c}
\end{aligned}
$$

93. The vectors

$$
\begin{aligned}
& \overrightarrow{B A}=\hat{i}-2 \hat{j}+6 \hat{k} \\
& \overrightarrow{C A}=(2-a) \hat{i}+2 \hat{j} \\
& \overrightarrow{C B}=(1-a) \hat{i}-6 \hat{j}
\end{aligned}
$$

imply

$$
\begin{aligned}
& \overrightarrow{C A} \cdot \overrightarrow{C B}=0 \\
& \Rightarrow \quad(2-a)(1-a)=0 \\
& \Rightarrow \quad a=2,1
\end{aligned}
$$

94. Given that $|2 \hat{u} \times 3 \hat{v}|=1$
$\Rightarrow 6|\hat{u}||\hat{v}||\sin \theta|=1$
$\Rightarrow \sin \theta=\frac{1}{6}$
Hence there is exactly one value of $\theta$ for which $2 \hat{u} \times 3 \hat{v}$ is a unit vector.
95. $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$

Therefore, $\left|\begin{array}{ccc}x & x-2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 2\end{array}\right|=0$
$\Rightarrow 3 x+2-x+2=0$
$\Rightarrow 2 x=-4$
$\Rightarrow x=-2$.
96. Let for some $\lambda, \vec{a}=\lambda(\vec{b}+\vec{c})$, then
$\alpha \hat{i}+2 \hat{j}+\beta \hat{k}=\lambda\left(\frac{\hat{i}+2 \hat{j}+\hat{k}}{\sqrt{2}}\right)$
$\Rightarrow \lambda=\sqrt{2} \alpha$ and $\lambda=\sqrt{2}$ and $\lambda=\sqrt{2} \beta$
$\Rightarrow \alpha=1$ and $\beta=1$.
97. Since $\vec{a}=8 \vec{b}, \vec{c}=-7 \vec{b}$.
$\therefore \quad \vec{a}$ and $\vec{b}$ are like vectors and $\vec{b}$ and $\vec{c}$ are unlike.
$\Rightarrow \vec{a}$ and $\vec{c}$ will be unlike
Hence, the angle between $\vec{a}$ and $\vec{c}=\pi$.
98. $\left(3 p^{2}-p q+2 q^{2}\right)[\vec{u} \quad \vec{v} \quad \vec{w}]=0$

But since $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar, $[\vec{u} \quad \vec{v} \quad \vec{w}] \neq 0$ which implies that
$3 p^{2}-p q+2 q^{2}=0$
$\Rightarrow 2 p^{2}+p^{2}-p q+\left(\frac{q}{2}\right)^{2}+\frac{7 q^{2}}{4}=0$
$\Rightarrow 2 p^{2}+\left(p-\frac{q}{2}\right)^{2}+\frac{7}{4} q^{2}=0=0$
$\Rightarrow p=0, q=0, p=\frac{q}{2}$
This is possible only when $p=0, q=0$ and therefore exactly one value of $(p, q)$.
99. Projection of a vector on coordinate axes are $x_{2}-x_{1}, y_{2}-y_{1}$, $z_{2}-z_{1}$
So, $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}=\sqrt{36+9+4}=7$
The D.C's of the vector are $\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$ respectively.
100. $\vec{c}=\vec{b} \times \vec{a}$
$\Rightarrow \vec{b} \cdot \vec{c}=0$
$\Rightarrow\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right) \cdot(\hat{i}-\hat{j}-\hat{k})=0$
$\Rightarrow b_{1}-b_{2}-b_{3}=0$
and $\vec{a} \cdot \vec{b}=3$
$\Rightarrow b_{2}-b_{3}=3$
$\Rightarrow b_{1}=b_{2}+b_{3}=3+2 b_{3}$
$\Rightarrow \vec{b}=\left(3+2 b_{3}\right) \hat{i}+\left(3+b_{3}\right) \hat{j}+b_{3} \hat{k}$
101. $\vec{a} \cdot \vec{b}=0$,
$\vec{b} \cdot \vec{c}=0$,
$\vec{c} \cdot \vec{a}=0$
$\Rightarrow 2 \lambda+4+\mu=0 \lambda-1+2 \mu=0$
Solving we get:
$\lambda=-3, \mu=2$.
102. $(2 \vec{a}-\vec{b}) \cdot\{(\vec{a} \times \vec{b}) \times(\vec{a}+2 \vec{b})\}$

$$
\begin{aligned}
& =(2 \vec{a}-\vec{b}) \cdot\{2 \vec{a}-\vec{b})\{[ \vec{a} \cdot(\vec{a}+2 \vec{b})] \vec{b}-[\vec{b} \cdot(\vec{a}+2 \vec{b}) \vec{a}]\} \\
& =-5(\vec{a}) 2(\vec{b}) 2+5(\vec{a} \vec{b}) 2=-5
\end{aligned}
$$

103. $\vec{b} \times c=\vec{b} \times \vec{d}$
$\Rightarrow \vec{a} \times(\vec{b} \times \vec{c})=\vec{a} \times(\vec{b} \times \vec{d})$
$\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{a} \cdot \vec{d}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{d}$
$\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) c=-(\vec{a} \cdot \vec{b}) \vec{d}$
$\therefore \vec{d}=\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

---

104. $\vec{c} \cdot \vec{d}=\overrightarrow{0} \Rightarrow 5|\vec{a}|^{2}+6 \vec{a} \cdot \vec{b}-8|\vec{b}|^{2}=0$
$\Rightarrow 6 \vec{a} \cdot \vec{b}=3 \Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \Rightarrow(\vec{a} \cdot \vec{b})=\frac{\pi}{3}$
105. $\overline{A E}=$ vector component of $\vec{q}$ on $\vec{p}$
$\overline{A E}=\frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{q})} \vec{p}$
$\therefore$ From $\triangle A B E ; \overline{A B}+\overline{B E}=\overline{A E}$
$\Rightarrow \vec{q}+\vec{r}=\frac{(\vec{p} \cdot \vec{q}) \vec{p}}{(\vec{p} \cdot \vec{q})}$
$\Rightarrow \vec{r}=-\vec{q}+\frac{(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{q})} \vec{p}$
106. Let $M$ be the median, then
$\overline{A M}=\frac{\overline{A B}+\overline{A C}}{2}$
$\overline{A M}=4 \hat{i}=-\hat{j}+4 \hat{k}$
$\therefore \mid \overline{A M} \mid=\sqrt{16+16+1}=\sqrt{33}$
107. $[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=\lambda[\vec{a} \vec{b} \vec{c}]^{2}$
$\lambda=1$
108. Since $(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$
$\Rightarrow-\vec{b} \cdot \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}|$
$\Rightarrow-|\vec{b}||\vec{c}| \cos \theta=\frac{1}{3}|\vec{b}||\vec{c}|$
$\Rightarrow \cos \theta=-\frac{1}{3}$
$\Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3}$.
109. Given expression is $\left(\vec{a} \cdot \vec{c}-\frac{\sqrt{3}}{2}\right) \vec{b}-\left(\vec{a} \cdot \vec{b}+\frac{\sqrt{3}}{2}\right) \vec{c}=0$
$\Rightarrow \vec{a} \cdot \vec{b}=\cos \theta=-\sqrt{3} / 2 \Rightarrow \theta=5 \pi / 6$