Chapter 9 Differential Equations

NCERT Solutions Class 12 Maths Chapter 10 Vector Algebra

Class 12 Maths Chapter 10, Vector Algebra, introduces quantities that have both magnitude and direction. The chapter covers important topics such as types of vectors, vector addition, scalar (dot) product, vector (cross) product, projections, and triple products. These concepts are important for understanding three-dimensional geometry and solving vector-based questions in Class 12 Maths.

The CBSE-compliant NCERT Solutions for Class 12 Maths Chapter 10 by ALLEN covers the questions in the NCERT textbook. Vector operations, dot products, cross products, projections, and triple products are all explained in detail in the solutions. By studying these problems, students can improve their understanding of Vector Algebra, improve their accuracy, and get ready for competitive and CBSE board exams.

1.0Key Concepts of Class 12 Maths Chapter 10 Vector Algebra

Class 12 Maths Chapter 10, Vector Algebra, introduces vectors and explains their different types, operations, and applications. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 10 include:

Vectors and Scalars: Understand the difference between scalar and vector quantities with suitable examples.
Types of Vectors: Learn about zero vectors, unit vectors, equal vectors, collinear vectors, and other common types of vectors.
Vector Addition and Subtraction: Study vector addition and subtraction using the triangle law and parallelogram law.
Scalar (Dot) Product: Understand the dot product, its properties, and its use in finding angles and projections of vectors.
Vector (Cross) Product: Learn the cross product, its properties, and its applications in geometry and physics.
Vector Triple Product: Understand scalar and vector triple products and their important identities.

2.0NCERT Class 12 Maths Chapter 10 Vector Algebra : Detailed Solutions

EXERCISE - 10.1

Represent graphically a displacement of 40 km, 30° east of north. Sol. Here, $OP$ vector represents the displacement of 40 km, 30° East of North.

Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 m north-west (iii) 40° (iv) 40 watt (v) $1 0^{- 19}$ coulomb (vi) $20 m / s^{2}$

Sol.

(i) 10 kg is a scalar quantity because it involves only magnitude. (ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction. (iii) $4 0^{\circ}$ is a scalar quantity as it involves only magnitude. (iv) 40 watts is a scalar quantity as it involves only magnitude. (v) $1 0^{- 19}$ coulomb is a scalar quantity as it involves only magnitude. (vi) $20 m / s^{2}$ is a vector quantity as it involves magnitude as well as direction.

Classify the following as scalar and vector quantities. (i) time period (ii) distance (iii) force (iv) velocity (v) work done Sol. (i) Time period is a scalar quantity as it involves only magnitude. (ii) Distance is a scalar quantity as it involves only magnitude.

(iii) Force is a vector quantity as it involves both magnitude and direction. (iv) Velocity is a vector quantity as it involves both magnitude as well as direction. (v) Work done is a scalar quantity as it involves only magnitude.

In Figure, identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal

Sol.

(i) Vectors $a$ and $d$ are coinitial because they have the same initial point. (ii) Vectors $b$ and $d$ are equal because they have the same magnitude and direction. (iii) Vectors $a$ and $c$ are collinear but not equal. This is because although they are parallel, their directions are not the same.

Answer the following as true or false. (i) $a$ and $- a$ are collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal.

Sol.

(i) True, Vectors $a$ and $- a$ are parallel to the same line. (ii) False, Collinear vectors are those vectors that are parallel to the same line. (iii) False, Two vectors having the same magnitude need not necessarily be parallel to the same line. (iv) False, Only if the magnitude and direction of two vectors are the same regard less of the positions of their initial points, the two vectors are said to be equal.

EXERCISE 10.2

Compute the magnitude of the following vectors:

$a = \hat{i} + \hat{j} + \hat{k}; b = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}; c = \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} - \frac{1}{3} \hat{k}$

Sol. The given vectors are: $a = \hat{i} + \hat{j} + \hat{k}$ ;

$b ∣ a ∣ ∣ b ∣ ∣ c ∣ = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}; c = \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} - \frac{1}{3} \hat{k} = (1)^{2} + (1)^{2} + (1)^{2} = 3 = (2)^{2} + (- 7)^{2} + (- 3)^{2} = 4 + 49 + 9 = 62 = (\frac{1}{3})^{2} + (\frac{1}{3})^{2} + (- \frac{1}{3})^{2} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

Write two different vectors having same magnitude. Sol. Consider

$a = (\hat{i} - 2 \hat{j} + 3 \hat{k}) and b = (2 \hat{i} + \hat{j} - 3 \hat{k}) .$

It can be observed that

$∣ a ∣ = 1^{2} + (- 2)^{2} + 3^{2} = 1 + 4 + 9 = 14 and ∣ b ∣ = 2^{2} + 1^{2} + (- 3)^{2} = 4 + 1 + 9 = 14$

Hence, $a$ and $b$ are two different vectors having the same magnitude. The vectors are different because they have different directions.

Write two different vectors having same direction. Sol. Consider $a = (\hat{i} + \hat{j} + \hat{k})$ and $b = (2 \hat{i} + 2 \hat{j} + 2 \hat{k})$ . The direction consines of $a$ are given by,

$l = \frac{1}{1 ^{2} + 1 ^{2} + 1 ^{2}} = \frac{1}{3} m = \frac{1}{1 ^{2} + 1 ^{2} + 1 ^{2}} = \frac{1}{3}$

$n = \frac{1}{1 ^{2} + 1 ^{2} + 1 ^{2}} = \frac{1}{3} .$

The direction cosines of $b$ are given by

$l = \frac{2}{2 ^{2} + 2 ^{2} + 2 ^{2}} = \frac{2}{2 3} = \frac{1}{3} m = \frac{2}{2 ^{2} + 2 ^{2} + 2 ^{2}} = \frac{2}{2 3} = \frac{1}{3} n = \frac{2}{2 ^{2} + 2 ^{2} + 2 ^{2}} = \frac{2}{2 3} = \frac{1}{3}$

The direction cosines of $a$ and $b$ are the same. Hence, the two vectors have the same direction.

Find the values of x and y so that the vectors $2 \hat{i} + 3 \hat{j}$ and $x \hat{i} + y \hat{j}$ are equal. Sol. The two vectors $2 \hat{i} + 3 \hat{j}$ and $x \hat{i} + y \hat{j}$ will be equal if their corresponding scalar components are equal. Hence, the required values of x and y are 2 and 3 respectively.
Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point (-5, 7). Sol. The vector with the initial point P $(2, 1)$ and terminal point $Q (- 5, 7)$ can be given by, $PQ = (- 5 - 2) \hat{i} + (7 - 1) \hat{j} \Rightarrow PQ = - 7 \hat{i} + 6 \hat{j}$ Hence, the required scalar components are -7 and 6 while the vector components are $- 7 \hat{i}$ and $6 \hat{j}$ .
Find the sum of the vectors $a = \hat{i} - 2 \hat{j} + \hat{k}$ , $\hat{b} = - 2 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $c = \hat{i} - 6 \hat{j} - 7 \hat{k}$ . Sol. The given vectors are $a = \hat{i} - 2 \hat{j} + \hat{k}$ , $\hat{b} = - 2 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $c = \hat{i} - 6 \hat{j} - 7 \hat{k}$ . $∴ a + b + c$

$= (1 - 2 + 1) \hat{i} + (- 2 + 4 - 6) \hat{j} + (1 + 5 - 7)) \hat{k} = 0 \hat{i} - 4 \hat{j} - 1 \hat{k} = - 4 \hat{j} - \hat{k}$

Find the unit vector in the direction of the vector $a = \hat{i} + \hat{j} + 2 \hat{k}$ . Sol. The unit vector in the direction of vector $a = \hat{i} + \hat{j} + 2 \hat{k}$ is given by $\overset{a}{^} = \frac{a}{∣ a ∣}$ . $∣ a ∣ = 1^{2} + 1^{2} + 2^{2} = 1 + 1 + 4 = 6$ $∴ \overset{a}{^} = \frac{a}{∣ a ∣} = \frac{i ^ + j ^ + 2 k ^}{6} = \frac{1}{6} \hat{i} + \frac{1}{6} \hat{j} + \frac{2}{6} \hat{k}$
Find the unit vector in the direction of vector $PQ$ , where P and Q are the points (1,2,3) and (4, 5, 6), respectively. Sol. The given point are $P (1, 2, 3)$ and $Q (4, 5, 6)$ . $∴ PQ = (4 - 1) \hat{i} + (5 - 2) \hat{j} + (6 - 3) \hat{k} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ ∴ Magnitude of given vector,

$∣ PQ ∣ = 3^{2} + 3^{2} + 3^{2} = 9 + 9 + 9 = 27 = 33$

Hence, the unit vector in the direction of $PQ$ is $\frac{PQ}{∣ PQ ∣} = \frac{3 i ^ + 3 j ^ + 3 k ^}{3 3} = \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}$

For given vectors, $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $b = - \hat{i} + \hat{j} - \hat{k}$ , find the unit vector in the direction of the vector $a + b$ . Sol. The given vectors are $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $b = - \hat{i} + \hat{j} - \hat{k}$ . $∴ a + b = (2 - 1) \hat{i} + (- 1 + 1) \hat{j} + (2 - 1) \hat{k}$

$= 1 \hat{i} + + 0 \hat{j} + 1 \hat{k} = \hat{i} + \hat{k}$

$∣ a + b ∣ = 1^{2} + 1^{2} = 2$ Hence, the unit vector in the direction of $(a + b)$ is $\frac{( a + b )}{∣ a + b ∣} = \frac{i ^ + k ^}{2} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{k}$ 10. Find a vector in the direction of vector $5 \hat{i} - \hat{j} + 2 \hat{k}$ which has magnitude 8 units. Sol. Let $a = 5 \hat{i} - \hat{j} + 2 \hat{k}$ $∴ ∣ a ∣ = 5^{2} + (- 1)^{2} + 2^{2} = 25 + 1 + 4 = 30$ $∴ \overset{a}{^} = \frac{a}{∣ a ∣} = \frac{5 i ^ - j ^ + 2 k ^}{30}$ Hence, the vector in the direction of vector $5 \hat{i} - \hat{j} + 2 \hat{k}$ which has magnitude 8 units is given by,

$8 \overset{a}{^} = 8 (\frac{5 i ^ - j ^ + 2 k ^}{30}) = \frac{40}{30} \hat{i} - \frac{8}{30} \hat{j} + \frac{16}{30} \hat{k}$

Show that the vectors $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $- 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$ are collinear. Sol. Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$ It is observed that

$b = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k} = - 2 (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = - 2 a$

$∴ b = λ a$

where, $λ = - 2$ , Hence, the given vectors are collinear.

Find the direction cosines of the vector $\hat{i} + 2 \hat{j} + 3 \hat{k}$ Sol. Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$ $∴ ∣ a ∣ = 1^{2} + 2^{2} + 3^{2} = 1 + 4 + 9 = 14$ $∴ \overset{a}{^} = \frac{a}{∣ a ∣}$

Hence, the direction cosines of $\overset{a}{^}$ are $\frac{1}{14}, \frac{2}{14}, \frac{3}{14}$ .

Find the direction cosines of the vector joining the points $A (1, 2, - 3)$ and B(-1, -2, 1), directed from A to B. Sol. The given points are A (1, 2, -3) and B (-1, -2, 1). $∴ AB = (- 1 - 1) \hat{i} + (- 2 - 2) \hat{j} + {1 - (- 3)} \hat{k}$ $\Rightarrow AB = - 2 \hat{i} - 4 \hat{j} + 4 \hat{k}$

Direction ratios of $AB$ are $a = - 2, b = - 4, c = 4$ Now direction cosines of $AB$ are :

$ℓ = \frac{- 2}{( - 2 ) ^{2} + ( - 4 ) ^{2} + ( 4 ) ^{2}} = \frac{- 2}{6} = - \frac{1}{3}, m = \frac{- 4}{( - 2 ) ^{2} + ( - 4 ) ^{2} + ( 4 ) ^{2}} = \frac{- 4}{6} = - \frac{2}{3} n = \frac{4}{( - 2 ) ^{2} + ( - 4 ) ^{2} + ( 4 ) ^{2}} = \frac{4}{6} = \frac{2}{3}$

Hence, the direction cosines of $AB$ are $- \frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$ .

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ. Sol. Let $a = \hat{i} + \hat{j} + \hat{k}$ . Direction ratios of $a$ are $a = b = c = 1$ Now, direction cosines are

$ℓ = \frac{1}{( 1 ) ^{2} + ( 1 ) ^{2} + ( 1 ) ^{2}} = \frac{1}{3} = m = n$

Therefore, the direction cosines of $a$ are $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$ . Now, let $α, β$ and $γ$ be the angles formed by $a$ with the positive directions of x, y and z axes. Then, we have

$cos α = \frac{1}{3}, cos β = \frac{1}{3}, cos γ = \frac{1}{3}, α = β = γ = cos^{- 1} (\frac{1}{3})$

Hence, the given vector is equally inclined to axes OX, OY and OZ.

Find the position vector of a point R which divides the line joining two points P and $Q$ whose position vectors are $\hat{i} + 2 \hat{j} - \hat{k}$ and $- \hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (i) Internally (ii) Externally Sol. Here, $OP = \hat{i} + 2 \hat{j} - \hat{k} = a$ (let) and

$OQ = - \hat{i} + \hat{j} + \hat{k} = b (let),$

Also $m = 2, n = 1$ when R divides PQ internally in the ratio $2 : 1$ , then (i) P.V. of $R = \frac{m b + n a}{m + n}$

$= \frac{2 ( - i ^ + j ^ + k ^ ) + 1 ( i ^ + 2 j ^ - k ^ )}{2 + 1} = \frac{( - 2 i ^ + 2 j ^ + 2 k ^ ) + ( i ^ + 2 j ^ - k ^ )}{3} = \frac{- i ^ + 4 j ^ + k ^}{3} = - \frac{1}{3} \hat{i} + \frac{4}{3} \hat{j} + \frac{1}{3} \hat{k}$

(ii) when R divides PQ externally in the ratio 2 : 1 then,

$P . V . of R = \frac{m b - n a}{m - n} = \frac{2 ( - i ^ + j ^ + k ^ ) - 1 ( i ^ + 2 j ^ - k ^ )}{2 - 1} = - 3 \hat{i} + 3 \hat{k}$

Find the position vector of the mid-point of the vector joining the points $P (2, 3, 4)$ and Q (4, 1, -2). Sol. The position vector of P and Q are given by $p = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} & q = 4 \hat{i} + \hat{j} - 2 \hat{k}$ respectively

$∴$ P.V. of midpoint of $PQ = \frac{1}{2} (p + q)$

$= \frac{( 2 i ^ + 3 j ^ + 4 k ^ ) + ( 4 i ^ + j ^ - 2 k ^ )}{2} = \frac{6 i ^ + 4 j ^ + 2 k ^}{2} = 3 \hat{i} + 2 \hat{j} + \hat{k}$

Show that the points $A, B$ and C with position vectors, $a = 3 \hat{i} - 4 \hat{j} - 4 \hat{k}, b = 2 \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - 3 \hat{j} - 5 \hat{k}$ , respectively form the vertices of a right angled triangle. Sol. Position vectors of points A, B, and C are respectively given as:

$a = 3 \hat{i} - 4 \hat{j} - 4 \hat{k}, b = 2 \hat{i} - \hat{j} + \hat{k}$

and $c = \hat{i} - 3 \hat{j} - 5 \hat{k}$ $∴ AB = b - a$

$= (2 - 3) \hat{i} + (- 1 + 4) \hat{j} + (1 + 4) \hat{k} = - \hat{i} + 3 \hat{j} + 5 \hat{k}$

$BC = c - b = (1 - 2) \hat{i} + (- 3 + 1) \hat{j} + (- 5 - 1) \hat{k} = - \hat{i} - 2 \hat{j} - 6 \hat{k}$

$CA = a - c = (3 - 1) \hat{i} + (- 4 + 3) \hat{j} + (- 4 + 5) \hat{k} = 2 \hat{i} - \hat{j} + \hat{k}$

Now,

$AB + BC + CA = (- \hat{i} + 3 \hat{j} + 5 \hat{k}) + (- \hat{i} - 2 \hat{j} - 6 \hat{k}) + (2 \hat{i} - \hat{j} + \hat{k}) = 0$

$∴ A, B, C$ are vertices of triangle Now,

$∣ AB ∣^{2} = (- 1)^{2} + 3^{2} + 5^{2} = 1 + 9 + 25 = 35 ∣ BC ∣^{2} = (- 1)^{2} + (- 2)^{2} + (- 6)^{2} = 1 + 4 + 36 = 41 ∣ CA ∣^{2} = 2^{2} + (- 1)^{2} + 1^{2} = 4 + 1 + 1 = 6$

$∴ ∣ AB ∣^{2} + ∣ CA ∣^{2} = ∣ BC ∣^{2} = 35 + 6 = 41$

Hence, $A, B$ and C are vertices of right angled triangle.

In triangle ABC which of the following is not true:

(A) $AB + BC + CA = 0$ (B) $AB + BC - AC = 0$ (C) $AB + BC - CA = 0$ (D) $AB - CB + CA = 0$ Sol. On applying the triangle law of addition in the given triangle, we have:

$AB + BC = AC$

$\Rightarrow AB + BC = - CA$

$\Rightarrow AB + BC + CA = 0$

∴ The equation given in alternative A is true.

$AB + BC = AC$

$\Rightarrow AB + BC - AC = 0$ ∴ The equation given in alternative B is true. From equation (2), we have :

$AB - CB + CA = 0$

∴ The equation given in alternative D is true. Now, consider the equation given in alternative C :

$AB + BC - CA = 0$

$\Rightarrow AB + BC = CA$ From equation (1) and (3), we have :

$AC = CA \Rightarrow AC = - AC$

$\Rightarrow AC + AC = 0 \Rightarrow 2 AC = 0$ $\Rightarrow AC = 0$ , which is not true. Hence, the equation given in alternative C is incorrect. The correct answer is (C).

If $a$ and $b$ are two collinear vectors, then which of the following are incorrect: (A) $b = λ a$ , for some scalar $λ$ (B) $a = \pm b$ (C) the respective components of $a$ and $b$ are proportional (D) both the vectors $a$ and $b$ have same direction, but different magnitudes Sol. If $a$ and $b$ are two collinear vector, then they are parallel. Therefore, we have : $b = λ a$ (For some scalar $λ$ ) If $λ = \pm 1$ , then $a = \pm b$ . If $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $b = λ a$ . $\Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = λ (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k})$ $\Rightarrow b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = (λ a_{1}) \hat{i} + (λ a_{2}) \hat{j} + (λ a_{3}) \hat{k}$ $\Rightarrow b_{1} = λ a_{1}, b_{2} = λ a_{2}, b_{3} = λ a_{3}$ $\Rightarrow \frac{b _{1}}{a _{1}} = \frac{b _{2}}{a _{2}} = \frac{b _{3}}{a _{3}} = λ$

Thus, the respective scalar components of $a$ and $b$ are proportional. However, vectors $a$ and $b$ can have different directions. Hence, the statement given in D is incorrect. The correct answer is (D).

EXERCISE 10.3

Find the angle between two vectors $a$ and $b$ with magnitude $3$ and 2, respectively having $a \cdot b = 6$ . Sol. It is given that, $∣ a ∣ = 3, ∣ b ∣ = 2$ and, $a \cdot b = 6$ Now, we know that $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ . $∴ 6 = 3 \times 2 \times cos θ$ $\Rightarrow cos θ = \frac{6}{3 \times 2}$ $\Rightarrow cos θ = \frac{1}{2}$ $\Rightarrow θ = \frac{π}{4}$ Hence, the angle between the given vectors $a$ and $b$ is $\frac{π}{4}$ .
Find the angle between the vectors $\hat{i} - 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} - 2 \hat{j} + \hat{k}$ . Sol. Let $a = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + \hat{k}$ . $∣ a ∣ = 1^{2} + (- 2)^{2} + 3^{2} = 1 + 4 + 9 = 14$ $∣ b ∣ = 3^{2} + (- 2)^{2} + 1^{2} = 9 + 4 + 1 = 14$ Now, $a \cdot b = (\hat{i} - 2 \hat{j} + 3 \hat{k}) \cdot (3 \hat{i} - 2 \hat{j} + \hat{k})$ $= 1 \cdot 3 + (- 2) (- 2) + 3 \cdot 1$ $= 3 + 4 + 3$ $= 10$ Also, we know that $a \cdot b = ∣ a ∣ \cdot ∣ b ∣ cos θ$ $∴ 10 = 1414 cos θ$ $\Rightarrow cos θ = \frac{10}{14}$ $\Rightarrow θ = cos^{- 1} (\frac{5}{7})$
Find the projection of the vector $\hat{i} - \hat{j}$ on the vector $\hat{i} + \hat{j}$ . Sol. Let $a = \hat{i} - \hat{j}$ and $b = \hat{i} + \hat{j}$ Now, projection of vector $a$ on $b$ is given by, $\frac{1}{∣ b ∣} (a \cdot b) = \frac{1}{1 + 1} {1 \cdot 1 + (- 1) (1)} = \frac{1}{2} (1 - 1) = 0$ Hence, the projection of vector $a$ on $b$ is 0 .
Find the projection of the vector $\hat{i} + 3 \hat{j} + 7 \hat{k}$ on the vector $7 \hat{i} - \hat{j} + 8 \hat{k}$ . Sol. Let $a = \hat{i} + 3 \hat{j} + 7 \hat{k}$ and $\hat{b} = 7 \hat{i} - \hat{j} + 8 \hat{k}$ . Now, projection of vector $a$ on $b$ is given by

$\frac{1}{∣ b ∣} (a \cdot b) = \frac{1}{7 ^{2} + ( - 1 ) ^{2} + 8 ^{2}} {1 (7) + 3 (- 1) + 7 (8)} = \frac{7 - 3 + 56}{49 + 1 + 64} + = \frac{60}{114}$

Hence, the projection of vector $a$ on $b$ is $\frac{60}{114}$

Show that each of the given three vectors is a unit vector:

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

Also, show that they mutually perpendicular to each other. Sol. Let

$a \Rightarrow ∣ a ∣ b \Rightarrow ∣ b ∣ c \Rightarrow ∣ c ∣ = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) = \frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} = (\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{6}{7})^{2} = \frac{4}{49} + \frac{9}{49} + \frac{36}{49} = 1 = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k} = (\frac{3}{7})^{2} + (- \frac{6}{7})^{2} + (\frac{2}{7})^{2} = \frac{9}{49} + \frac{36}{49} + \frac{4}{49} = 1 = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k}) = \frac{6}{7} \hat{i} + \frac{2}{7} \hat{j} - \frac{3}{7} \hat{k} = (\frac{6}{7})^{2} + (\frac{2}{7})^{2} + (- \frac{3}{7})^{2} = \frac{36}{49} + \frac{4}{49} + \frac{9}{49} = 1$

Thus, each of the given three vectors is a unit vector.

$a \cdot b b \cdot c c \cdot a = \frac{2}{7} \times \frac{3}{7} + \frac{3}{7} \times (\frac{- 6}{7}) + \frac{6}{7} \times \frac{2}{7} = \frac{6}{49} - \frac{18}{49} + \frac{12}{49} = 0 = \frac{3}{7} \times \frac{6}{7} + (\frac{- 6}{7}) \times \frac{2}{7} + \frac{2}{7} \times (\frac{- 3}{7}) = \frac{18}{49} - \frac{12}{49} - \frac{6}{49} = 0 = \frac{6}{7} \times \frac{2}{7} + \frac{2}{7} \times \frac{3}{7} + (\frac{- 3}{7}) \times \frac{6}{7} = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = 0$

Hence, the given three vectors are mutually perpendicular to each other.

Find $∣ a ∣$ and $∣ b ∣$ , if $(a + b) \cdot (a - b) = 8$ and $∣ a ∣ = 8∣ b ∣$ .

Sol. $(a + b) \cdot (a - b) = 8$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \frac{8}{63} a \cdot a - a \cdot b + b \cdot a - b \cdot b = 8 ∣ a ∣^{2} - ∣ b ∣^{2} = 8 (8∣ b ∣^{2} - ∣ b ∣^{2} = 8 64∣ b ∣^{2} - ∣ b ∣^{2} = 8 ∣ b ∣^{2} = \frac{8}{63} \Rightarrow [∵ a \cdot a = ∣ a ∣^{2}] [∵ ∣ a ∣ = 8∣ b ∣] \Rightarrow ∣ b ∣^{2} = 8$

[Magnitude of a vector is non-negative]

$\Rightarrow ∴ ∣ b ∣ = \frac{2 2}{3 7} ∣ a ∣ = 8∣ b ∣ = \frac{8 \times 2 2}{3 7} = \frac{16 2}{3 7}$

Evaluate the product $(3 a - 5 b) \cdot (2 a + 7 b)$ . Sol. $(3 a - 5 b) \cdot (2 a + 7 b)$

$= 6 (a \cdot a) + 21 (a \cdot b) - 10 (b \cdot a) - 35 (b \cdot b) = 6 (a \cdot a) + 21 (a \cdot b) - 10 (a \cdot b) - 35 (b \cdot b) = 6∣ a ∣^{2} + 11 a \cdot b - 35∣ b ∣^{2} (∵ a \cdot a = ∣ a ∣^{2} and a \cdot b = b \cdot a)$

Find the magnitude of two vectors $a$ and $b$ , having the same magnitude and such that the angle between them is 60° and their scalar product is $\frac{1}{2}$ . Sol. Let $θ$ be the angle between the vectors $a$ and $b$ . It is given that $∣ a ∣ = ∣ b ∣, a . b = \frac{1}{2}$ and $θ = 6 0^{\circ}$ We know that $a . b = ∣ a ∣∣ b ∣ cos θ$ . $∴ \frac{1}{2} = ∣ a ∣∣ a ∣ cos 6 0^{\circ}$ [Using (1)] $\Rightarrow \frac{1}{2} = ∣ a ∣^{2} \times \frac{1}{2}$ $\Rightarrow ∣ a ∣^{2} = 1 \Rightarrow ∣ a ∣ = 1$ $\Rightarrow ∣ a ∣ = ∣ b ∣ = 1$
Find| $x ∣$ , if for a unit vector

$a, (x - a) \cdot (x + a) = 12.$

Sol. $(x - a) \cdot (x + a) = 12$ $\Rightarrow x \cdot x + x \cdot a - a \cdot x - a \cdot a = 12$ $\Rightarrow ∣ x ∣^{2} - ∣ a ∣^{2} = 12 (∵ a \cdot a = ∣ a ∣^{2}$ and $a \cdot b = b \cdot a)$ $\Rightarrow ∣ x ∣^{2} - 1 = 12 [∣ a ∣ = 1$ as $a$ is a unit vector] $\Rightarrow ∣ x ∣^{2} = 13$ $∴ ∣ x ∣ = 13$

If $a = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, b = - \hat{i} + 2 \hat{j} + \hat{k}$ and $c = 3 \hat{i} + \hat{j}$ are such that $a + λ b$ is perpendicular to $c$ , then find the value of $λ$ . Sol. The given vectors are $a = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $b = - \hat{i} + 2 \hat{j} + \hat{k}$ and $c = 3 \hat{i} + \hat{j}$ . Now,

$a + λ b = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- \hat{i} + 2 \hat{j} + \hat{k}) = (2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}$

If $(a + λ b)$ is perpendicular to $c$ , then $(a + λ b) \cdot c = 0$ $\Rightarrow [(2 - λ) \hat{i} + (2 + 2 λ) \hat{j} + (3 + λ) \hat{k}] \cdot (3 \hat{i} + \hat{j}) = 0$ $\Rightarrow (2 - λ) 3 + (2 + 2 λ) 1 + (3 + λ) 0 = 0$ $\Rightarrow 6 - 3 λ + 2 + 2 λ = 0$ $\Rightarrow - λ + 8 = 0$ $\Rightarrow λ = 8$

Hence, the required value of $λ$ is 8 .

Show that $∣ a ∣ b + ∣ b ∣ a$ is perpendicular to $∣ a ∣ b - ∣ b ∣ a$ , for any two non-zero vectors $a$ and $b$ . Sol. $(∣ a ∣ b + ∣ b ∣ a) \cdot (∣ a ∣ b - ∣ b ∣ a)$

$= ∣ a ∣^{2} (b \cdot b) - ∣ a ∣∣ b ∣ (b \cdot a) + ∣ b ∣∣ a ∣ (a \cdot b) - ∣ b ∣^{2} (a \cdot a) = ∣ a ∣^{2} ∣ b ∣^{2} - ∣ b ∣^{2} ∣ a ∣^{2} = 0$

$(∵ a \cdot a = ∣ a ∣^{2}$ and $a \cdot b = b \cdot a)$ Hence, $∣ a ∣ b + ∣ b ∣ a$ and $∣ a ∣ b - ∣ b ∣ a$ are perpendicular to each other for any two nonzero vectors $a$ and $b$ .

If $a \cdot a = 0$ and $a \cdot b = 0$ , then what can be concluded about the vector $b$ ? Sol. It is given that $a \cdot a = 0$ and $a \cdot b = 0$ . Now, $a \cdot a = 0 \Rightarrow ∣ a ∣^{2} = 0 \Rightarrow ∣ a ∣ = 0$ $∴ a$ is a zero vector. Hence, vector $b$ satisfying $a \cdot b = 0$ can be any vector.
f $a, b, c$ are unit vectors such that $a + b + c = 0$ find the value of $a \cdot b + b \cdot c + c \cdot a$ . Sol. Given $a, b, c$ are unit vectors, therefore

$∣ a ∣ = 1, ∣ b ∣ = 1 and ∣ c ∣ = 1.$

Again given $a + b + c = 0$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow ∣ a + b + c ∣ = 0 (a + b + c) \cdot (a + b + c) = 0 a \cdot a + b \cdot b + c \cdot c + 2 (a \cdot b + b \cdot c + c \cdot a) = 0 ∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a) = 0 [∵ a \cdot a = ∣ a ∣^{2}] 1 + 1 + 1 + 2 (a \cdot b + b \cdot c + c \cdot a) = 0$

[Using equation (1)]

$\Rightarrow \Rightarrow 2 (a \cdot b + b \cdot c + c \cdot a) = - 3 a \cdot b + b \cdot c + c \cdot a = \frac{- 3}{2}$

If either vector $a = 0$ or $b = 0$ , then $a \cdot b = 0$ . But the converse need not be true. Justify your answer with an example. Sol. Consider $a = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ and $b = 3 \hat{i} + 3 \hat{j} - 6 \hat{k}$ . Then, $a \cdot b = 2 \cdot 3 + 4 \cdot 3 + 3 (- 6) = 6 + 12 - 18 = 0$ We now observe that:

$∣ a ∣ = 2^{2} + 4^{2} + 3^{2} = 29 ∣ b ∣ = 3^{2} + 3^{2} + (- 6)^{2} = 54 ∴ a \neq = 0 ∴ b \neq = 0$

Hence, the converse of the given statement need not be true.

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2), respectively, then find $∠ ABC$ . [ $∠ ABC$ is the angle between the vectors $BA$ and $BC]$ . Sol. The vertices of $△ ABC$ are given as $A (1, 2, 3)$ , B (-1, 0, 0), and C (0, 1, 2). Also, it is given that $∠ ABC$ is the angle between the vectors $BA$ and $BC$ .

$BA = (1 - (- 1)) \hat{i} + (2 - 0) \hat{j} + (3 - 0) \hat{k} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$

$∴ BA \cdot BC = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$

$∣ BA ∣ = 2^{2} + 2^{2} + 3^{2} = 4 + 4 + 9 = 17 ∣ BC ∣ = 1 + 1 + 2^{2} = 6$

Now, it is known that :

$\Rightarrow \Rightarrow \Rightarrow BA \cdot BC = ∣ BA ∣∣ BC ∣ cos (∠ ABC) 10 = 17 \times 6 cos (∠ ABC) cos (∠ ABC) = \frac{10}{17 \times 6} ∠ ABC = cos^{- 1} (\frac{10}{102})$

Show that the points $A (1, 2, 7), B (2, 6, 3)$ and C (3, 10, -1) are collinear. Sol. The given points are $A (1, 2, 7), B (2, 6, 3)$ , and C (3, 10, -1). Position vectors of points $A, B$ and C are $a = \hat{i} + 2 \hat{j} + 7 \hat{k}, b = 2 \hat{i} + 6 \hat{j} + 3 \hat{k}$ and $c = 3 \hat{i} + 10 \hat{j} - \hat{k}$ respectively

$∴ AB = b - a = (2 - 1) \hat{i} + (6 - 2) \hat{j} + (3 - 7) \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k}$

$BC = c - b = (3 - 2) \hat{i} + (10 - 6) \hat{j} + (- 1 - 3) \hat{k} = \hat{i} + 4 \hat{j} - 4 \hat{k}$

Since $AB = BC \Rightarrow AB ∥ BC$ Here, point B is common in $AB$ and $BC$ So, $AB$ and $BC$ are collinear. Hence, $A, B, C$ are collinear

Show that the vectors $2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ form the vertices of a right angled triangle. Sol. Let vectors $2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ be position vectors of points $A, B$ and C respectively.

$i.e., OC ∴ AB OA = 2 \hat{i} - \hat{j} + \hat{k}, OB = \hat{i} - 3 = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} = (1 - 2) \hat{i} + (- 3 + 1) \hat{j} + (- 5 - 1) \hat{k} = - \hat{i} - 2 \hat{j} - 6 \hat{k}$

$BC AC = (3 - 1) \hat{i} + (- 4 + 3) \hat{j} + (- 4 + 5) \hat{k} = 2 \hat{i} - \hat{j} + \hat{k} = (3 - 2) \hat{i} + (- 4 + 1) \hat{j} + (- 4 - 1) \hat{k} = \hat{i} - 3 \hat{j} - 5 \hat{k}$

Now,

$AB + BC = (- \hat{i} - 2 \hat{j} + 6 \hat{k}) + (2 \hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3 \hat{j} - 5 \hat{k} = A C$

$∴ A, B$ and C are vertices of a triangle. Now,

$∣ AB ∣ ∣ BC ∣ ∣ AC ∣ ∴ ∣ BC ∣^{2} = (- 1)^{2} + (- 2)^{2} + (- 6)^{2} = 1 + 4 + 36 = 41 = 2^{2} + (- 1)^{2} + 1^{2} = 4 + 1 + 1 = 6 = (1)^{2} + (- 3)^{2} + (- 5)^{2} = 1 + 9 + 25 = 35 + ∣ AC ∣^{2} = 6 + 35 = 41 = ∣ AB ∣^{2}$

Hence, $A, B$ and C are vertices of a right angle triangle.

If $a$ is a nonzero vector of magnitude 'a' and $λ$ a nonzero scalar, then $λ a$ is unit vector if (A) $λ = 1$ (B) $λ = - 1$ (C) $a = ∣ λ ∣$ (D) $a = \frac{1}{∣ λ ∣}$ Sol. (D) Given $λ a$ is unit vector.

$∴ \Rightarrow ∣ λ a ∣ = 1 ∣ a ∣ = \frac{1}{∣ λ ∣} \Rightarrow ∣ λ ∣∣ a ∣ = 1 [λ \neq = 0]$

Hence, vector $λ a$ is a unit vector if

$a = \frac{1}{∣ λ ∣} . (λ \neq = 0)$

The correct answer is (D).

EXERCISE 10.4

Find $∣ a \times b ∣$ , if $a = \hat{i} - 7 \hat{j} + 7 \hat{k}$ and

$b = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} .$

Sol. We have, $a = \hat{i} - 7 \hat{j} + 7 \hat{k}$ and $b = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$

$a \times b = \hat{i} 13 \hat{j} - 7 - 2 \hat{k} 72 = \hat{i} (- 14 + 14) - \hat{j} (2 - 21) + \hat{k} (- 2 + 21) = 0 \hat{i} + 19 \hat{j} + 19 \hat{k}$

$∴ ∣ a \times b ∣ = (19)^{2} + (19)^{2} = 2 \times (19)^{2} = 192$

Find a unit vector perpendicular to each of the vectors $a + b$ and $a - b$ , where $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ . Sol. We have, $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$

$∴ a + b = 4 \hat{i} + 4 \hat{j}, a - b = 2 \hat{i} + 4 \hat{k} (a + b) \times (a - b) = \hat{i} 42 \hat{j} 40 \hat{k} 04 = \hat{i} (16) - \hat{j} (16) + \hat{k} (- 8) = 16 \hat{i} - 16 \hat{j} - 8 \hat{k} ∴ ∣ (a + b) \times (a - b) ∣ = 1 6^{2} + (- 16)^{2} + (- 8)^{2} = 8 2^{2} + 2^{2} + 1 = 89 = 8 \times 3 = 24$

Hence, the unit vector perpendicular to each of the vectors $a + b$ and $a - b$ is given by the relation,

$\pm \frac{( a + b ) \times ( a - b )}{∣ ( a + b ) \times ( a - b ) ∣} = \pm \frac{( 16 i ^ - 16 j ^ - 8 k ^ )}{24} = \pm \frac{( 2 i ^ - 2 j ^ - k ^ )}{3}$

Required vector is $\frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3} \hat{k}$ or

$- \frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}$

If a unit vector a makes an angle $\frac{π}{3}$ with $\hat{i}, \frac{π}{4}$ with $\hat{j}$ and an acute angle $θ$ with $\hat{k}$ , then find $θ$ and hence, the components of $a$ . Sol. Let unit vector $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ Since $a$ is a unit vector, $∣ a ∣ = 1$ . Also, it is given that $a$ makes angle $\frac{π}{3}$ with $\hat{i}$ , $\frac{π}{4}$ with $\hat{j}$ and an acute angle $θ$ with $\hat{k}$ . Then, we have:

$\Rightarrow cos \frac{π}{3} = \frac{a _{1}}{∣ a ∣} \frac{1}{2} = a_{1} [∣ a ∣ = 1]$

Also, $cos \frac{π}{4} = \frac{a _{2}}{∣ a ∣}$

$\Rightarrow \frac{1}{2} = a_{2} [∣ a ∣ = 1]$

Now, $∣ a ∣ = 1$

$\Rightarrow a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 1 \Rightarrow (\frac{1}{2})^{2} + (\frac{1}{2})^{2} + cos^{2} θ = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + cos^{2} θ = 1 \Rightarrow \frac{3}{4} + cos^{2} θ = 1 \Rightarrow cos^{2} θ = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3} ∴ a_{3} = cos \frac{π}{3} = \frac{1}{2} Hence, θ = \frac{π}{3} and the components of a are \frac{1}{2}, \frac{1}{2}, \frac{1}{2} .$

Show that $(a - b) \times (a + b) = 2 (a \times b)$ Sol. L.H.S. $(a - b) \times (a + b)$

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

[By distributivity of vector product over addition]

$= a \times a - b \times a + a \times b - b \times b$

[Again, by distributivity of vector product over addition]

$= 0 + a \times b + a \times b - 0 = 2 (a \times b) R.H.S.$

Find $λ$ and $μ$ if

$(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 .$

Sol. $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0$

$\Rightarrow \hat{i} 21 \hat{j} 6 λ \hat{k} 27 μ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$

$\Rightarrow \hat{i} (6 μ - 27 λ) - \hat{j} (2 μ - 27) + \hat{k} (2 λ - 6) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$

On comparing the corresponding components, we have :

$6 μ - 27 λ = 0 2 μ - 27 = 0 2 λ - 6 = 0$

Now, from equation (3), $2 λ - 6 = 0 \Rightarrow λ = 3$ from equation (2), $2 μ - 27 = 0 \Rightarrow μ = \frac{27}{2}$ Here, values of $λ$ and $μ$ satisfy to equation (1) Hence, $λ = 3$ and $μ = \frac{27}{2}$

Given that $a \cdot b = 0$ and $a \times b = 0$ . What can you conclude about the vectors $a$ and $b$ ? Sol. $a . b = 0$ Either $∣ a ∣ = 0$ or $∣ b ∣ = 0$ or $a ⊥ b$ $a \times b = 0$ Then, either $∣ a ∣ = 0$ or $∣ b ∣ = 0$ or $a ∥ b$ But, $a$ and $b$ cannot be perpendicular and parallel simultaneously. Hence, $∣ a ∣ = 0$ or $∣ b ∣ = 0$ . i.e. either $a = 0$ or $b = 0$
Let the vectors $a, b, c$ are given as $a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ respectively. Then show that $a \times (b + c) = a \times b + a \times c$ . Sol. We have

$a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} (b + c) = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k}$

$a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3}$ $= \hat{i} [a_{2} b_{3} - a_{3} b_{2}] + \hat{j} [a_{3} b_{1} - a_{1} b_{3}] + \hat{k} [a_{1} b_{2} - a_{2} b_{1}]$ $a \times c = \hat{i} a_{1} c_{1} \hat{j} a_{2} c_{2} \hat{k} a_{3} c_{3}$ $= \hat{i} [a_{2} c_{3} - a_{3} c_{2}] + \hat{j} [a_{3} c_{1} - a_{1} c_{3}] + \hat{k} [a_{1} c_{2} - a_{2} c_{1}]$ On adding (2) and (3), we get :

$(a \times b) + (a \times c) + + = \hat{i} [a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}] \hat{j} [b_{1} a_{3} + a_{3} c_{1} - a_{1} b_{3} - a_{1} c_{3}] \hat{k} [a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}]$

Now, from (1) and (4), we have: $a \times (b + c) = a \times b + a \times c$

If either $a = 0$ or $b = 0$ , then $a \times b = 0$ . Is the converse true? Justify your answer with an example. Sol. Take any parallel non-zero vectors so that $a \times b = 0$ . Let $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, b = 4 \hat{i} + 6 \hat{j} + 8 \hat{k}$ . Then, $a \times b = \hat{i} 24 \hat{j} 36 \hat{k} 48$ $= \hat{i} (24 - 24) - \hat{j} (16 - 16) + \hat{k} (12 - 12)$ $= 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = 0$ It can now be observed that : $∣ a ∣ = 2^{2} + 3^{2} + 4^{2} = 29 ∴ a \neq = 0$ $∣ b ∣ = 4^{2} + 6^{2} + 8^{2} = 116 ∴ b \neq = 0$ Hence, the converse of the given statement need not be true.
Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5). Sol. The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and C (1, 5, 5). Position vector of points $A, B$ and C are

$a = \hat{i} + \hat{j} + 2 \hat{k}, b = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

and $c = \hat{i} + 5 \hat{j} + 5 \hat{k}$ respectively. The adjacent sides $AB$ and $BC$ of $△ ABC$ are given as: $AB = b - a$ $\Rightarrow AB = (2 - 1) \hat{i} + (3 - 1) \hat{j} + (5 - 2) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $BC = c - b$ $\Rightarrow BC = (1 - 2) \hat{i} + (5 - 3) \hat{j} + (5 - 5) \hat{k} = - \hat{i} + 2 \hat{j}$ $AB \times BC = \hat{i} 1 - 1 \hat{j} 22 \hat{k} 30$ $= \hat{i} (- 6) - \hat{j} (3) + \hat{k} (2 + 2) = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k}$

∴

$∣ AB \times BC ∣ = (- 6)^{2} + (- 3)^{2} + 4^{2} = 36 + 9 + 16 = 61$

Area of $Δ ABC = \frac{1}{2} ∣ AB \times BC ∣ = \frac{1}{2} 61$ Hence, the area of $△ ABC$ is $\frac{61}{2}$ square units.

Find the area of the parallelogram whose adjacent sides are determined by the vector $a = \hat{i} - \hat{j} + 3 \hat{k}$ and $b = 2 \hat{i} - 7 \hat{j} + \hat{k}$ . Sol. The area of the parallelogram whose adjacent sides are $a$ and $b$ is $∣ a \times b ∣$ . Adjacent sides are given as: $a = \hat{i} - \hat{j} + 3 \hat{k}$ and $b = 2 \hat{i} - 7 \hat{j} + \hat{k}$ $∴ a \times b = \hat{i} 12 \hat{j} - 1 - 7 \hat{k} 31$

$= \hat{i} (- 1 + 21) - \hat{j} (1 - 6) + \hat{k} (- 7 + 2) = 20 \hat{i} + 5 \hat{j} - 5 \hat{k}$

$\Rightarrow ∣ a \times b ∣ = 2 0^{2} + 5^{2} + 5^{2}$

$= 400 + 25 + 25 = 152$

Hence, the area of the given parallelogram is $152$ square units.

Choose the correct answer in the following Q. 11 and 12

Let the vectors $a$ and $b$ be such that $∣ a ∣ = 3$ and $∣ b ∣ = \frac{2}{3}$ , then $a \times b$ is a unit vector, if the angle between $a$ and $b$ is (A) $\frac{π}{6}$ (B) $\frac{π}{4}$ (C) $\frac{π}{3}$ (D) $\frac{π}{2}$ Sol. (B) It is given that $∣ a ∣ = 3$ and $∣ b ∣ = \frac{2}{3}$ . We know that $a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}$ , where $\overset{n}{^}$ is a unit vector perpendicular to both $a$ and $b$ and $θ$ is the angle between $a$ and $b$ .Now, $a \times b$ is a unit vector if $∣ a \times b ∣ = 1$

$∣ a \times b ∣ = 1 \Rightarrow ∣∣ a ∣∣ b ∣ sin θ ∣ = 1$

$\Rightarrow 3 \times \frac{2}{3} \times sin θ = 1 \Rightarrow sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{4}$ Hence, $a \times b$ is a unit vector if the angle between $a$ and $b$ is $\frac{π}{4}$ . The correct answer is B.

12. Area of a rectangle having vertices A, B, C, and D with position vectors

$- \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} respectively is:$

(A) $\frac{1}{2}$ (B) 1 (C) 2 (D) 4 Sol. (C), The position vectors of vertices A, B, C, and $D$ of rectangle $A BC D$ are given as:

$OA = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k} OB = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k} OC = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} OD = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k}$

The adjacent sides $AB$ and $BC$ of the given rectangle are given as:

$AB = OB - OA = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i}$

$BC = OC - OB = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = - \hat{j}$

$∴ AB \times BC = \hat{i} 20 \hat{j} 0 - 1 \hat{k} 00 = \hat{k} (- 2) = - 2 \hat{k}$ $\Rightarrow ∣ AB \times BC ∣ = (- 2)^{2} = 2$

Now, it is known that the area of a rectangle whose adjacent sides are $a$ and $b$ is $∣ a \times b ∣$ Hence, the area of the given rectangle is $∣ AB \times BC ∣ = 2$ square units. The correct answer is C.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Maths chapter-wise, with exercise answers, key formulas, and step-by-step solutions to build understanding and improve question-solving practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 10 Vector Algebra Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 10.1	5 Questions & Solutions	Direction cosines of vectors and finding vectors from given direction cosines
Exercise 10.2	19 Questions & Solutions	Magnitude and direction of vectors, vector addition, subtraction and scalar multiplication
Exercise 10.3	18 Questions & Solutions	Scalar product and vector product of two vectors
Exercise 10.4	12 Questions & Solutions	Scalar triple product and vector triple product

5.0Key Features and Benefits of Class 12 Maths Chapter 10 Vector Algebra

NCERT-Based Exercises: The solutions follow the NCERT syllabus and cover question types relevant to CBSE board exam preparation.
Step-by-Step Working: Detailed steps make it easier to apply vector laws, properties, and identities correctly while solving questions.
Better Concept Understanding: The solutions help students understand how vectors are used in physics and coordinate geometry.
Improved Accuracy: Regular practice can improve spatial understanding and accuracy while solving vector-based problems.
Useful for Competitive Exams: Strong fundamentals in vector algebra can support preparation for Mathematics Olympiads and other competitive examinations.

Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra?

What is the difference between scalar and vector quantities in NCERT Class 12 Maths Chapter 10?

How do NCERT Solutions for Class 12 Maths Chapter 10 explain vector addition and subtraction?

Which exercises cover dot product and cross product in NCERT Class 12 Maths Chapter 10?

Why are scalar and vector triple products included in NCERT Solutions for Class 12 Maths Chapter 10?

How can NCERT Solutions for Class 12 Maths Chapter 10 help me prepare for exams?

Join ALLEN!

Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra?

What is the difference between scalar and vector quantities in NCERT Class 12 Maths Chapter 10?

How do NCERT Solutions for Class 12 Maths Chapter 10 explain vector addition and subtraction?

Which exercises cover dot product and cross product in NCERT Class 12 Maths Chapter 10?

Why are scalar and vector triple products included in NCERT Solutions for Class 12 Maths Chapter 10?

How can NCERT Solutions for Class 12 Maths Chapter 10 help me prepare for exams?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 10 Vector Algebra

1.0Key Concepts of Class 12 Maths Chapter 10 Vector Algebra

2.0NCERT Class 12 Maths Chapter 10 Vector Algebra : Detailed Solutions

EXERCISE - 10.1

EXERCISE 10.2

EXERCISE 10.3

EXERCISE 10.4

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 10 Vector Algebra Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 10 Vector Algebra

Table of Contents