NCERT Solutions for Class 12 Maths Chapter 10: Vector Algebra (NCERT 2026–27)

These Class 12 Maths Chapter 10 solutions cover Vector Algebra from the NCERT textbook (Reprint 2026–27). Every question of Exercise 10.1, 10.2, 10.3, 10.4 and the Miscellaneous Exercise on Chapter 10 is reproduced exactly as in the book and solved step by step — scalars and vectors, magnitude and unit vectors, the section formula, the scalar (dot) product and the vector (cross) product — with every answer cross-checked against the NCERT answer key.

Class: 12 Subject: Mathematics Chapter: 10 – Vector Algebra Exercises: 10.1, 10.2, 10.3, 10.4 + Miscellaneous Book: Mathematics Part II Session: 2026–27
On this page

Chapter 10 Overview

Chapter 10, Vector Algebra, introduces quantities that have both magnitude and direction. It begins with basic concepts — position vectors, direction cosines and direction ratios — and the different types of vectors (zero, unit, collinear, equal and negative vectors). It then develops the addition of vectors (triangle and parallelogram laws), multiplication of a vector by a scalar, the component form xî + yĵ + z&kcirc;, the vector joining two points and the section formula. The second half studies the two products of vectors: the scalar (dot) product, used to find angles, projections and to test perpendicularity, and the vector (cross) product, used to find areas of triangles and parallelograms and a vector perpendicular to two given vectors. The Class 12 Maths Chapter 10 solutions below work through every exercise and the Miscellaneous Exercise question by question.

Key Concepts & Definitions

Vector: a quantity that has both magnitude and direction (e.g. displacement, velocity, force). A scalar has only magnitude (e.g. time, mass, speed).

Position vector: for a point P(x, y, z), the vector OP = xî + yĵ + z&kcirc;, with magnitude |OP| = √(x2 + y2 + z2).

Direction cosines (l, m, n): cosines of the angles a vector makes with the x, y, z axes; l2 + m2 + n2 = 1. The components a, b, c are the direction ratios.

Unit vector: a vector of magnitude 1; the unit vector in the direction of a is â = a/|a|.

Collinear / equal vectors: vectors parallel to the same line are collinear; equal vectors have the same magnitude and direction.

Scalar (dot) product: a·b = |a||b| cosθ, a scalar; zero for perpendicular vectors.

Vector (cross) product: a×b = |a||b| sinθ n̂, a vector perpendicular to both; zero for parallel vectors.

Important Formulas (Chapter 10)

Magnitude: for a = a1î + a2ĵ + a3&kcirc;, |a| = √(a12 + a22 + a32).

Vector joining two points: PQ = (x2 − x1)î + (y2 − y1)ĵ + (z2 − z1)&kcirc;.

Section formula (internal, ratio m : n): r = (mb + na)/(m + n); (external) r = (mb − na)/(m − n); midpoint r = (a + b)/2.

Dot product: a·b = a1b1 + a2b2 + a3b3; angle cosθ = (a·b)/(|a||b|).

Projection of a on b = (a·b)/|b|.

Cross product: a×b = (a2b3 − a3b2)î − (a1b3 − a3b1)ĵ + (a1b2 − a2b1)&kcirc;.

Area: triangle = ½|a×b|; parallelogram (adjacent sides a, b) = |a×b|.

Exercise 10.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

1. Represent graphically a displacement of 40 km, 30° east of north.

SOLUTION Draw a vector OP starting from the origin O. Take ON pointing due north (upward). Rotate 30° from ON towards the east (to the right) and draw OP along this direction. Choose a scale, say 1 cm = 10 km, so OP is 4 cm long. Then the directed segment OP of length 40 km making 30° with north (towards east) represents the required displacement.

2. Classify the following measures as scalars and vectors. (i) 10 kg   (ii) 2 meters north-west   (iii) 40° (iv) 40 watt   (v) 10–19 coulomb   (vi) 20 m/s2

SOLUTION (i) 10 kg → mass → scalar. (ii) 2 meters north-west → has a direction → vector. (iii) 40° → angle → scalar. (iv) 40 watt → power → scalar. (v) 10–19 coulomb → electric charge → scalar. (vi) 20 m/s2 → acceleration (has direction) → vector.

3. Classify the following as scalar and vector quantities. (i) time period   (ii) distance   (iii) force (iv) velocity   (v) work done

SOLUTION (i) time period → scalar. (ii) distance → scalar. (iii) force → vector. (iv) velocity → vector. (v) work done → scalar.

4. In Fig 10.6 (a square), identify the following vectors. (i) Coinitial   (ii) Equal   (iii) Collinear but not equal

SOLUTION (In the square of Fig 10.6 with the four labelled vectors a, b, c, d.) (i) Coinitial vectors are those with the same initial point: a and d are coinitial. (ii) Equal vectors have the same magnitude and direction: b and d are equal. (iii) Collinear but not equal: a and c are parallel to the same line but point in opposite directions, so they are collinear but not equal.

5. 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.

SOLUTION (i) True. a and –a are parallel to the same line, hence collinear. (ii) False. Collinear vectors are parallel but may have different magnitudes. (iii) False. Equal magnitudes do not force them to be parallel; their directions may differ. (iv) False. They could point in opposite directions (e.g. a and –a), so they need not be equal.

Exercise 10.2 Solutions

1. Compute the magnitude of the following vectors: a = î + ĵ + &kcirc;;   b = 2î − 7ĵ − 3&kcirc;;   c = (1/√3)î + (1/√3)ĵ − (1/√3)&kcirc;

SOLUTION |a| = √(12 + 12 + 12) = √3. |b| = √(22 + (−7)2 + (−3)2) = √(4 + 49 + 9) = √62. |c| = √(1/3 + 1/3 + 1/3) = √1 = 1. So all three magnitudes are √3, √62 and 1 respectively.

2. Write two different vectors having same magnitude.

SOLUTION Take a = î + 2ĵ + 3&kcirc; and b = 3î + 2ĵ + &kcirc;. |a| = √(1 + 4 + 9) = √14 and |b| = √(9 + 4 + 1) = √14. They have the same magnitude √14 but different directions (an infinite number of such pairs exist).

3. Write two different vectors having same direction.

SOLUTION Take a = î + ĵ + &kcirc; and b = 2î + 2ĵ + 2&kcirc; = 2a. Both have the same unit vector (1/√3)(î + ĵ + &kcirc;), so they point in the same direction but have different magnitudes (an infinite number of such pairs exist).

4. Find the values of x and y so that the vectors 2î + 3ĵ and xî + yĵ are equal.

SOLUTION Two vectors are equal iff their corresponding components are equal. Comparing: x = 2 and y = 3. ∴ x = 2, y = 3.

5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

SOLUTION Vector = (terminal) − (initial) = (−5 − 2)î + (7 − 1)ĵ = −7î + 6ĵ. Scalar components: −7 and 6. Vector components: −7î and 6ĵ.

6. Find the sum of the vectors a = î − 2ĵ + &kcirc;, b = −2î + 4ĵ + 5&kcirc; and c = î − 6ĵ − 7&kcirc;.

SOLUTION Add the components: (1 − 2 + 1)î + (−2 + 4 − 6)ĵ + (1 + 5 − 7)&kcirc;. = 0î − 4ĵ − &kcirc; = −4ĵ − &kcirc;.

7. Find the unit vector in the direction of the vector a = î + ĵ + 2&kcirc;.

SOLUTION |a| = √(1 + 1 + 4) = √6. â = a/|a| = (1/√6)î + (1/√6)ĵ + (2/√6)&kcirc;.

8. 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.

SOLUTION PQ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)&kcirc; = 3î + 3ĵ + 3&kcirc;. |PQ| = √(9 + 9 + 9) = √27 = 3√3. Unit vector = (1/3√3)(3î + 3ĵ + 3&kcirc;) = (1/√3)î + (1/√3)ĵ + (1/√3)&kcirc;.

9. For given vectors, a = 2î − ĵ + 2&kcirc; and b = −î + ĵ − &kcirc;, find the unit vector in the direction of the vector a + b.

SOLUTION a + b = (2 − 1)î + (−1 + 1)ĵ + (2 − 1)&kcirc; = î + 0ĵ + &kcirc; = î + &kcirc;. |a + b| = √(1 + 0 + 1) = √2. Unit vector = (1/√2)î + (1/√2)&kcirc;.

10. Find a vector in the direction of vector 5î − ĵ + 2&kcirc; which has magnitude 8 units.

SOLUTION |5î − ĵ + 2&kcirc;| = √(25 + 1 + 4) = √30. Required vector = 8 × unit vector = (8/√30)(5î − ĵ + 2&kcirc;). = (40/√30)î − (8/√30)ĵ + (16/√30)&kcirc;.

11. Show that the vectors 2î − 3ĵ + 4&kcirc; and −4î + 6ĵ − 8&kcirc; are collinear.

SOLUTION Let a = 2î − 3ĵ + 4&kcirc; and b = −4î + 6ĵ − 8&kcirc;. b = −2(2î − 3ĵ + 4&kcirc;) = −2a. Since b is a scalar multiple of a (λ = −2), the two vectors are collinear.

12. Find the direction cosines of the vector î + 2ĵ + 3&kcirc;.

SOLUTION Magnitude = √(1 + 4 + 9) = √14. Direction cosines = (1/√14, 2/√14, 3/√14).

13. Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

SOLUTION AB = (−1 − 1)î + (−2 − 2)ĵ + (1 − (−3))&kcirc; = −2î − 4ĵ + 4&kcirc;. |AB| = √(4 + 16 + 16) = √36 = 6. Direction cosines = (−2/6, −4/6, 4/6) = (−1/3, −2/3, 2/3).

14. Show that the vector î + ĵ + &kcirc; is equally inclined to the axes OX, OY and OZ.

SOLUTION |î + ĵ + &kcirc;| = √3. Direction cosines = (1/√3, 1/√3, 1/√3), so cosα = cosβ = cosγ = 1/√3. Equal direction cosines mean α = β = γ (= cos−1(1/√3)), so the vector is equally inclined to all three axes.

15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are î + 2ĵ − &kcirc; and –î + ĵ + &kcirc; respectively, in the ratio 2 : 1 (i) internally   (ii) externally

SOLUTION Here a = î + 2ĵ − &kcirc; (for P) and b = −î + ĵ + &kcirc; (for Q); m : n = 2 : 1. (i) Internal: r = (mb + na)/(m + n) = [2(−î + ĵ + &kcirc;) + 1(î + 2ĵ − &kcirc;)]/3. = [(−2 + 1)î + (2 + 2)ĵ + (2 − 1)&kcirc;]/3 = (−î + 4ĵ + &kcirc;)/3 = (−1/3)î + (4/3)ĵ + (1/3)&kcirc;. (ii) External: r = (mb − na)/(m − n) = [2(−î + ĵ + &kcirc;) − (î + 2ĵ − &kcirc;)]/1. = (−2 − 1)î + (2 − 2)ĵ + (2 + 1)&kcirc; = −3î + 3&kcirc;.

16. Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

SOLUTION Midpoint position vector = (OP + OQ)/2. = [(2 + 4)î + (3 + 1)ĵ + (4 − 2)&kcirc;]/2 = (6î + 4ĵ + 2&kcirc;)/2. = 3î + 2ĵ + &kcirc;.

17. Show that the points A, B and C with position vectors a = 3î − 4ĵ − 4&kcirc;, b = 2î − ĵ + &kcirc; and c = î − 3ĵ − 5&kcirc;, respectively form the vertices of a right angled triangle.

SOLUTION AB = b − a = (2 − 3)î + (−1 + 4)ĵ + (1 + 4)&kcirc; = −î + 3ĵ + 5&kcirc;. BC = c − b = (1 − 2)î + (−3 + 1)ĵ + (−5 − 1)&kcirc; = −î − 2ĵ − 6&kcirc;. CA = a − c = (3 − 1)î + (−4 + 3)ĵ + (−4 + 5)&kcirc; = 2î − ĵ + &kcirc;. |AB|2 = 1 + 9 + 25 = 35; |BC|2 = 1 + 4 + 36 = 41; |CA|2 = 4 + 1 + 1 = 6. Since |AB|2 + |CA|2 = 35 + 6 = 41 = |BC|2, the converse of Pythagoras’ theorem holds — so the triangle is right angled (at A).

18. In triangle ABC (Fig 10.18), 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

SOLUTION By the triangle law, AB + BC = AC, i.e. AB + BC + CA = 0 (A is true) and AB + BC − AC = 0 (B is true). (D) AB − CB + CA = AB + BC + CA = 0, which is true. (C) AB + BC − CA = AC − CA = AC + AC = 2AC ≠ 0, so it is not true. ∴ the correct option is (C).

19. If a and b are two collinear vectors, then which of the following are incorrect: (A) b = λa, for some scalar λ   (B) a = ±b (C) the respective components of a and b are not proportional (D) both the vectors a and b have same direction, but different magnitudes.

SOLUTION Collinear vectors satisfy b = λa for some scalar λ, so (A) is correct. (B) a = ±b only holds when magnitudes are equal — not in general → incorrect. (C) For collinear vectors the components are proportional, so saying they are “not proportional” is incorrect. (D) Collinear vectors may have opposite directions, so claiming the same direction is incorrect. ∴ the incorrect statements are (B), (C) and (D).

Exercise 10.3 Solutions

1. Find the angle between two vectors a and b with magnitudes √3 and 2, respectively having a·b = √6.

SOLUTION cosθ = (a·b)/(|a||b|) = √6/(√3 × 2) = √6/(2√3) = √2/2 = 1/√2. ∴ θ = π/4.

2. Find the angle between the vectors î − 2ĵ + 3&kcirc; and 3î − 2ĵ + &kcirc;.

SOLUTION a·b = (1)(3) + (−2)(−2) + (3)(1) = 3 + 4 + 3 = 10. |a| = √(1 + 4 + 9) = √14; |b| = √(9 + 4 + 1) = √14. cosθ = 10/(√14 × √14) = 10/14 = 5/7. ∴ θ = cos−1(5/7).

3. Find the projection of the vector î − ĵ on the vector î + ĵ.

SOLUTION Projection of a on b = (a·b)/|b|. a·b = (1)(1) + (−1)(1) = 0; |b| = √2. Projection = 0/√2 = 0.

4. Find the projection of the vector î + 3ĵ + 7&kcirc; on the vector 7î − ĵ + 8&kcirc;.

SOLUTION a·b = (1)(7) + (3)(−1) + (7)(8) = 7 − 3 + 56 = 60. |b| = √(49 + 1 + 64) = √114. Projection = 60/√114.

5. Show that each of the given three vectors is a unit vector: (1/7)(2î + 3ĵ + 6&kcirc;),   (1/7)(3î − 6ĵ + 2&kcirc;),   (1/7)(6î + 2ĵ − 3&kcirc;) Also, show that they are mutually perpendicular to each other.

SOLUTION Let a = (1/7)(2î + 3ĵ + 6&kcirc;), b = (1/7)(3î − 6ĵ + 2&kcirc;), c = (1/7)(6î + 2ĵ − 3&kcirc;). |a| = (1/7)√(4 + 9 + 36) = (1/7)√49 = 1. Similarly |b| = (1/7)√(9 + 36 + 4) = 1 and |c| = (1/7)√(36 + 4 + 9) = 1. So all are unit vectors. a·b = (1/49)[(2)(3) + (3)(−6) + (6)(2)] = (1/49)(6 − 18 + 12) = 0. b·c = (1/49)[(3)(6) + (−6)(2) + (2)(−3)] = (1/49)(18 − 12 − 6) = 0. c·a = (1/49)[(6)(2) + (2)(3) + (−3)(6)] = (1/49)(12 + 6 − 18) = 0. Each pair has zero dot product, so the three vectors are mutually perpendicular.

6. Find |a| and |b|, if (a + b)·(a − b) = 8 and |a| = 8|b|.

SOLUTION (a + b)·(a − b) = |a|2 − |b|2 = 8. With |a| = 8|b|: (8|b|)2 − |b|2 = 8 ⇒ 64|b|2 − |b|2 = 63|b|2 = 8. |b|2 = 8/63 ⇒ |b| = √(8/63) = 2√2/(3√7) = 16/(3√7)? Let us keep it tidy: |b| = (2√2)/(3√7). ∴ |b| = (2√2)/(3√7) and |a| = 8|b| = (16√2)/(3√7).

7. Evaluate the product (3a − 5b)·(2a + 7b).

SOLUTION Expand using distributivity: (3a − 5b)·(2a + 7b) = 6(a·a) + 21(a·b) − 10(b·a) − 35(b·b). Since a·a = |a|2, b·b = |b|2 and a·b = b·a: = 6|a|2 + (21 − 10)(a·b) − 35|b|2 = 6|a|2 + 11(a·b) − 35|b|2.

8. 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 ½.

SOLUTION Let |a| = |b| = t. Then a·b = |a||b| cos60° = t2(1/2) = ½. So t2 = 1 ⇒ t = 1. ∴ |a| = |b| = 1.

9. Find |x|, if for a unit vector a, (x − a)·(x + a) = 12.

SOLUTION (x − a)·(x + a) = |x|2 − |a|2 = 12. Since a is a unit vector, |a| = 1, so |x|2 − 1 = 12 ⇒ |x|2 = 13. ∴ |x| = √13.

10. If a = 2î + 2ĵ + 3&kcirc;, b = −î + 2ĵ + &kcirc; and c = 3î + ĵ are such that a + λb is perpendicular to c, then find the value of λ.

SOLUTION a + λb = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)&kcirc;. Perpendicular to c = 3î + ĵ means (a + λb)·c = 0. 3(2 − λ) + 1(2 + 2λ) + 0(3 + λ) = 6 − 3λ + 2 + 2λ = 8 − λ = 0. ∴ λ = 8.

11. Show that |a|b + |b|a is perpendicular to |a|b − |b|a, for any two nonzero vectors a and b.

SOLUTION Compute the dot product (|a|b + |b|a)·(|a|b − |b|a). = |a|2(b·b) − |a||b|(b·a) + |a||b|(a·b) − |b|2(a·a). = |a|2|b|2 − |b|2|a|2 = 0 (the middle terms cancel since a·b = b·a). A zero dot product proves the two vectors are perpendicular.

12. If a·a = 0 and a·b = 0, then what can be concluded about the vector b?

SOLUTION a·a = |a|2 = 0 ⇒ |a| = 0, so a is the zero vector. Then a·b = 0 is automatically satisfied for every b. ∴ b can be any vector.

13. If a, b, c are unit vectors such that a + b + c = 0, find the value of a·b + b·c + c·a.

SOLUTION Square the relation: |a + b + c|2 = 0. |a|2 + |b|2 + |c|2 + 2(a·b + b·c + c·a) = 0. Since each is a unit vector, 1 + 1 + 1 + 2(a·b + b·c + c·a) = 0. ⇒ 2(…) = −3 ⇒ a·b + b·c + c·a = −3/2.

14. If either vector a = 0 or b = 0, then a·b = 0. But the converse need not be true. Justify your answer with an example.

SOLUTION Take any two non-zero, perpendicular vectors, e.g. a = î and b = ĵ. a·b = (1)(0) + (0)(1) + 0 = 0, yet neither a nor b is the zero vector. So a·b = 0 does not imply a = 0 or b = 0; the converse is not true.

15. 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].

SOLUTION BA = A − B = (1 + 1)î + (2 − 0)ĵ + (3 − 0)&kcirc; = 2î + 2ĵ + 3&kcirc;. BC = C − B = (0 + 1)î + (1 − 0)ĵ + (2 − 0)&kcirc; = î + ĵ + 2&kcirc;. BA·BC = 2 + 2 + 6 = 10; |BA| = √(4 + 4 + 9) = √17; |BC| = √(1 + 1 + 4) = √6. cos(∠ABC) = 10/(√17 · √6) = 10/√102. ∴ ∠ABC = cos−1(10/√102).

16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

SOLUTION AB = (2 − 1)î + (6 − 2)ĵ + (3 − 7)&kcirc; = î + 4ĵ − 4&kcirc;. BC = (3 − 2)î + (10 − 6)ĵ + (−1 − 3)&kcirc; = î + 4ĵ − 4&kcirc;. Since AB = BC (same vector) and they share point B, A, B, C lie on one line, i.e. they are collinear.

17. Show that the vectors 2î − ĵ + &kcirc;, î − 3ĵ − 5&kcirc; and 3î − 4ĵ − 4&kcirc; form the vertices of a right angled triangle.

SOLUTION Let A = 2î − ĵ + &kcirc;, B = î − 3ĵ − 5&kcirc;, C = 3î − 4ĵ − 4&kcirc; be the position vectors. AB = B − A = −î − 2ĵ − 6&kcirc;; BC = C − B = 2î − ĵ + &kcirc;; CA = A − C = −î + 3ĵ + 5&kcirc;. AB·BC = (−1)(2) + (−2)(−1) + (−6)(1) = −2 + 2 − 6 = −6 (not zero), but check the squares: |AB|2 = 1 + 4 + 36 = 41; |BC|2 = 4 + 1 + 1 = 6; |CA|2 = 1 + 9 + 25 = 35. |BC|2 + |CA|2 = 6 + 35 = 41 = |AB|2. By the converse of Pythagoras, the triangle is right angled (at C).

18. If a is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λa is unit vector if (A) λ = 1   (B) λ = –1   (C) a = |λ|   (D) a = 1/|λ|

SOLUTION λa is a unit vector means |λa| = |λ| |a| = |λ| a = 1. So a = 1/|λ|. ∴ the correct option is (D).

Exercise 10.4 Solutions

1. Find |a × b|, if a = î − 7ĵ + 7&kcirc; and b = 3î − 2ĵ + 2&kcirc;.

SOLUTION a × b = |î ĵ &kcirc;; 1 −7 7; 3 −2 2|. î-term: (−7)(2) − (7)(−2) = −14 + 14 = 0. ĵ-term: −[(1)(2) − (7)(3)] = −(2 − 21) = 19. &kcirc;-term: (1)(−2) − (−7)(3) = −2 + 21 = 19. So a × b = 0î + 19ĵ + 19&kcirc;, |a × b| = √(0 + 361 + 361) = √722 = 19√2.

2. Find a unit vector perpendicular to each of the vector a + b and a − b, where a = 3î + 2ĵ + 2&kcirc; and b = î + 2ĵ − 2&kcirc;.

SOLUTION a + b = 4î + 4ĵ + 0&kcirc;; a − b = 2î + 0ĵ + 4&kcirc;. (a + b) × (a − b) = |î ĵ &kcirc;; 4 4 0; 2 0 4|. î: (4)(4) − (0)(0) = 16; ĵ: −[(4)(4) − (0)(2)] = −16; &kcirc;: (4)(0) − (4)(2) = −8. = 16î − 16ĵ − 8&kcirc;; magnitude = √(256 + 256 + 64) = √576 = 24. Unit vector = (1/24)(16î − 16ĵ − 8&kcirc;) = ±[(2/3)î − (2/3)ĵ − (1/3)&kcirc;].

3. If a unit vector a makes angles π/3 with î, π/4 with ĵ and an acute angle θ with &kcirc;, then find θ and hence, the components of a.

SOLUTION For a unit vector, l2 + m2 + n2 = 1, where l = cos(π/3) = 1/2, m = cos(π/4) = 1/√2, n = cosθ. (1/2)2 + (1/√2)2 + cos2θ = 1 ⇒ 1/4 + 1/2 + cos2θ = 1 ⇒ cos2θ = 1/4. θ acute ⇒ cosθ = 1/2 ⇒ θ = π/3. Components of a = (l, m, n) = (1/2, 1/√2, 1/2).

4. Show that (a − b) × (a + b) = 2(a × b).

SOLUTION Expand using distributivity of the cross product: (a − b) × (a + b) = a×a + a×b − b×a − b×b. a×a = 0, b×b = 0 and −b×a = +a×b. = a×b + a×b = 2(a × b). Hence proved.

5. Find λ and μ if (2î + 6ĵ + 27&kcirc;) × (î + λĵ + μ&kcirc;) = 0.

SOLUTION The cross product is zero iff the two vectors are parallel, i.e. their components are proportional. 2/1 = 6/λ = 27/μ. From 2/1 = 6/λ: λ = 6/2 = 3. From 2/1 = 27/μ: μ = 27/2. ∴ λ = 3, μ = 27/2.

6. Given that a·b = 0 and a × b = 0. What can you conclude about the vectors a and b?

SOLUTION a·b = 0 ⇒ either a vector is zero, or a ⊥ b. a × b = 0 ⇒ either a vector is zero, or a ∥ b. A non-zero pair cannot be both perpendicular and parallel, so both conditions hold only if at least one of a, b is the zero vector (i.e. either a = 0 or b = 0).

7. Let the vectors a, b, c be given as a1î + a2ĵ + a3&kcirc;, b1î + b2ĵ + b3&kcirc;, c1î + c2ĵ + c3&kcirc;. Then show that a × (b + c) = a × b + a × c.

SOLUTION b + c = (b1 + c1)î + (b2 + c2)ĵ + (b3 + c3)&kcirc;. The î-component of a × (b + c) = a2(b3 + c3) − a3(b2 + c2) = (a2b3 − a3b2) + (a2c3 − a3c2). This is exactly the î-component of a × b plus that of a × c. The ĵ- and &kcirc;-components split the same way. Hence a × (b + c) = a × b + a × c. Proved.

8. If either a = 0 or b = 0, then a × b = 0. Is the converse true? Justify your answer with an example.

SOLUTION The converse is not true. Take any two non-zero collinear (parallel) vectors. e.g. a = 2î + 3ĵ + 4&kcirc; and b = 4î + 6ĵ + 8&kcirc; = 2a. Then a × b = 0 (parallel vectors), yet neither a nor b is the zero vector. So a × b = 0 does not imply a = 0 or b = 0.

9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

SOLUTION AB = (1, 2, 3); AC = (0, 4, 3). AB × AC = |î ĵ &kcirc;; 1 2 3; 0 4 3| = (2·3 − 3·4)î − (1·3 − 3·0)ĵ + (1·4 − 2·0)&kcirc;. = (6 − 12)î − (3)ĵ + (4)&kcirc; = −6î − 3ĵ + 4&kcirc;. |AB × AC| = √(36 + 9 + 16) = √61. Area = ½|AB × AC| = (√61)/2 square units.

10. Find the area of the parallelogram whose adjacent sides are determined by the vectors a = î − ĵ + 3&kcirc; and b = 2î − 7ĵ + &kcirc;.

SOLUTION a × b = |î ĵ &kcirc;; 1 −1 3; 2 −7 1|. î: (−1)(1) − (3)(−7) = −1 + 21 = 20. ĵ: −[(1)(1) − (3)(2)] = −(1 − 6) = 5. &kcirc;: (1)(−7) − (−1)(2) = −7 + 2 = −5. a × b = 20î + 5ĵ − 5&kcirc;; |a × b| = √(400 + 25 + 25) = √450 = 15√2 square units.

11. Let the vectors a and b be such that |a| = 3 and |b| = √2/3, then a × b is a unit vector, if the angle between a and b is (A) π/6   (B) π/4   (C) π/3   (D) π/2

SOLUTION |a × b| = |a||b| sinθ = 1 (unit vector). 3 × (√2/3) × sinθ = √2 sinθ = 1 ⇒ sinθ = 1/√2. ⇒ θ = π/4. ∴ the correct option is (B).

12. Area of a rectangle having vertices A, B, C and D with position vectors −î + ½ĵ + 4&kcirc;, î + ½ĵ + 4&kcirc;, î − ½ĵ + 4&kcirc; and −î − ½ĵ + 4&kcirc;, respectively is (A) ½   (B) 1   (C) 2   (D) 4

SOLUTION AB = B − A = (1 − (−1))î + (½ − ½)ĵ + 0&kcirc; = 2î, so length AB = 2. AD = D − A = (−1 − (−1))î + (−½ − ½)ĵ + 0&kcirc; = −ĵ, so length AD = 1. Area of rectangle = AB × AD = 2 × 1 = 2 square units. ∴ the correct option is (C).

Miscellaneous Exercise on Chapter 10 — Solutions

1. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

SOLUTION A unit vector in the XY-plane at angle θ with the x-axis is cosθ î + sinθ ĵ. With θ = 30°: cos30° = √3/2 and sin30° = 1/2. ∴ required vector = (√3/2)î + (1/2)ĵ.

2. Find the scalar components and magnitude of the vector joining the points P(x1, y1, z1) and Q(x2, y2, z2).

SOLUTION PQ = (x2 − x1)î + (y2 − y1)ĵ + (z2 − z1)&kcirc;. Scalar components: (x2 − x1), (y2 − y1), (z2 − z1). Magnitude: |PQ| = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2].

3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

SOLUTION Take east as +î and north as +ĵ. First walk: 4 km west = −4î. Second walk: 3 km, 30° east of north → components (3 sin30°)î + (3 cos30°)ĵ = (3/2)î + (3√3/2)ĵ. Net displacement = (−4 + 3/2)î + (3√3/2)ĵ = (−5/2)î + (3√3/2)ĵ. ∴ displacement = (−5/2)î + (3√3/2)ĵ (matching the NCERT answer).

4. If a = b + c, then is it true that |a| = |b| + |c|? Justify your answer.

SOLUTION No, it is not true in general. By the triangle inequality, |b + c| ≤ |b| + |c|, with equality only when b and c are in the same direction. Example: let b, c and a represent the sides of a triangle. Take b = î, c = ĵ, then a = î + ĵ, |a| = √2. But |b| + |c| = 1 + 1 = 2 ≠ √2, so |a| ≠ |b| + |c|.

5. Find the value of x for which x(î + ĵ + &kcirc;) is a unit vector.

SOLUTION |x(î + ĵ + &kcirc;)| = |x|√(1 + 1 + 1) = |x|√3 = 1. ⇒ |x| = 1/√3. ∴ x = ±1/√3.

6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a = 2î + 3ĵ − &kcirc; and b = î − 2ĵ + &kcirc;.

SOLUTION Resultant c = a + b = (2 + 1)î + (3 − 2)ĵ + (−1 + 1)&kcirc; = 3î + ĵ. |c| = √(9 + 1) = √10. Required vector = 5 × (c/|c|) = (5/√10)(3î + ĵ) = (3√10/2)î + (√10/2)ĵ.

7. If a = î + ĵ + &kcirc;, b = 2î − ĵ + 3&kcirc; and c = î − 2ĵ + &kcirc;, find a unit vector parallel to the vector 2a − b + 3c.

SOLUTION 2a = 2î + 2ĵ + 2&kcirc;; −b = −2î + ĵ − 3&kcirc;; 3c = 3î − 6ĵ + 3&kcirc;. Sum = (2 − 2 + 3)î + (2 + 1 − 6)ĵ + (2 − 3 + 3)&kcirc; = 3î − 3ĵ + 2&kcirc;. Magnitude = √(9 + 9 + 4) = √22. Unit vector = (3/√22)î − (3/√22)ĵ + (2/√22)&kcirc;.

8. Show that the points A(1, –2, –8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

SOLUTION AB = (5 − 1)î + (0 + 2)ĵ + (−2 + 8)&kcirc; = 4î + 2ĵ + 6&kcirc;. BC = (11 − 5)î + (3 − 0)ĵ + (7 + 2)&kcirc; = 6î + 3ĵ + 9&kcirc; = (3/2)(4î + 2ĵ + 6&kcirc;) = (3/2)AB. Since BC is a scalar multiple of AB and they share B, the points are collinear. AB : BC = 1 : (3/2) = 2 : 3, so B divides AC in the ratio 2 : 3.

9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a + b) and (a − 3b) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

SOLUTION Position vector of P = 2a + b, of Q = a − 3b; ratio m : n = 1 : 2 (external). External division: R = (m·Q − n·P)/(m − n) = [1(a − 3b) − 2(2a + b)]/(1 − 2). = [(a − 3b) − (4a + 2b)]/(−1) = (−3a − 5b)/(−1) = 3a + 5b. Midpoint of RQ = (R + Q)/2 = [(3a + 5b) + (a − 3b)]/2 = (4a + 2b)/2 = 2a + b = P. So P is the midpoint of RQ. Proved.

10. The two adjacent sides of a parallelogram are 2î − 4ĵ + 5&kcirc; and î − 2ĵ − 3&kcirc;. Find the unit vector parallel to its diagonal. Also, find its area.

SOLUTION Diagonal = sum of adjacent sides = (2 + 1)î + (−4 − 2)ĵ + (5 − 3)&kcirc; = 3î − 6ĵ + 2&kcirc;. |diagonal| = √(9 + 36 + 4) = √49 = 7. Unit vector = (1/7)(3î − 6ĵ + 2&kcirc;). For area, take a = 2î − 4ĵ + 5&kcirc;, b = î − 2ĵ − 3&kcirc;. a × b = |î ĵ &kcirc;; 2 −4 5; 1 −2 −3|. î: (−4)(−3) − (5)(−2) = 12 + 10 = 22; ĵ: −[(2)(−3) − (5)(1)] = −(−6 − 5) = 11; &kcirc;: (2)(−2) − (−4)(1) = −4 + 4 = 0. a × b = 22î + 11ĵ + 0&kcirc;; |a × b| = √(484 + 121) = √605 = 11√5. ∴ area = 11√5 square units.

11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are ±(1/√3, 1/√3, 1/√3).

SOLUTION Let the equal angle be α, so l = m = n = cosα. Using l2 + m2 + n2 = 1: 3cos2α = 1 ⇒ cosα = ±1/√3. ∴ the direction cosines are ±(1/√3, 1/√3, 1/√3). Proved.

12. Let a = î + 4ĵ + 2&kcirc;, b = 3î − 2ĵ + 7&kcirc; and c = 2î − ĵ + 4&kcirc;. Find a vector d which is perpendicular to both a and b, and c·d = 15.

SOLUTION d perpendicular to both a and b ⇒ d is parallel to a × b. a × b = |î ĵ &kcirc;; 1 4 2; 3 −2 7| = (28 + 4)î − (7 − 6)ĵ + (−2 − 12)&kcirc; = 32î − ĵ − 14&kcirc;. Let d = λ(32î − ĵ − 14&kcirc;). Then c·d = λ[(2)(32) + (−1)(−1) + (4)(−14)] = λ(64 + 1 − 56) = 9λ = 15. ⇒ λ = 5/3, so d = (5/3)(32î − ĵ − 14&kcirc;) = (1/3)(160î − 5ĵ − 70&kcirc;).

13. The scalar product of the vector î + ĵ + &kcirc; with a unit vector along the sum of vectors 2î + 4ĵ − 5&kcirc; and λî + 2ĵ + 3&kcirc; is equal to one. Find the value of λ.

SOLUTION Sum = (2 + λ)î + 6ĵ − 2&kcirc;; magnitude = √[(2 + λ)2 + 36 + 4] = √[(2 + λ)2 + 40]. Scalar product of (î + ĵ + &kcirc;) with the unit vector = [(2 + λ) + 6 − 2]/√[(2 + λ)2 + 40] = (λ + 6)/√[(2 + λ)2 + 40] = 1. ⇒ (λ + 6)2 = (2 + λ)2 + 40 ⇒ λ2 + 12λ + 36 = λ2 + 4λ + 4 + 40. ⇒ 12λ + 36 = 4λ + 44 ⇒ 8λ = 8 ⇒ λ = 1.

14. If a, b, c are mutually perpendicular vectors of equal magnitudes, show that the vector a + b + c is equally inclined to a, b and c.

SOLUTION Let |a| = |b| = |c| = k. Mutually perpendicular ⇒ a·b = b·c = c·a = 0. |a + b + c|2 = |a|2 + |b|2 + |c|2 = 3k2, so |a + b + c| = √3 k. (a + b + c)·a = |a|2 + 0 + 0 = k2; similarly the dot products with b and c equal k2. cos(angle with a) = k2/(√3 k · k) = 1/√3, and the same for b and c. Equal cosines ⇒ equal angles, so a + b + c is equally inclined to a, b, c. Proved.

15. Prove that (a + b)·(a + b) = |a|2 + |b|2, if and only if a, b are perpendicular, given a ≠ 0, b ≠ 0.

SOLUTION (a + b)·(a + b) = a·a + 2(a·b) + b·b = |a|2 + 2(a·b) + |b|2. This equals |a|2 + |b|2 iff 2(a·b) = 0, i.e. a·b = 0. Since a, b are non-zero, a·b = 0 ⇔ a ⊥ b. Proved.

16. If θ is the angle between two vectors a and b, then a·b ≥ 0 only when (A) 0 < θ < π/2   (B) 0 ≤ θ ≤ π/2   (C) 0 < θ < π   (D) 0 ≤ θ ≤ π

SOLUTION a·b = |a||b| cosθ ≥ 0 requires cosθ ≥ 0. cosθ ≥ 0 for 0 ≤ θ ≤ π/2 (including the endpoints, where the product is exactly 0). ∴ the correct option is (B).

17. Let a and b be two unit vectors and θ is the angle between them. Then a + b is a unit vector if (A) θ = π/4   (B) θ = π/3   (C) θ = π/2   (D) θ = 2π/3

SOLUTION |a + b|2 = |a|2 + |b|2 + 2a·b = 1 + 1 + 2cosθ = 2 + 2cosθ. For a unit vector, |a + b|2 = 1 ⇒ 2 + 2cosθ = 1 ⇒ cosθ = −1/2. ⇒ θ = 2π/3. ∴ the correct option is (D).

18. The value of î·(ĵ × &kcirc;) + ĵ·(î × &kcirc;) + &kcirc;·(î × ĵ) is (A) 0   (B) –1   (C) 1   (D) 3

SOLUTION ĵ × &kcirc; = î, so î·(ĵ × &kcirc;) = î·î = 1. î × &kcirc; = −ĵ, so ĵ·(î × &kcirc;) = ĵ·(−ĵ) = −1. î × ĵ = &kcirc;, so &kcirc;·(î × ĵ) = &kcirc;·&kcirc; = 1. Sum = 1 − 1 + 1 = 1. ∴ the correct option is (C).

19. If θ is the angle between any two vectors a and b, then |a·b| = |a × b| when θ is equal to (A) 0   (B) π/4   (C) π/2   (D) π

SOLUTION |a·b| = |a||b||cosθ| and |a × b| = |a||b| sinθ. Equal ⇒ |cosθ| = sinθ ⇒ tanθ = 1 ⇒ θ = π/4. ∴ the correct option is (B).

Common Mistakes to Avoid

Watch out for these

  • Confusing the dot and cross products: the dot product gives a scalar, the cross product gives a vector. Use dot for angles/projections, cross for areas/perpendiculars.
  • Forgetting the minus sign on the ĵ-term when expanding the 3×3 determinant for a × b.
  • Mixing up internal and external section formulae — internal uses (mb + na)/(m + n), external uses (mb − na)/(m − n).
  • Treating “a·b = 0” as proof that a vector is zero — it only means the vectors are perpendicular (or one is zero).
  • For projection, dividing by |a| instead of |b| — projection of a on b divides by |b|.
  • Dropping the ± for unit vectors perpendicular to a plane — there are always two opposite directions.

Practice MCQs & Assertion–Reason

1. The magnitude of the vector 2î − 3ĵ + 6&kcirc; is:

(a) 5    (b) 7    (c) 11    (d) 49

2. The dot product î·ĵ equals:

(a) 1    (b) 0    (c) &kcirc;    (d) −1

3. If a·b = 0 for non-zero vectors, then a and b are:

(a) parallel    (b) equal    (c) perpendicular    (d) collinear

4. The value of î × ĵ is:

(a) 0    (b) &kcirc;    (c) −&kcirc;    (d) î

5. The area of a triangle with adjacent sides a and b is:

(a) |a × b|    (b) ½|a × b|    (c) a·b    (d) ½(a·b)

6. The unit vector in the direction of î + ĵ + 2&kcirc; is:

(a) (î + ĵ + 2&kcirc;)/√6    (b) (î + ĵ + 2&kcirc;)/6    (c) (î + ĵ + 2&kcirc;)/2    (d) î + ĵ + 2&kcirc;

7. If a = 2î − 3ĵ + 4&kcirc; and b = −4î + 6ĵ − 8&kcirc;, then a and b are:

(a) perpendicular    (b) collinear    (c) equal    (d) coinitial

8. The projection of î − ĵ on the vector î + ĵ is:

(a) 1    (b) √2    (c) 0    (d) 2

9. Direction cosines (l, m, n) of any vector satisfy:

(a) l + m + n = 1    (b) l2 + m2 + n2 = 1    (c) l = m = n    (d) lmn = 1

10. The midpoint of the segment joining P(2, 3, 4) and Q(4, 1, −2) has position vector:

(a) 3î + 2ĵ + &kcirc;    (b) 6î + 4ĵ + 2&kcirc;    (c) î − ĵ − 3&kcirc;    (d) 2î + 2ĵ + 6&kcirc;

Answer key: 1-(b), 2-(b), 3-(c), 4-(b), 5-(b), 6-(a), 7-(b), 8-(c), 9-(b), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The vectors 2î − 3ĵ + 4&kcirc; and −4î + 6ĵ − 8&kcirc; are collinear.

Reason: Two vectors are collinear if one is a scalar multiple of the other.

A-R 2. Assertion: For non-zero vectors, a·b = 0 implies a and b are perpendicular.

Reason: a·b = |a||b| cosθ, and cos90° = 0.

A-R 3. Assertion: The cross product a × b is commutative.

Reason: a × b = −(b × a).

A-R 4. Assertion: The direction cosines of î + ĵ + &kcirc; are (1/√3, 1/√3, 1/√3).

Reason: The vector î + ĵ + &kcirc; is equally inclined to all three coordinate axes.

A-R 5. Assertion: The area of a triangle with adjacent sides a and b is ½|a × b|.

Reason: |a × b| equals the area of the parallelogram with adjacent sides a and b.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • A vector has magnitude and direction; |a| = √(a12 + a22 + a32) and the unit vector is a/|a|.
  • Direction cosines satisfy l2 + m2 + n2 = 1; the components a, b, c are the direction ratios.
  • Section formula: internal (mb + na)/(m + n), external (mb − na)/(m − n), midpoint (a + b)/2.
  • Dot product a·b = a1b1 + a2b2 + a3b3 = |a||b|cosθ; it is 0 for perpendicular vectors.
  • Projection of a on b = (a·b)/|b|.
  • Cross product is a vector perpendicular to both; it is 0 for parallel vectors; a × b = −(b × a).
  • Area: triangle = ½|a × b|; parallelogram = |a × b|.

How to score full marks in this chapter

State the formula you use (dot for angles/projections, cross for areas/perpendiculars) before substituting numbers. Keep the ± whenever a unit vector perpendicular to a plane is asked. For section-formula and collinearity questions, write the position vectors explicitly and show that one vector is a scalar multiple of another. Always simplify magnitudes (e.g. √450 = 15√2) and double-check the ĵ-term sign in every cross-product determinant — that single sign is the most common slip in board exams.

Frequently Asked Questions

What is Class 12 Maths Chapter 10 Vector Algebra about?

Chapter 10, Vector Algebra, covers scalars and vectors, position vectors, direction cosines and ratios, types of vectors, addition of vectors, multiplication by a scalar, the component form, the section formula, and the two products of vectors — the scalar (dot) product and the vector (cross) product with their applications to angles, projections and areas.

How many exercises are there in Class 12 Maths Chapter 10?

There are four exercises — Exercise 10.1, 10.2, 10.3 and 10.4 — plus a Miscellaneous Exercise on Chapter 10. Every question of all five is solved step by step on this page.

What is the difference between the dot product and the cross product?

The dot product a·b = |a||b|cosθ gives a scalar and is zero when the vectors are perpendicular. The cross product a × b = |a||b|sinθ n̂ gives a vector perpendicular to both and is zero when the vectors are parallel.

Are these Class 12 Maths Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

← Chapter 9: Differential Equations Class 12 Maths Chapter 11: Three Dimensional Geometry →
Scroll to Top