Class 12 Maths NCERT Solutions Chapter 10 Ex 10.3 – Scalar (Dot) Product | Boundless Maths
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Ex 10.3 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 10 Ex 10.3 – Scalar (Dot) Product

This Class 12 Maths NCERT Solutions Chapter 10 Ex 10.3 page covers all 18 questions, solved step-by-step, on the scalar (dot) product — the operation \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta that turns two vectors into a single number, and quietly does the heavy lifting behind angle, projection and perpendicularity questions throughout the rest of the chapter.

Questions 1–9 are the exercise's computational backbone: finding the angle between two vectors from their dot product, computing a projection, and using the identity \vec{a}\cdot\vec{a}=|\vec{a}|^2 to expand expressions like (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) — a technique CBSE reuses in nearly every 2019–2026 sample paper. Questions 10–14 shift to the perpendicularity condition \vec{a}\cdot\vec{b}=0, including the "converse need not be true" justification that examiners specifically like testing conceptually rather than numerically. Questions 15–17 apply the dot product to triangles — finding an angle from position vectors, proving three points collinear, and proving a right angle using the Pythagoras converse — question types that also show up, almost unchanged, in Chapter 11 (Three Dimensional Geometry). The exercise closes with an MCQ on unit vectors (18).

18Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 10 Ex 10.3 — All 18 Questions

1

Find the angle between two vectors \vec{a} and \vec{b} with magnitudes \sqrt3 and 2, respectively, having \vec{a}\cdot\vec{b}=\sqrt6.

Easy +
Solution

\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\sqrt6}{\sqrt3\cdot 2} = \dfrac{\sqrt6}{2\sqrt3} = \dfrac{\sqrt2}{2} = \dfrac{1}{\sqrt2}.

Answer: θ = π/4 (45°).
2

Find the angle between the vectors \hat{i}-2\hat{j}+3\hat{k} and 3\hat{i}-2\hat{j}+\hat{k}.

Easy +
Solution

Let \vec{a}=\hat{i}-2\hat{j}+3\hat{k} and \vec{b}=3\hat{i}-2\hat{j}+\hat{k}.

\vec{a}\cdot\vec{b} = 1(3)+(-2)(-2)+3(1) = 3+4+3 = 10.

|\vec{a}| = \sqrt{1+4+9}=\sqrt{14}, |\vec{b}| = \sqrt{9+4+1}=\sqrt{14}.

\cos\theta = \dfrac{10}{\sqrt{14}\cdot\sqrt{14}} = \dfrac{10}{14} = \dfrac57.

Answer: θ = cos⁻¹(5/7).
3

Find the projection of the vector \hat{i}-\hat{j} on the vector \hat{i}+\hat{j}.

Easy +
Solution

Projection of \vec{a} on \vec{b} is \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}.

Here \vec{a}\cdot\vec{b} = (1)(1)+(-1)(1) = 1-1 = 0.

Answer: the projection is 0.
4

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}.

Easy +
Solution

\vec{a}\cdot\vec{b} = 1(7)+3(-1)+7(8) = 7-3+56 = 60.

|\vec{b}| = \sqrt{49+1+64} = \sqrt{114}.

Answer: the projection is 60/√114.
5

Show that each of the given three vectors is a unit vector: \dfrac17(2\hat{i}+3\hat{j}+6\hat{k}), \dfrac17(3\hat{i}-6\hat{j}+2\hat{k}), \dfrac17(6\hat{i}+2\hat{j}-3\hat{k}). Also, show that they are mutually perpendicular to each other.

Medium +
Solution

Let \vec{a}=\frac17(2\hat{i}+3\hat{j}+6\hat{k}), \vec{b}=\frac17(3\hat{i}-6\hat{j}+2\hat{k}), \vec{c}=\frac17(6\hat{i}+2\hat{j}-3\hat{k}).

|\vec{a}| = \frac17\sqrt{4+9+36}=\frac17\sqrt{49}=\frac77=1. Similarly |\vec{b}|=\frac17\sqrt{9+36+4}=1 and |\vec{c}|=\frac17\sqrt{36+4+9}=1. All three are unit vectors.

\vec{a}\cdot\vec{b} = \frac{1}{49}[2(3)+3(-6)+6(2)] = \frac{1}{49}[6-18+12] = 0.

\vec{b}\cdot\vec{c} = \frac{1}{49}[3(6)+(-6)(2)+2(-3)] = \frac{1}{49}[18-12-6] = 0.

\vec{a}\cdot\vec{c} = \frac{1}{49}[2(6)+3(2)+6(-3)] = \frac{1}{49}[12+6-18] = 0.

Answer: all three are unit vectors, and every pair has dot product 0, so they are mutually perpendicular.
6

Find |\vec{a}| and |\vec{b}|, if (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=8 and |\vec{a}|=8|\vec{b}|.

Medium +
Solution

(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2-|\vec{b}|^2 = 8.

Substituting |\vec{a}|=8|\vec{b}|: 64|\vec{b}|^2-|\vec{b}|^2=8 \Rightarrow 63|\vec{b}|^2=8 \Rightarrow |\vec{b}|^2=\dfrac{8}{63}.

|\vec{b}| = \sqrt{\dfrac{8}{63}} = \dfrac{2\sqrt{14}}{21}, and |\vec{a}| = 8|\vec{b}| = \dfrac{16\sqrt{14}}{21}.

Answer: |b| = 2√14/21 and |a| = 16√14/21.
7

Evaluate the product (3\vec{a}-5\vec{b})\cdot(2\vec{a}+7\vec{b}).

Easy +
Solution

Expanding using distributivity: (3\vec{a}-5\vec{b})\cdot(2\vec{a}+7\vec{b}) = 6(\vec{a}\cdot\vec{a})+21(\vec{a}\cdot\vec{b})-10(\vec{b}\cdot\vec{a})-35(\vec{b}\cdot\vec{b}).

Since \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}, the middle terms combine to 11(\vec{a}\cdot\vec{b}).

Answer: 6|a|² + 11(a·b) − 35|b|².
8

Find the magnitude of two vectors \vec{a} and \vec{b}, having the same magnitude and such that the angle between them is 60° and their scalar product is \dfrac12.

Easy +
Solution

Let |\vec{a}|=|\vec{b}|=x. Then \vec{a}\cdot\vec{b} = x\cdot x\cdot\cos60^\circ = x^2\cdot\dfrac12 = \dfrac12.

So x^2=1 \Rightarrow x=1 (magnitude is non-negative).

Answer: |a| = |b| = 1.
9

Find |\vec{x}|, if for a unit vector \vec{a}, (\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12.

Easy +
Solution

(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = |\vec{x}|^2-|\vec{a}|^2 = 12.

Since \vec{a} is a unit vector, |\vec{a}|^2=1, so |\vec{x}|^2 = 13.

Answer: |x| = √13.
10

If \vec{a}=2\hat{i}+2\hat{j}+3\hat{k}, \vec{b}=-\hat{i}+2\hat{j}+\hat{k} and \vec{c}=3\hat{i}+\hat{j} are such that \vec{a}+\lambda\vec{b} is perpendicular to \vec{c}, then find the value of \lambda.

Medium +
Solution

Perpendicularity requires (\vec{a}+\lambda\vec{b})\cdot\vec{c}=0, i.e. \vec{a}\cdot\vec{c}+\lambda(\vec{b}\cdot\vec{c})=0.

\vec{a}\cdot\vec{c} = 2(3)+2(1)+3(0) = 8.

\vec{b}\cdot\vec{c} = -1(3)+2(1)+1(0) = -1.

So 8-\lambda=0.

Answer: λ = 8.
11

Show that |\vec{a}|\vec{b}+|\vec{b}|\vec{a} is perpendicular to |\vec{a}|\vec{b}-|\vec{b}|\vec{a}, for any two nonzero vectors \vec{a} and \vec{b}.

Medium +
Solution

(|\vec{a}|\vec{b}+|\vec{b}|\vec{a})\cdot(|\vec{a}|\vec{b}-|\vec{b}|\vec{a}) = |\vec{a}|^2(\vec{b}\cdot\vec{b}) - |\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a}) + |\vec{a}||\vec{b}|(\vec{a}\cdot\vec{b}) - |\vec{b}|^2(\vec{a}\cdot\vec{a}).

= |\vec{a}|^2|\vec{b}|^2 - |\vec{a}|^2|\vec{b}|^2 = 0 (using \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}, so the two middle terms cancel and the remaining terms are equal and opposite).

Answer: the dot product is 0, so the two vectors are perpendicular, proved.
12

If \vec{a}\cdot\vec{a}=0 and \vec{a}\cdot\vec{b}=0, then what can be concluded about the vector \vec{b}?

Easy +
Solution

\vec{a}\cdot\vec{a}=0 \Rightarrow |\vec{a}|^2=0 \Rightarrow \vec{a}=\vec{0}, the zero vector.

Since the zero vector's dot product with any vector is automatically 0, the condition \vec{a}\cdot\vec{b}=0 holds no matter what \vec{b} is.

Answer: no conclusion can be drawn about b — it can be any vector.
13

If \vec{a},\vec{b},\vec{c} are unit vectors such that \vec{a}+\vec{b}+\vec{c}=\vec{0}, find the value of \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}.

Medium +
Solution

Squaring: (\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = 0.

|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0.

Since each is a unit vector, 1+1+1+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0.

Answer: a·b + b·c + c·a = −3/2.
14

If either vector \vec{a}=\vec{0} or \vec{b}=\vec{0}, then \vec{a}\cdot\vec{b}=0. But the converse need not be true. Justify your answer with an example.

Easy +
Solution

Take \vec{a}=\hat{i} and \vec{b}=\hat{j}, both nonzero vectors.

\vec{a}\cdot\vec{b} = \hat{i}\cdot\hat{j} = 0, since they are mutually perpendicular unit vectors.

Answer: here a·b = 0 even though neither a nor b is the zero vector — so a·b = 0 does not imply a = 0 or b = 0; it can also mean the two vectors are perpendicular.
15

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (−1, 0, 0), (0, 1, 2), respectively, then find \angle ABC. [∠ABC is the angle between the vectors \vec{BA} and \vec{BC}].

Medium +
Solution

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

\vec{BC} = (0-(-1))\hat{i}+(1-0)\hat{j}+(2-0)\hat{k} = \hat{i}+\hat{j}+2\hat{k}.

\vec{BA}\cdot\vec{BC} = 2(1)+2(1)+3(2) = 2+2+6=10.

|\vec{BA}| = \sqrt{4+4+9}=\sqrt{17}, |\vec{BC}| = \sqrt{1+1+4}=\sqrt6.

\cos(\angle ABC) = \dfrac{10}{\sqrt{17}\cdot\sqrt6} = \dfrac{10}{\sqrt{102}}.

Answer: ∠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.

Medium +
Solution

\vec{AB} = (2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k} = \hat{i}+4\hat{j}-4\hat{k}, so |\vec{AB}|=\sqrt{1+16+16}=\sqrt{33}.

\vec{BC} = (3-2)\hat{i}+(10-6)\hat{j}+(-1-3)\hat{k} = \hat{i}+4\hat{j}-4\hat{k}, so |\vec{BC}|=\sqrt{33}.

\vec{AC} = (3-1)\hat{i}+(10-2)\hat{j}+(-1-7)\hat{k} = 2\hat{i}+8\hat{j}-8\hat{k}, so |\vec{AC}|=\sqrt{4+64+64}=\sqrt{132}=2\sqrt{33}.

Since |\vec{AC}| = |\vec{AB}|+|\vec{BC}| (as \sqrt{33}+\sqrt{33}=2\sqrt{33}), the three points lie on the same straight line.

Answer: AC = AB + BC in magnitude, so A, B, C are collinear.
17

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.

Hard +
Solution

Let these be the position vectors of A, B, C respectively: \vec{a}=2\hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}-5\hat{k}, \vec{c}=3\hat{i}-4\hat{j}-4\hat{k}.

\vec{AB}=\vec{b}-\vec{a}=-\hat{i}-2\hat{j}-6\hat{k}, so |\vec{AB}|^2=1+4+36=41.

\vec{BC}=\vec{c}-\vec{b}=2\hat{i}-\hat{j}+\hat{k}, so |\vec{BC}|^2=4+1+1=6.

\vec{CA}=\vec{a}-\vec{c}=-\hat{i}+3\hat{j}+5\hat{k}, so |\vec{CA}|^2=1+9+25=35.

Since |\vec{BC}|^2+|\vec{CA}|^2 = 6+35 = 41 = |\vec{AB}|^2, the converse of the Pythagoras theorem is satisfied, with the right angle at the vertex common to BC and CA.

Answer: |BC|² + |CA|² = |AB|², so the triangle is right angled at C.
18

If \vec{a} is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then \lambda\vec{a} is a unit vector if
(A) \lambda=1
(B) \lambda=-1
(C) a=|\lambda|
(D) a=1/|\lambda|

Easy +
Solution

\lambda\vec{a} is a unit vector when |\lambda\vec{a}|=1, i.e. |\lambda|\cdot|\vec{a}|=1, i.e. |\lambda|\cdot a=1.

This gives a=\dfrac{1}{|\lambda|}.

Answer: (D).

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Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 10 Ex 10.3

How many questions are there in Exercise 10.3?

Exercise 10.3 has 18 questions on the scalar (dot) product — finding the angle between two vectors, projection of one vector on another, checking perpendicularity, and applying the dot product to triangles and collinear points.

What concept does Exercise 10.3 test?

It tests the scalar product formula a·b = |a||b|cosθ, the projection formula, and the key perpendicularity test a·b = 0, along with properties such as a·a = |a|² that come up repeatedly in proof-based questions.

Where can I find the official NCERT textbook for this exercise?

Exercise 10.3 is from Chapter 10, Vector Algebra, in the NCERT Class 12 Mathematics textbook (Part I), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

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