Class 12 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise – Vector Algebra | Boundless Maths
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Miscellaneous Exercise Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise – Vector Algebra

This Class 12 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise page covers all 19 questions, solved step-by-step. Unlike Exercises 10.1–10.4, which isolate one idea at a time, this exercise deliberately mixes them — a single question might need vector addition, the section formula and a dot product in the same solution, which is exactly how vectors show up in an actual CBSE paper.

Questions 1–10 revisit the toolkit from earlier exercises in combined, real-world framing: a unit vector at a given angle, a girl's net displacement after walking in two directions (a recurring board-exam favourite), collinearity with a ratio of division, and a parallelogram's diagonal and area together in one question. Questions 11–15 lean harder into proof: showing a vector is equally inclined to the axes, constructing a vector perpendicular to two others under an extra dot-product constraint, and proving the "if and only if" identity connecting (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2 to perpendicularity — a proof-writing style CBSE tests directly in 3- and 5-mark questions. The exercise closes with four MCQs (16–19) that quietly test the dot and cross product formulas from the opposite direction — given the answer, find the angle.

19Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise — All 19 Questions

1

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

Easy +
Solution

A unit vector in the XY-plane making angle θ with the positive x-axis is \cos\theta\,\hat{i}+\sin\theta\,\hat{j}.

For \theta=30^\circ: \cos30^\circ=\dfrac{\sqrt3}{2}, \sin30^\circ=\dfrac12.

Answer: (√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).

Easy +
Solution

\vec{PQ} = (x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}.

Magnitude: |\vec{PQ}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.

Answer: scalar components are (x₂−x₁), (y₂−y₁), (z₂−z₁); magnitude as above.
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.

Medium +
Solution

Take east as \hat{i} and north as \hat{j}. The first walk is \vec{OA} = -4\hat{i} (4 km west).

The second walk is 3 km at 30° east of north: \vec{AB} = 3\sin30^\circ\,\hat{i} + 3\cos30^\circ\,\hat{j} = \dfrac32\hat{i}+\dfrac{3\sqrt3}{2}\hat{j}.

Net displacement \vec{OB} = \vec{OA}+\vec{AB} = \left(-4+\dfrac32\right)\hat{i}+\dfrac{3\sqrt3}{2}\hat{j} = -\dfrac52\hat{i}+\dfrac{3\sqrt3}{2}\hat{j}.

|\vec{OB}| = \sqrt{\left(\dfrac52\right)^2+\left(\dfrac{3\sqrt3}{2}\right)^2} = \sqrt{\dfrac{25}{4}+\dfrac{27}{4}} = \sqrt{13}.

Answer: displacement = −(5/2)î + (3√3/2)ĵ km, magnitude √13 km.
4

If \vec{a}=\vec{b}+\vec{c}, then is it true that |\vec{a}|=|\vec{b}|+|\vec{c}|? Justify your answer.

Easy +
Solution

No, this is not true in general. By the triangle inequality, |\vec{a}|\le|\vec{b}|+|\vec{c}|, with equality only when \vec{b} and \vec{c} are collinear and in the same direction.

Counter-example: let \vec{b}=\hat{i}, \vec{c}=\hat{j}. Then \vec{a}=\hat{i}+\hat{j}, so |\vec{a}|=\sqrt2, but |\vec{b}|+|\vec{c}|=1+1=2.

Answer: false in general — it holds only when b and c point in exactly the same direction.
5

Find the value of x for which x(\hat{i}+\hat{j}+\hat{k}) is a unit vector.

Easy +
Solution

|x(\hat{i}+\hat{j}+\hat{k})| = |x|\sqrt3 = 1 \Rightarrow |x|=\dfrac{1}{\sqrt3}.

Answer: x = ±1/√3.
6

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \vec{a}=2\hat{i}+3\hat{j}-\hat{k} and \vec{b}=\hat{i}-2\hat{j}+\hat{k}.

Easy +
Solution

Resultant \vec{a}+\vec{b} = 3\hat{i}+\hat{j}+0\hat{k}. Magnitude = \sqrt{9+1}=\sqrt{10}.

Required vector = \dfrac{5}{\sqrt{10}}(3\hat{i}+\hat{j}) = \dfrac{15}{\sqrt{10}}\hat{i}+\dfrac{5}{\sqrt{10}}\hat{j}, which simplifies to \dfrac{3\sqrt{10}}{2}\hat{i}+\dfrac{\sqrt{10}}{2}\hat{j}.

Answer: (3√10/2)î + (√10/2)ĵ.
7

If \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2\hat{i}-\hat{j}+3\hat{k} and \vec{c}=\hat{i}-2\hat{j}+\hat{k}, find a unit vector parallel to the vector 2\vec{a}-\vec{b}+3\vec{c}.

Medium +
Solution

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

Sum: 2\vec{a}-\vec{b}+3\vec{c} = (2-2+3)\hat{i}+(2+1-6)\hat{j}+(2-3+3)\hat{k} = 3\hat{i}-3\hat{j}+2\hat{k}.

Magnitude = \sqrt{9+9+4}=\sqrt{22}.

Answer: (3î − 3ĵ + 2k̂)/√22.
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.

Medium +
Solution

\vec{AB} = 4\hat{i}+2\hat{j}+6\hat{k}, so |\vec{AB}|=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}.

\vec{BC} = 6\hat{i}+3\hat{j}+9\hat{k}, so |\vec{BC}|=\sqrt{36+9+81}=\sqrt{126}=3\sqrt{14}.

\vec{AC} = 10\hat{i}+5\hat{j}+15\hat{k}, so |\vec{AC}|=\sqrt{100+25+225}=\sqrt{350}=5\sqrt{14}.

Since |\vec{AC}|=|\vec{AB}|+|\vec{BC}| (as 2\sqrt{14}+3\sqrt{14}=5\sqrt{14}), the three points are collinear.

Also, \vec{AB} and \vec{BC} are in the same direction (each is a positive multiple of 2\hat{i}+\hat{j}+3\hat{k}), so B lies between A and C, dividing AC in the ratio |AB|:|BC| = 2\sqrt{14}:3\sqrt{14} = 2:3.

Answer: A, B, C are collinear, and B divides AC internally 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 (2\vec{a}+\vec{b}) and (\vec{a}-3\vec{b}) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

Hard +
Solution

For external division in ratio m : n = 1 : 2, \vec{OR} = \dfrac{m\vec{OQ}-n\vec{OP}}{m-n}.

\vec{OR} = \dfrac{1(\vec{a}-3\vec{b})-2(2\vec{a}+\vec{b})}{1-2} = \dfrac{\vec{a}-3\vec{b}-4\vec{a}-2\vec{b}}{-1} = \dfrac{-3\vec{a}-5\vec{b}}{-1} = 3\vec{a}+5\vec{b}.

Midpoint of R and Q: \dfrac{\vec{OR}+\vec{OQ}}{2} = \dfrac{(3\vec{a}+5\vec{b})+(\vec{a}-3\vec{b})}{2} = \dfrac{4\vec{a}+2\vec{b}}{2} = 2\vec{a}+\vec{b}, which is exactly \vec{OP}.

Answer: R = 3a + 5b; since the midpoint of RQ equals P's position vector, P is the midpoint of RQ.
10

The two adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}+5\hat{k} and \hat{i}-2\hat{j}-3\hat{k}. Find the unit vector parallel to its diagonal. Also, find its area.

Medium +
Solution

Let \vec{a}=2\hat{i}-4\hat{j}+5\hat{k}, \vec{b}=\hat{i}-2\hat{j}-3\hat{k}. Diagonal = \vec{a}+\vec{b} = 3\hat{i}-6\hat{j}+2\hat{k}.

|\vec{a}+\vec{b}| = \sqrt{9+36+4}=\sqrt{49}=7. Unit vector = \dfrac37\hat{i}-\dfrac67\hat{j}+\dfrac27\hat{k}.

Area of parallelogram = |\vec{a}\times\vec{b}|. \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-4&5\\1&-2&-3\end{vmatrix} = \hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)

= 22\hat{i}+11\hat{j}+0\hat{k}, so |\vec{a}\times\vec{b}|=\sqrt{484+121}=\sqrt{605}=11\sqrt5.

Answer: unit vector = (3/7)î − (6/7)ĵ + (2/7)k̂; area = 11√5 sq units.
11

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \pm\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right).

Easy +
Solution

If the vector is equally inclined to all three axes, its direction cosines satisfy l=m=n.

Using l^2+m^2+n^2=1: 3l^2=1 \Rightarrow l=\pm\dfrac{1}{\sqrt3}.

Answer: the direction cosines are ±(1/√3, 1/√3, 1/√3).
12

Let \vec{a}=\hat{i}+4\hat{j}+2\hat{k}, \vec{b}=3\hat{i}-2\hat{j}+7\hat{k} and \vec{c}=2\hat{i}-\hat{j}+4\hat{k}. Find a vector \vec{d} which is perpendicular to both \vec{a} and \vec{b}, and \vec{c}\cdot\vec{d}=15.

Hard +
Solution

Since \vec{d} is perpendicular to both \vec{a} and \vec{b}, it must be parallel to \vec{a}\times\vec{b}, so \vec{d}=t(\vec{a}\times\vec{b}) for some scalar t.

\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&4&2\\3&-2&7\end{vmatrix} = \hat{i}(28+4)-\hat{j}(7-6)+\hat{k}(-2-12) = 32\hat{i}-\hat{j}-14\hat{k}.

\vec{d}=t(32\hat{i}-\hat{j}-14\hat{k}). Then \vec{c}\cdot\vec{d} = t[2(32)+(-1)(-1)+4(-14)] = t[64+1-56]=9t.

Setting 9t=15 gives t=\dfrac53.

Answer: d = (160/3)î − (5/3)ĵ − (70/3)k̂.
13

The scalar product of the vector \hat{i}+\hat{j}+\hat{k} with a unit vector along the sum of vectors 2\hat{i}+4\hat{j}-5\hat{k} and \lambda\hat{i}+2\hat{j}+3\hat{k} is equal to one. Find the value of λ.

Hard +
Solution

Sum of the two vectors: (2+\lambda)\hat{i}+6\hat{j}-2\hat{k}.

Its magnitude: \sqrt{(2+\lambda)^2+36+4} = \sqrt{(2+\lambda)^2+40}.

The condition gives \dfrac{(\hat{i}+\hat{j}+\hat{k})\cdot[(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}]}{\sqrt{(2+\lambda)^2+40}} = 1, i.e. \dfrac{(2+\lambda)+6-2}{\sqrt{(2+\lambda)^2+40}}=1.

6+\lambda = \sqrt{(2+\lambda)^2+40}. Squaring: 36+12\lambda+\lambda^2 = 4+4\lambda+\lambda^2+40.

36+12\lambda = 44+4\lambda \Rightarrow 8\lambda=8.

Answer: λ = 1.
14

If \vec{a},\vec{b},\vec{c} are mutually perpendicular vectors of equal magnitude, show that the vector \vec{a}+\vec{b}+\vec{c} is equally inclined to \vec{a},\vec{b} and \vec{c}.

Hard +
Solution

Let |\vec{a}|=|\vec{b}|=|\vec{c}|=k, with \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0 (mutual perpendicularity).

(\vec{a}+\vec{b}+\vec{c})\cdot\vec{a} = \vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{a}+\vec{c}\cdot\vec{a} = k^2+0+0 = k^2, and identically (\vec{a}+\vec{b}+\vec{c})\cdot\vec{b}=k^2, (\vec{a}+\vec{b}+\vec{c})\cdot\vec{c}=k^2.

Also |\vec{a}+\vec{b}+\vec{c}|^2 = |\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}) = 3k^2, so |\vec{a}+\vec{b}+\vec{c}|=k\sqrt3.

The cosine of the angle with each of \vec{a},\vec{b},\vec{c} is \dfrac{k^2}{k\sqrt3\cdot k} = \dfrac{1}{\sqrt3} in every case — the same value.

Answer: since (a+b+c) makes the same angle cos⁻¹(1/√3) with each of a, b and c, it is equally inclined to all three.
15

Prove that (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2, if and only if \vec{a},\vec{b} are perpendicular, given \vec{a}\neq\vec{0}, \vec{b}\neq\vec{0}.

Medium +
Solution

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

This equals |\vec{a}|^2+|\vec{b}|^2 if and only if 2(\vec{a}\cdot\vec{b})=0, i.e. \vec{a}\cdot\vec{b}=0.

Since \vec{a} and \vec{b} are both nonzero, \vec{a}\cdot\vec{b}=0 holds exactly when \vec{a}\perp\vec{b}.

Answer: proved both ways — the identity holds precisely when a and b are perpendicular.
16

If θ is the angle between two vectors \vec{a} and \vec{b}, then \vec{a}\cdot\vec{b}\ge0 only when
(A) 0<\theta<\dfrac{\pi}{2}
(B) 0\le\theta\le\dfrac{\pi}{2}
(C) 0<\theta<\pi
(D) 0\le\theta\le\pi

Easy +
Solution

\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \ge 0 \iff \cos\theta\ge0, which happens exactly when 0\le\theta\le\dfrac{\pi}{2}.

Answer: (B).
17

Let \vec{a} and \vec{b} be two unit vectors and θ is the angle between them. Then \vec{a}+\vec{b} is a unit vector if
(A) \theta=\dfrac{\pi}{4}
(B) \theta=\dfrac{\pi}{3}
(C) \theta=\dfrac{\pi}{2}
(D) \theta=\dfrac{2\pi}{3}

Medium +
Solution

|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+2(\vec{a}\cdot\vec{b})+|\vec{b}|^2 = 1+2\cos\theta+1 = 2+2\cos\theta.

Setting this equal to 1 (for a unit vector): 2+2\cos\theta=1 \Rightarrow \cos\theta=-\dfrac12.

Answer: (D) θ = 2π/3.
18

The value of \hat{i}\cdot(\hat{j}\times\hat{k})+\hat{j}\cdot(\hat{i}\times\hat{k})+\hat{k}\cdot(\hat{i}\times\hat{j}) is
(A) 0
(B) −1
(C) 1
(D) 3

Easy +
Solution

\hat{j}\times\hat{k}=\hat{i}, so \hat{i}\cdot(\hat{j}\times\hat{k}) = \hat{i}\cdot\hat{i}=1.

\hat{i}\times\hat{k}=-\hat{j}, so \hat{j}\cdot(\hat{i}\times\hat{k}) = \hat{j}\cdot(-\hat{j})=-1.

\hat{i}\times\hat{j}=\hat{k}, so \hat{k}\cdot(\hat{i}\times\hat{j}) = \hat{k}\cdot\hat{k}=1.

Sum = 1-1+1=1.

Answer: (C) 1.
19

If θ is the angle between any two vectors \vec{a} and \vec{b}, then |\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}| when θ is equal to
(A) 0
(B) \dfrac{\pi}{4}
(C) \dfrac{\pi}{2}
(D) π

Medium +
Solution

|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}|\cos\theta and |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta.

Setting these equal: \cos\theta=\sin\theta \Rightarrow \tan\theta=1.

Answer: (B) θ = π/4.

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Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 10 Miscellaneous Exercise

How many questions are there in the Miscellaneous Exercise?

The Miscellaneous Exercise has 19 questions (15 regular questions plus 4 MCQs) that combine ideas from all four exercises of Chapter 10 — vector addition, the section formula, the dot product, and the cross product.

Is the Miscellaneous Exercise important for the CBSE board exam?

Yes. Because it mixes concepts rather than drilling one at a time, it is the closest thing in the chapter to an actual exam question, and CBSE has drawn board-exam questions directly from this exercise in past years — particularly the girl's-walk displacement problem, the section formula proof, and the mutually-perpendicular-vectors question.

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

The Miscellaneous Exercise 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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