CBSE Class 12 · Chapter 11 · Vectors and Three-Dimensional Geometry · 2026-27

Three Dimensional Geometry Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceDirection cosines and ratios

A line passes through $P (2, 3, - 1)$ and $Q (4, 4, 1)$ . Which triple gives the direction cosines of $\vec{P Q}$ ?

Show answer

Answer: (a)

\frac{2}{3}, \frac{1}{3}, \frac{2}{3}

Why: Direction ratios 2, 1, 2 have magnitude 3.

$\vec{P Q} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , $| \vec{P Q} | = \sqrt{4 + 1 + 4} = 3$ . Direction cosines $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ .

Q2

·1 mark·Multiple choiceDirection cosines and ratios

If $l, m, n$ are the direction cosines of a line, then $l^{2} + m^{2} + n^{2}$ equals

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Answer: (b)

1

Why: cos²α + cos²β + cos²γ = 1.

$l, m, n$ are the components of a unit vector along the line, so $l^{2} + m^{2} + n^{2} = 1$ .

Q3

·1 mark·Multiple choiceEquation of a line

The vector equation of the line through $(1, - 1, 2)$ parallel to $3 \hat{i} + \hat{j} - \hat{k}$ is

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Answer: (c)

\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + \hat{j} - \hat{k})

Why: r = (position vector of the point) + λ(direction vector).

$\vec{r} = \vec{a} + λ \vec{b}$ with $\vec{a} = \hat{i} - \hat{j} + 2 \hat{k}$ (a point on the line) and $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$ (its direction).

Q4

·1 mark·Multiple choiceEquation of a line

The Cartesian equations of the line through $(2, 0, - 3)$ with direction ratios $1, - 2, 4$ are

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Answer: (d)

\frac{x - 2}{1} = \frac{y}{- 2} = \frac{z + 3}{4}

Why: (x − x₁)/a = (y − y₁)/b = (z − z₁)/c.

$\frac{x - 2}{1} = \frac{y - 0}{- 2} = \frac{z - (- 3)}{4}$ , i.e. $\frac{x - 2}{1} = \frac{y}{- 2} = \frac{z + 3}{4}$ .

Q5

·1 mark·Multiple choiceAngle between two lines

The angle between two lines with direction ratios $1, 2, 2$ and $2, - 2, 1$ is

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Answer: (a)

\frac{π}{2}

Why: 1·2 + 2·(−2) + 2·1 = 0.

$\cos θ = \frac{| 2 - 4 + 2 |}{3 \cdot 3} = 0 \Rightarrow θ = \frac{π}{2}$ .

Q6

·1 mark·Multiple choiceAngle between two lines

The angle between two lines with direction ratios $1, 0, 1$ and $0, 1, 1$ is

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Answer: (b)

\frac{π}{3}

Why: cos θ = 1/(√2 · √2) = 1/2.

$\cos θ = \frac{0 + 0 + 1}{\sqrt{2} \sqrt{2}} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$ .

Q7

·1 mark·Multiple choiceAngle between two lines

The lines $\frac{x - 1}{2} = \frac{y}{k} = \frac{z}{3}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{- 2}$ are perpendicular when $k$ equals

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Answer: (c)

2

Why: a₁a₂ + b₁b₂ + c₁c₂ = 2 + 2k − 6 = 0.

$2 (1) + k (2) + 3 (- 2) = 0 \Rightarrow 2 k - 4 = 0 \Rightarrow k = 2$ .

Q8

·1 mark·Multiple choiceAngle between two lines

The lines with direction ratios $2, - 1, 4$ and $- 4, 2, μ$ are parallel when $μ$ equals

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Answer: (d)

- 8

Why: Parallel lines have proportional direction ratios: −4/2 = 2/(−1) = μ/4.

$\frac{- 4}{2} = \frac{2}{- 1} = - 2$ , so $\frac{μ}{4} = - 2 \Rightarrow μ = - 8$ .

Q9

·1 mark·Multiple choiceEquation of a line

Which of the following points lies on the line $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{- 2}$ ?

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Answer: (a)

(4, 3, - 1)

Why: General point (2 + λ, −1 + 2λ, 3 − 2λ); λ = 2 gives (4, 3, −1).

A general point is $(2 + λ, - 1 + 2 λ, 3 - 2 λ)$ . $λ = 2$ gives $(4, 3, - 1)$ . For the others the three ratios differ (e.g. $(3, 1, 2)$ : $1, 1, \frac{1}{2}$ ).

Q10

·1 mark·Multiple choiceShortest distance between two lines

The shortest distance between the $x$ -axis and the line $\vec{r} = (2 \hat{j} + 3 \hat{k}) + μ \hat{k}$ is

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Answer: (b)

2

Why: The x-axis is r = λî; b₁ × b₂ = î × k̂ = −ĵ.

${\vec{b}}_{1} \times {\vec{b}}_{2} = \hat{i} \times \hat{k} = - \hat{j}$ ; ${\vec{a}}_{2} - {\vec{a}}_{1} = 2 \hat{j} + 3 \hat{k}$ . SD $= | (2 \hat{j} + 3 \hat{k}) \cdot (- \hat{j}) | = 2$ . (The second line is the vertical line $x = 0$ , $y = 2$ .)

Q11

·1 mark·Multiple choiceShortest distance between two lines

The distance between the parallel lines $\vec{r} = (\hat{i} + 2 \hat{j}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + 2 \hat{j}) + μ (2 \hat{i} - 2 \hat{j} + 2 \hat{k})$ is

Show answer

Answer: (a)

\frac{\sqrt{6}}{3}

Why: For parallel lines, d = |(a₂ − a₁) × b|/|b|.

$\vec{b} = \hat{i} - \hat{j} + \hat{k}$ , ${\vec{a}}_{2} - {\vec{a}}_{1} = \hat{i}$ . $\hat{i} \times \vec{b} = - \hat{j} - \hat{k}$ , magnitude $\sqrt{2}$ . $d = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$ .

Q12

·1 mark·Multiple choiceDirection cosines and ratios

The direction ratios of the line $\frac{x - 1}{3} = \frac{2 - y}{4} = \frac{z}{5}$ are

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Answer: (b)

3, - 4, 5

Why: Rewrite (2 − y)/4 as (y − 2)/(−4).

$\frac{2 - y}{4} = \frac{y - 2}{- 4}$ , so the line is $\frac{x - 1}{3} = \frac{y - 2}{- 4} = \frac{z - 0}{5}$ : direction ratios $3, - 4, 5$ .

Q13

·1 mark·Multiple choiceDirection cosines and ratios

A line makes angles of $90^{\circ}$ , $60^{\circ}$ and $30^{\circ}$ with the positive $x$ -, $y$ - and $z$ -axes. Its direction cosines are

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Answer: (c)

0, \frac{1}{2}, \frac{\sqrt{3}}{2}

Why: cos 90° = 0, cos 60° = 1/2, cos 30° = √3/2.

$(\cos 90^{\circ}, \cos 60^{\circ}, \cos 30^{\circ}) = (0, \frac{1}{2}, \frac{\sqrt{3}}{2})$ ; check: $0 + \frac{1}{4} + \frac{3}{4} = 1$ .

Q14

·1 mark·Multiple choiceDirection cosines and ratios

A line makes an acute angle $α$ with the $x$ -axis, $45^{\circ}$ with the $y$ -axis and $60^{\circ}$ with the $z$ -axis. Then $\cos α$ is

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Answer: (c)

\frac{1}{2}

Why: cos²α + cos²45° + cos²60° = 1 gives cos²α = 1/4.

$\cos^{2} α + \frac{1}{2} + \frac{1}{4} = 1 \Rightarrow \cos^{2} α = \frac{1}{4}$ ; $α$ is acute, so $\cos α = \frac{1}{2}$ ( $α = 60^{\circ}$ ). ( $\frac{1}{4}$ is $\cos^{2} α$ , not $\cos α$ .)

Q15

·1 mark·Multiple choiceShortest distance between two lines

The shortest distance between the lines $\vec{r} = λ (\hat{i} + \hat{j})$ and $\vec{r} = 3 \hat{k} + μ (\hat{i} - \hat{j})$ is

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Answer: (d)

3

Why: b₁ × b₂ = −2k; (a₂ − a₁)·(b₁ × b₂) = 3k·(−2k) = −6.

${\vec{b}}_{1} \times {\vec{b}}_{2} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = - 2 \hat{k}$ . SD $= \frac{| 3 \hat{k} \cdot (- 2 \hat{k}) |}{2} = \frac{6}{2} = 3$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Zip-line cable (4 marks)

An adventure park stretches a straight zip-line cable from the top of a tower at $A (2, 1, 6)$ to a landing post at $B (6, 3, 2)$ , with coordinates in metres and the $z$ -axis vertical.

(i) Find the direction cosines of

\vec{A B}

. [1 mark]

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Answer:

\frac{2}{3}, \frac{1}{3}, - \frac{2}{3}

Direction ratios

4, 2, - 4

, magnitude

6

: direction cosines

\frac{2}{3}, \frac{1}{3}, - \frac{2}{3}

. A1

(ii) Find the length of the cable. [1 mark]

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Answer:

6

| A B | = \sqrt{16 + 4 + 16} = 6

m. A1

(iii) Write a vector equation of the cable's line and find the point of the cable at height

4

m. [2 marks]

Show answer

Answer:

\vec{r} = (2 \hat{i} + \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})

;

(4, 2, 4)

\vec{r} = (2 \hat{i} + \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})

M1;

6 - 2 λ = 4 \Rightarrow λ = 1

, point

(4, 2, 4)

. A1

OR Show that if the cable were extended beyond

B

, it would reach the ground (

z = 0

) at

(8, 4, 0)

. [2 marks]

Show answer

Answer:

λ = 3

gives

(8, 4, 0)

\vec{r} = (2 \hat{i} + \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})

z = 0 \Rightarrow λ = 3

M1, giving

(2 + 6, 1 + 3, 0) = (8, 4, 0)

. A1

Case study 2: Two drone flight paths (4 marks)

Two delivery drones fly along straight paths. Drone $P$ follows $\vec{r} = (\hat{i} + 2 \hat{j}) + s (\hat{i} + \hat{k})$ and drone $Q$ follows $\vec{r} = 3 \hat{k} + t (\hat{j} + \hat{k})$ (units: hundreds of metres). Air-traffic rules require the paths to stay well apart.

(i) Find the angle between the two paths. [1 mark]

Show answer

Answer:

\frac{π}{3}

\cos θ = \frac{| 0 + 0 + 1 |}{\sqrt{2} \sqrt{2}} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}

. A1

(ii) Find

{\vec{b}}_{1} \times {\vec{b}}_{2}

for the two direction vectors. [1 mark]

Show answer

Answer:

- \hat{i} - \hat{j} + \hat{k}

(\hat{i} + \hat{k}) \times (\hat{j} + \hat{k}) = - \hat{i} - \hat{j} + \hat{k}

. A1

(iii) Find the shortest distance between the two paths. [2 marks]

Show answer

Answer:

2 \sqrt{3}

{\vec{a}}_{2} - {\vec{a}}_{1} = - \hat{i} - 2 \hat{j} + 3 \hat{k}

;

({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = 1 + 2 + 3 = 6

M1; SD

= \frac{6}{\sqrt{3}} = 2 \sqrt{3}

(hundred metres). A1

OR Show that the two paths do not meet. [2 marks]

Show answer

Answer: The equations are inconsistent

Equate points:

1 + s = 0

2 = t

s = 3 + t

M1. The first two give

s = - 1

t = 2

, but then

s = 3 + t = 5 \neq - 1

: no common point. A1

Next steps for Three Dimensional Geometry

This free set is separate from the chapter's question bank. On the Three Dimensional Geometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 46 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Three Dimensional Geometry: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.