CBSE Class 12 · Chapter 10 · Vectors and Three-Dimensional Geometry · 2026-27

Vector Algebra 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 choiceMagnitude, direction and components

The magnitude of $4 \hat{i} - 4 \hat{j} + 7 \hat{k}$ is

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Answer: (a)

9

Why: √(16 + 16 + 49) = √81.

$\sqrt{4^{2} + (- 4)^{2} + 7^{2}} = \sqrt{81} = 9$ .

Q2

·1 mark·Multiple choiceMagnitude, direction and components

The unit vector in the direction of $\hat{i} + 2 \hat{j} - 2 \hat{k}$ is

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Answer: (b)

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

Why: Divide by the magnitude √(1 + 4 + 4) = 3.

$| \vec{a} | = \sqrt{1 + 4 + 4} = 3$ , so $\hat{a} = \frac{1}{3} (\hat{i} + 2 \hat{j} - 2 \hat{k})$ .

Q3

·1 mark·Multiple choiceScalar (dot) product

If $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b} = 2 \hat{i} - \hat{j} + \hat{k}$ , then $\vec{a} \cdot \vec{b}$ is

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Answer: (c)

3

Why: Multiply matching components and add: 2 − 2 + 3.

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

Q4

·1 mark·Multiple choiceScalar (dot) product

The angle between $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$ is

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Answer: (d)

\frac{π}{3}

Why: cos θ = (a·b)/(|a||b|) = 1/(√2 · √2).

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

Q5

·1 mark·Multiple choiceScalar (dot) product

For which value of $λ$ is $3 \hat{i} + λ \hat{j} - 2 \hat{k}$ perpendicular to $2 \hat{i} + 4 \hat{j} + \hat{k}$ ?

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Answer: (a)

- 1

Why: Dot product zero: 6 + 4λ − 2 = 0.

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

Q6

·1 mark·Multiple choiceScalar (dot) product

The scalar projection of $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ on $\vec{b} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ is

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Answer: (b)

\frac{4}{3}

Why: Projection = a·b/|b| = 4/3.

$\vec{a} \cdot \vec{b} = 2 - 2 + 4 = 4$ , $| \vec{b} | = 3$ . Projection $= \frac{4}{3}$ .

Q7

·1 mark·Multiple choiceVector (cross) product

$\hat{k} \times \hat{i}$ equals

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Answer: (d)

\hat{j}

Why: Cyclic order i → j → k → i: k × i = j.

In the cyclic order $\hat{i}, \hat{j}, \hat{k}$ : $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

Q8

·1 mark·Multiple choiceVector (cross) product

If $| \vec{a} | = 3$ , $| \vec{b} | = 4$ and the angle between them is $\frac{π}{6}$ , then $| \vec{a} \times \vec{b} |$ is

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Answer: (c)

6

Why: |a × b| = |a||b| sin θ = 12 × 1/2.

$| \vec{a} \times \vec{b} | = 3 \times 4 \times \sin \frac{π}{6} = 6$ . ( $6 \sqrt{3}$ would be $\vec{a} \cdot \vec{b}$ .)

Q9

·1 mark·Multiple choiceVector (cross) product

$(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j})$ equals

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Answer: (b)

- 2 \hat{k}

Why: Only the k-component survives: (1)(−1) − (1)(1) = −2.

$| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & - 1 & 0 \end{matrix} | = \hat{i} (0) - \hat{j} (0) + \hat{k} (- 1 - 1) = - 2 \hat{k}$ .

Q10

·1 mark·Multiple choiceVector (cross) product

The area of the parallelogram with adjacent sides $\vec{a} = 2 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 3 \hat{j}$ is

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Answer: (d)

5

sq units

Why: Area = |a × b| = |(6 − 1)k|.

$\vec{a} \times \vec{b} = (2 \cdot 3 - 1 \cdot 1) \hat{k} = 5 \hat{k}$ , so the area is $5$ sq units.

Q11

·1 mark·Multiple choiceMagnitude, direction and components

The direction cosines of the vector $2 \hat{i} - \hat{j} + 2 \hat{k}$ are

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Answer: (a)

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

Why: Divide each component by the magnitude 3.

$| \vec{a} | = \sqrt{4 + 1 + 4} = 3$ : direction cosines $\frac{2}{3}, - \frac{1}{3}, \frac{2}{3}$ (check: $\frac{4}{9} + \frac{1}{9} + \frac{4}{9} = 1$ ).

Q12

·1 mark·Multiple choiceSection formula and collinearity

$A$ and $B$ have position vectors $\hat{i} + \hat{j}$ and $4 \hat{i} + 7 \hat{j}$ . The point $P$ dividing $A B$ internally in the ratio $1 : 2$ has position vector

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Answer: (c)

2 \hat{i} + 3 \hat{j}

Why: Section formula: (m·b + n·a)/(m + n) with m : n = 1 : 2.

$\vec{p} = \frac{1 (4 \hat{i} + 7 \hat{j}) + 2 (\hat{i} + \hat{j})}{3} = \frac{6 \hat{i} + 9 \hat{j}}{3} = 2 \hat{i} + 3 \hat{j}$ .

Q13

·1 mark·Multiple choiceScalar (dot) product

If $| \vec{a} | = 2$ , $| \vec{b} | = 3$ and $\vec{a} \cdot \vec{b} = 3$ , then the angle between $\vec{a}$ and $\vec{b}$ is

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Answer: (b)

\frac{π}{3}

Why: cos θ = 3/(2 × 3) = 1/2.

$\cos θ = \frac{3}{2 \times 3} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$ .

Q14

·1 mark·Multiple choiceSection formula and collinearity

The vectors $2 \hat{i} - \hat{j} + 3 \hat{k}$ and $- 4 \hat{i} + 2 \hat{j} + μ \hat{k}$ are collinear. Then $μ$ equals

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Answer: (d)

- 6

Why: Collinear means proportional components: −4/2 = 2/(−1) = μ/3 = −2.

$\frac{- 4}{2} = \frac{2}{- 1} = - 2$ , so $\frac{μ}{3} = - 2 \Rightarrow μ = - 6$ .

Q15

·1 mark·Multiple choiceScalar (dot) product

$\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $θ$ ( $0 \leq θ \leq π$ ). Then $| \hat{a} - \hat{b} |$ equals

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Answer: (b)

2 \sin \frac{θ}{2}

Why: |a − b|² = 1 + 1 − 2cos θ = 4 sin²(θ/2).

$| \hat{a} - \hat{b} |^{2} = | \hat{a} |^{2} + | \hat{b} |^{2} - 2 \hat{a} \cdot \hat{b} = 2 - 2 \cos θ = 4 \sin^{2} \frac{θ}{2}$ . As $\sin \frac{θ}{2} \geq 0$ , $| \hat{a} - \hat{b} | = 2 \sin \frac{θ}{2}$ .

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: Pushing a trolley (4 marks)

In a warehouse, a trolley is pushed in a straight line from the point $A (1, 2, 0)$ to the point $B (5, 4, 3)$ (coordinates in metres) by a constant force $\vec{F} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ newtons. The work done is $W = \vec{F} \cdot \vec{d}$ , where $\vec{d} = \vec{A B}$ .

(i) Find the displacement vector

\vec{d}

. [1 mark]

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Answer:

4 \hat{i} + 2 \hat{j} + 3 \hat{k}

\vec{d} = (5 - 1) \hat{i} + (4 - 2) \hat{j} + (3 - 0) \hat{k} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}

. A1

(ii) Find the distance moved. [1 mark]

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Answer:

\sqrt{29}

| \vec{d} | = \sqrt{16 + 4 + 9} = \sqrt{29}

m. A1

(iii) Find the work done by the force. [2 marks]

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Answer:

17

W = \vec{F} \cdot \vec{d} = 2 (4) + 3 (2) + 1 (3)

= 17

J. A1

OR Find the scalar projection of

\vec{F}

\vec{d}

. [2 marks]

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Answer:

\frac{17}{\sqrt{29}}

\frac{\vec{F} \cdot \vec{d}}{| \vec{d} |}

= \frac{17}{\sqrt{29}}

N. A1

Case study 2: Sail of a model boat (4 marks)

For a science project, Kavya makes a triangular sail for a model boat. In a coordinate frame (units: dm), its corners are $A (1, 0, 1)$ , $B (3, 2, 2)$ and $C (1, 3, 5)$ .

(i) Find the length of the edge

A B

. [1 mark]

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Answer:

3

\vec{A B} = 2 \hat{i} + 2 \hat{j} + \hat{k}

| \vec{A B} | = \sqrt{4 + 4 + 1} = 3

dm. A1

(ii) Find

\vec{A B} \cdot \vec{A C}

. [1 mark]

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Answer:

10

\vec{A C} = 3 \hat{j} + 4 \hat{k}

;

\vec{A B} \cdot \vec{A C} = 0 + 6 + 4 = 10

. A1

(iii) Find the area of the sail. [2 marks]

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Answer:

\frac{5 \sqrt{5}}{2} {dm}^{2}

\vec{A B} \times \vec{A C} = 5 \hat{i} - 8 \hat{j} + 6 \hat{k}

M1; area

= \frac{1}{2} \sqrt{25 + 64 + 36} = \frac{1}{2} \sqrt{125} = \frac{5 \sqrt{5}}{2} {dm}^{2}

. A1

OR Find a unit vector perpendicular to the plane of the sail. [2 marks]

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Answer:

\pm \frac{1}{5 \sqrt{5}} (5 \hat{i} - 8 \hat{j} + 6 \hat{k})

\vec{A B} \times \vec{A C} = 5 \hat{i} - 8 \hat{j} + 6 \hat{k}

, magnitude

5 \sqrt{5}

M1; unit normal

\pm \frac{1}{5 \sqrt{5}} (5 \hat{i} - 8 \hat{j} + 6 \hat{k})

. A1

Next steps for Vector Algebra

This free set is separate from the chapter's question bank. On the Vector Algebra chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 49 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.

Vector Algebra: 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.