The magnitude of \(4\hat{i} - 4\hat{j} + 7\hat{k}\) is
- (a)\(9\)
- (b)\(7\)
- (c)\(81\)
- (d)\(\sqrt{15}\)
Show answer
Answer: (a) \(9\)
Why: √(16 + 16 + 49) = √81.
\(\sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{81} = 9\).
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)
- No calculator needed
- Free, no sign-in
Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The unit vector in the direction of \(\hat{i} + 2\hat{j} - 2\hat{k}\) is
- (a)\(\hat{i} + 2\hat{j} - 2\hat{k}\)
- (b)\(\dfrac13\left(\hat{i} + 2\hat{j} - 2\hat{k}\right)\)
- (c)\(\dfrac19\left(\hat{i} + 2\hat{j} - 2\hat{k}\right)\)
- (d)\(\dfrac{1}{\sqrt5}\left(\hat{i} + 2\hat{j} - 2\hat{k}\right)\)
Show answer
Answer: (b) \(\dfrac13\left(\hat{i} + 2\hat{j} - 2\hat{k}\right)\)
Why: Divide by the magnitude √(1 + 4 + 4) = 3.
\(|\vec a| = \sqrt{1 + 4 + 4} = 3\), so \(\hat a = \dfrac13\left(\hat{i} + 2\hat{j} - 2\hat{k}\right)\).
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
- (a)\(5\)
- (b)\(-1\)
- (c)\(3\)
- (d)\(7\)
Show answer
Answer: (c) \(3\)
Why: Multiply matching components and add: 2 − 2 + 3.
\(\vec a \cdot \vec b = 1(2) + 2(-1) + 3(1) = 3\).
The angle between \(\vec a = \hat{i} + \hat{j}\) and \(\vec b = \hat{j} + \hat{k}\) is
- (a)\(\dfrac{\pi}{2}\)
- (b)\(\dfrac{\pi}{6}\)
- (c)\(\dfrac{\pi}{4}\)
- (d)\(\dfrac{\pi}{3}\)
Show answer
Answer: (d) \(\dfrac{\pi}{3}\)
Why: cos θ = (a·b)/(|a||b|) = 1/(√2 · √2).
\(\cos\theta = \dfrac{0 + 1 + 0}{\sqrt2 \cdot \sqrt2} = \dfrac12 \Rightarrow \theta = \dfrac{\pi}{3}\).
For which value of \(\lambda\) is \(3\hat{i} + \lambda\hat{j} - 2\hat{k}\) perpendicular to \(2\hat{i} + 4\hat{j} + \hat{k}\)?
- (a)\(-1\)
- (b)\(1\)
- (c)\(2\)
- (d)\(-\dfrac12\)
Show answer
Answer: (a) \(-1\)
Why: Dot product zero: 6 + 4λ − 2 = 0.
\(3(2) + \lambda(4) + (-2)(1) = 4 + 4\lambda = 0 \Rightarrow \lambda = -1\).
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
- (a)\(\dfrac49\)
- (b)\(\dfrac43\)
- (c)\(4\)
- (d)\(\dfrac34\)
Show answer
Answer: (b) \(\dfrac43\)
Why: Projection = a·b/|b| = 4/3.
\(\vec a \cdot \vec b = 2 - 2 + 4 = 4\), \(|\vec b| = 3\). Projection \(= \dfrac{4}{3}\).
\(\hat{k} \times \hat{i}\) equals
- (a)\(\hat{i}\)
- (b)\(-\hat{j}\)
- (c)\(\vec 0\)
- (d)\(\hat{j}\)
Show answer
Answer: (d) \(\hat{j}\)
Why: Cyclic order i → j → k → i: k × i = j.
In the cyclic order \(\hat i, \hat j, \hat k\): \(\hat i \times \hat j = \hat k\), \(\hat j \times \hat k = \hat i\), \(\hat k \times \hat i = \hat j\).
If \(|\vec a| = 3\), \(|\vec b| = 4\) and the angle between them is \(\dfrac{\pi}{6}\), then \(|\vec a \times \vec b|\) is
- (a)\(6\sqrt3\)
- (b)\(12\)
- (c)\(6\)
- (d)\(3\)
Show answer
Answer: (c) \(6\)
Why: |a × b| = |a||b| sin θ = 12 × 1/2.
\(|\vec a \times \vec b| = 3 \times 4 \times \sin\dfrac{\pi}{6} = 6\). (\(6\sqrt3\) would be \(\vec a \cdot \vec b\).)
\((\hat{i} + \hat{j}) \times (\hat{i} - \hat{j})\) equals
- (a)\(2\hat{k}\)
- (b)\(-2\hat{k}\)
- (c)\(\vec 0\)
- (d)\(2\hat{i}\)
Show answer
Answer: (b) \(-2\hat{k}\)
Why: Only the k-component survives: (1)(−1) − (1)(1) = −2.
\(\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} = \hat i(0) - \hat j(0) + \hat k(-1 - 1) = -2\hat k\).
The area of the parallelogram with adjacent sides \(\vec a = 2\hat{i} + \hat{j}\) and \(\vec b = \hat{i} + 3\hat{j}\) is
- (a)\(7\) sq units
- (b)\(10\) sq units
- (c)\(\sqrt{50}\) sq units
- (d)\(5\) sq units
Show answer
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.
The direction cosines of the vector \(2\hat{i} - \hat{j} + 2\hat{k}\) are
- (a)\(\dfrac23, -\dfrac13, \dfrac23\)
- (b)\(2, -1, 2\)
- (c)\(\dfrac23, \dfrac13, \dfrac23\)
- (d)\(\dfrac{2}{\sqrt5}, -\dfrac{1}{\sqrt5}, \dfrac{2}{\sqrt5}\)
Show answer
Answer: (a) \(\dfrac23, -\dfrac13, \dfrac23\)
Why: Divide each component by the magnitude 3.
\(|\vec a| = \sqrt{4 + 1 + 4} = 3\): direction cosines \(\dfrac23, -\dfrac13, \dfrac23\) (check: \(\tfrac49 + \tfrac19 + \tfrac49 = 1\)).
\(A\) and \(B\) have position vectors \(\hat{i} + \hat{j}\) and \(4\hat{i} + 7\hat{j}\). The point \(P\) dividing \(AB\) internally in the ratio \(1 : 2\) has position vector
- (a)\(3\hat{i} + 5\hat{j}\)
- (b)\(\dfrac52\hat{i} + 4\hat{j}\)
- (c)\(2\hat{i} + 3\hat{j}\)
- (d)\(3\hat{i} + 6\hat{j}\)
Show answer
Answer: (c) \(2\hat{i} + 3\hat{j}\)
Why: Section formula: (m·b + n·a)/(m + n) with m : n = 1 : 2.
\(\vec p = \dfrac{1(4\hat{i} + 7\hat{j}) + 2(\hat{i} + \hat{j})}{3} = \dfrac{6\hat i + 9\hat j}{3} = 2\hat{i} + 3\hat{j}\).
If \(|\vec a| = 2\), \(|\vec b| = 3\) and \(\vec a \cdot \vec b = 3\), then the angle between \(\vec a\) and \(\vec b\) is
- (a)\(\dfrac{\pi}{6}\)
- (b)\(\dfrac{\pi}{3}\)
- (c)\(\dfrac{\pi}{2}\)
- (d)\(\dfrac{2\pi}{3}\)
Show answer
Answer: (b) \(\dfrac{\pi}{3}\)
Why: cos θ = 3/(2 × 3) = 1/2.
\(\cos\theta = \dfrac{3}{2 \times 3} = \dfrac12 \Rightarrow \theta = \dfrac{\pi}{3}\).
The vectors \(2\hat{i} - \hat{j} + 3\hat{k}\) and \(-4\hat{i} + 2\hat{j} + \mu\hat{k}\) are collinear. Then \(\mu\) equals
- (a)\(6\)
- (b)\(\dfrac32\)
- (c)\(-3\)
- (d)\(-6\)
Show answer
Answer: (d) \(-6\)
Why: Collinear means proportional components: −4/2 = 2/(−1) = μ/3 = −2.
\(\dfrac{-4}{2} = \dfrac{2}{-1} = -2\), so \(\dfrac{\mu}{3} = -2 \Rightarrow \mu = -6\).
\(\hat a\) and \(\hat b\) are unit vectors inclined at an angle \(\theta\) (\(0 \le \theta \le \pi\)). Then \(|\hat a - \hat b|\) equals
- (a)\(2\cos\dfrac{\theta}{2}\)
- (b)\(2\sin\dfrac{\theta}{2}\)
- (c)\(2\sin\theta\)
- (d)\(\sin\dfrac{\theta}{2}\)
Show answer
Answer: (b) \(2\sin\dfrac{\theta}{2}\)
Why: |a − b|² = 1 + 1 − 2cos θ = 4 sin²(θ/2).
\(|\hat a - \hat b|^2 = |\hat a|^2 + |\hat b|^2 - 2\hat a \cdot \hat b = 2 - 2\cos\theta = 4\sin^2\dfrac{\theta}{2}\). As \(\sin\tfrac{\theta}{2} \ge 0\), \(|\hat a - \hat b| = 2\sin\dfrac{\theta}{2}\).
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.
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 \(\vec d = \overrightarrow{AB}\).
(i) Find the displacement vector \(\vec d\). [1 mark]
Show answer
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]
Show answer
Answer: \(\sqrt{29}\) m
\(|\vec d| = \sqrt{16 + 4 + 9} = \sqrt{29}\) m. A1
(iii) Find the work done by the force. [2 marks]
Show answer
Answer: \(17\) J
\(W = \vec F \cdot \vec d = 2(4) + 3(2) + 1(3)\) M1 \(= 17\) J. A1
OR Find the scalar projection of \(\vec F\) on \(\vec d\). [2 marks]
Show answer
Answer: \(\dfrac{17}{\sqrt{29}}\) N
\(\dfrac{\vec F \cdot \vec d}{|\vec d|}\) M1 \(= \dfrac{17}{\sqrt{29}}\) 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 \(AB\). [1 mark]
Show answer
Answer: \(3\) dm
\(\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}\), \(|\overrightarrow{AB}| = \sqrt{4 + 4 + 1} = 3\) dm. A1
(ii) Find \(\overrightarrow{AB} \cdot \overrightarrow{AC}\). [1 mark]
Show answer
Answer: \(10\)
\(\overrightarrow{AC} = 3\hat{j} + 4\hat{k}\); \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 0 + 6 + 4 = 10\). A1
(iii) Find the area of the sail. [2 marks]
Show answer
Answer: \(\dfrac{5\sqrt5}{2}\ \text{dm}^2\)
\(\overrightarrow{AB} \times \overrightarrow{AC} = 5\hat{i} - 8\hat{j} + 6\hat{k}\) M1; area \(= \dfrac12\sqrt{25 + 64 + 36} = \dfrac12\sqrt{125} = \dfrac{5\sqrt5}{2}\ \text{dm}^2\). A1
OR Find a unit vector perpendicular to the plane of the sail. [2 marks]
Show answer
Answer: \(\pm\dfrac{1}{5\sqrt5}\left(5\hat{i} - 8\hat{j} + 6\hat{k}\right)\)
\(\overrightarrow{AB} \times \overrightarrow{AC} = 5\hat{i} - 8\hat{j} + 6\hat{k}\), magnitude \(5\sqrt5\) M1; unit normal \(\pm\dfrac{1}{5\sqrt5}\left(5\hat{i} - 8\hat{j} + 6\hat{k}\right)\). 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 Essentials adds all 39 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 and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
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 re-checked by computer algebra and 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.