MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2025: question 36

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedVector Algebra

Case Study – 1

Three friends A, B and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his predecided destination, following straight paths from A to C and B to C in such a way that $O A = a$ , $O B = b$ and $O C = 5 a - 2 b$ respectively.

Based upon the above information, answer the following questions :

(i)
Complete the given figure to explain their entire movement plan along the respective vectors.
1 mark
(ii)
Find vectors $A C$ and $B C$ .
1 mark
(iii)
(a) If $a \cdot b = 1$ , distance of O to A is 1 km and that from O to B is 2 km, then find the angle between $O A$ and $O B$ . Also, find $∣ a \times b ∣$ .
2 marks
OR
(iii)
(b) If $a = 2 \hat{i} - \hat{j} + 4 \hat{k}$ and $b = \hat{j} - \hat{k}$ , then find a unit vector perpendicular to $(a + b)$ and $(a - b)$ .
2 marks

Figure in the paper: Point O at left with three arrows from it: OA along vector a to point A (horizontal, right); OB along vector b to point B (upper right); OC along vector 5a − 2b to point C (lower right). Segments AC, BC, AB are not drawn (to be completed). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

A vector from point P to point Q equals (position vector of Q) − (position vector of P).

Hint 2

Use $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ and $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ .

Hint 3

A vector perpendicular to two vectors is their cross product; divide by its magnitude to make it a unit vector.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Complete the figure by drawing the straight path from A to C (arrow from A towards C) and from B to C (arrow from B towards C), so A moves $O \to A \to C$ and B moves $O \to B \to C$ , while C moves directly $O \to C$ along $5 a - 2 b$ ; the three meet at C.
1 mark
(ii) $A C = O C - O A = (5 a - 2 b) - a = 4 a - 2 b$ .
½ mark
(ii) $B C = O C - O B = (5 a - 2 b) - b = 5 a - 3 b$ .
½ mark
(iii)(a) Given $∣ a ∣ = 1$ , $∣ b ∣ = 2$ , $a \cdot b = 1$ . Angle $θ = cos^{- 1} (\frac{a \cdot b}{∣ a ∣∣ b ∣}) = cos^{- 1} (\frac{1}{1 \times 2}) = \frac{π}{3}$ .
1 mark
(iii)(a) $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ = 1 \times 2 \times \frac{3}{2} = 3$ .
1 mark
(iii)(b) $a + b = 2 \hat{i} + 0 \hat{j} + 3 \hat{k}$ and $a - b = 2 \hat{i} - 2 \hat{j} + 5 \hat{k}$ .
1 mark
(iii)(b) A vector perpendicular to both: $c = (a + b) \times (a - b) = \hat{i} 22 \hat{j} 0 - 2 \hat{k} 35 = (0 + 6) \hat{i} - (10 - 6) \hat{j} + (- 4 - 0) \hat{k} = 6 \hat{i} - 4 \hat{j} - 4 \hat{k}$ , with $∣ c ∣ = 36 + 16 + 16 = 68 = 217$ .
½ mark
(iii)(b) Required unit vector: $\overset{c}{^} = \frac{1}{2 17} (6 \hat{i} - 4 \hat{j} - 4 \hat{k}) = \frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ (its negative is also acceptable).
½ mark

Answer: (i) Draw arrows A→C and B→C; (ii) $A C = 4 a - 2 b$ , $B C = 5 a - 3 b$ ; (iii)(a) $θ = \frac{π}{3}$ , $∣ a \times b ∣ = 3$ ; (iii)(b) $\frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$

Where the marks are

(i) Completes figure with paths A→C and B→C (1 mark)
(ii) $A C = 4 a - 2 b$ (½ mark) and $B C = 5 a - 3 b$ (½ mark)
(iii)(a) Angle $\frac{π}{3}$ (1 mark); $∣ a \times b ∣ = 3$ (1 mark)
(iii)(b) Finds $a + b$ and $a - b$ (½ + ½ mark); cross product $6 \hat{i} - 4 \hat{j} - 4 \hat{k}$ with magnitude $68$ (½ mark); unit vector $\frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ (½ mark)

Revise the chapter

Chapter: Vector Algebra (notes, formulas, where marks are lost and every past question)
Topic: Section Formula and Collinearity of Vectors

Practise this chapterBoard-style questions from Vector Algebra, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2025: question 36

Hints

Step-by-step solution

Revise the chapter

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