If a+b+c=0, a =√(37), b =3 and c =4, then angle between b and c is (A) (π)/(6) (B) (π)/(4) (C) (π)/(3) (D) (π)/(2)

MathsCBSE Class 12 · Board exam 2025, set 65/1/1 · 1 mark · MCQ

CBSE Class 12 Mathematics 2025: question 11

Set in the CBSE Class 12 board exam of 2025, paper 65/1/1, section A, question 11, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q11 · 1 mark · MCQVector Algebra

If $a + b + c = 0$ , $∣ a ∣ = 37$ , $∣ b ∣ = 3$ and $∣ c ∣ = 4$ , then angle between $b$ and $c$ is

(A) $\frac{π}{6}$
(B) $\frac{π}{4}$
(C) $\frac{π}{3}$
(D) $\frac{π}{2}$

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

Hints

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

Hint 1

Write $b + c$ in terms of $a$ .

Hint 2

Take the magnitude squared of both sides and expand.

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.

$b + c = - a$ , so $∣ a ∣^{2} = ∣ b ∣^{2} + ∣ c ∣^{2} + 2∣ b ∣∣ c ∣ cos θ$ .
$37 = 9 + 16 + 24 cos θ \Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$ , option (C).
1 mark

Answer: $\frac{π}{3}$ , option (C)

Where the marks are

Squares $b + c = - a$ and gets $cos θ = \frac{1}{2}$ (1 mark)

Revise the chapter

Chapter: Vector Algebra (notes, formulas, where marks are lost and every past question)
Topic: Scalar (Dot) Product 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 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 11

Hints

Step-by-step solution

Revise the chapter

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