A vector A of magnitude 7√(2) unit, another vector B of magnitude 14 unit are inclined at an angle of 60°. The magnitude of the scalar product (A · B) is [cos (60)° = 1/2] (a) 49√(2) (b) 7√(2) (c) 14√(2) (d) 9√(2)

Answer: (a). A · B = ABcos θ = 7√(2) × 14 × 1/2 = 49√(2) So the answer is (a) 49√(2).

A vector A of magnitude 7√(2) unit, another vector B of magnitude 14… | MHT CET 2026 (19 May, Shift 2)

Q1MHT CET 2026 · 19 MAY · SHIFT 2Easy

Scalar (Dot) Product

A vector $\vec{A}$ of magnitude $7\sqrt{2}$ unit, another vector $\vec{B}$ of magnitude $14$ unit are inclined at an angle of $60^\circ$ . The magnitude of the scalar product $(\vec{A} \cdot \vec{B})$ is $\left[\cos(60)^\circ = \frac{1}{2}\right]$

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Answer: (a)

$\vec{A} \cdot \vec{B} = AB\cos\theta$
$= 7\sqrt{2} \times 14 \times \frac{1}{2} = 49\sqrt{2}$
So the answer is (a) $49\sqrt{2}$ .

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