For two vectors P and Q P · Q = |P × Q| The magnitude of R = P + Q is ( cos 45° = 1/√(2))? (a) √(P² + Q²) (b) (√(P² + Q²))/(√(2)) (c) √(P² + Q² + (PQ)/(√(2))) (d) √(P² + Q² + √(2) PQ)

Answer: (d). PQcos θ = PQsin θ ⇒ tan θ = 1 ⇒ θ = 45°. R² = P² + Q² + 2PQcos 45° = P² + Q² + √(2) PQ So the answer is (d) √(P² + Q² + √(2) PQ).

For two vectors P and Q P · Q = |P × Q| The magnitude of R = P + Q is… | MHT CET 2026 (19 Apr, Shift 1)

Q3MHT CET 2026 · 19 APR · SHIFT 1Medium

Vector (Cross) Product

For two vectors $\vec{P}$ and $\vec{Q} \quad \vec{P} \cdot \vec{Q} = |\vec{P} \times \vec{Q}|$

The magnitude of $\vec{R} = \vec{P} + \vec{Q}$ is ( $\cos 45^\circ = \frac{1}{\sqrt{2}}$ )?

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Answer: (d)

$PQ\cos\theta = PQ\sin\theta \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ$ .
$R^2 = P^2 + Q^2 + 2PQ\cos 45^\circ = P^2 + Q^2 + \sqrt{2}\,PQ$
So the answer is (d) $\sqrt{P^2 + Q^2 + \sqrt{2}\,PQ}$ .