The correct option is C sin( P+Q ) | O⃗X⃗×O⃗Y⃗| = |O⃗X⃗||O⃗Y⃗|sinθ , where θ is the angle between O⃗X⃗ and O⃗Y⃗ . |O⃗X⃗| = |O⃗Y⃗ ...

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Applications of Cross Product

| O⃗X⃗×O⃗Y⃗| ...

Question

$∣ ∣ \to O X \times \to O Y ∣ ∣ =$

sin (P + R)

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sin 2 R

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sin (Q + R)

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sin (P + Q)

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Solution

The correct option is C $sin (P + Q)$
$∣ ∣ \to O X \times \to O Y ∣ ∣ = ∣ ∣ \to O X ∣ ∣ ∣ ∣ \to O Y ∣ ∣ sin θ$ , where $θ$ is the angle between $\to O X$ and $\to O Y$ .

$∣ ∣ \to O X ∣ ∣ = ∣ ∣ \to O Y ∣ ∣ = 1$ since they are both unit vectors.
Also since $\to O X ∥ \to Q R$ and $\to O Y ∥ \to R P$ , $θ$ is also the angle between $\to Q R$ and $\to R P$ , i.e., $∠ R$

$∴ ∣ ∣ \to O X \times \to O Y ∣ ∣ = 1.1. sin R$
Now, since $P Q R$ is a triangle, $P + Q + R = π \Rightarrow R = π - (P + Q)$
$∴ ∣ ∣ \to O X \times \to O Y ∣ ∣$ $= sin (π - (P + Q)) = sin (P + Q)$

Hence, option D is correct.

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