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Physics · MOTION IN A PLANE
हिन्दीThe resultant of two vectors \( \mathbf{P} \) and \( \mathbf{Q} \) is perpendicular to \( \mathbf{P} \). If the magnitude of the resultant is equal to half of the magnitude of \( \mathbf{Q} \), then the angle between \( \mathbf{P} \) and \( \mathbf{Q} \) is:
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Correct answer · Option D
\( 150^\circ \)
Explanation
Since \( \mathbf{R} \perp \mathbf{P} \), \( P + Q\cos\theta = 0 \implies P = -Q\cos\theta \). In the right triangle, \( R^2 + P^2 = Q^2 \implies (Q/2)^2 + P^2 = Q^2 \implies P = \frac{\sqrt{3}}{2}Q \). Thus \( \cos\theta = -\frac{\sqrt{3}}{2} \implies \theta = 150^\circ \).
