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If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
A
90°
B
45°
C
180°
D
0°
Explanation
$|P+Q| = |P-Q|$ only when $\cos\theta = 0$.
Detailed Solution
$|\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}|$
Squaring: $P^2 + Q^2 + 2PQ\cos\theta = P^2 + Q^2 - 2PQ\cos\theta$
$\Rightarrow 4PQ\cos\theta = 0 \Rightarrow \cos\theta = 0$
$\theta = \frac{\pi}{2} = 90^\circ$
Squaring: $P^2 + Q^2 + 2PQ\cos\theta = P^2 + Q^2 - 2PQ\cos\theta$
$\Rightarrow 4PQ\cos\theta = 0 \Rightarrow \cos\theta = 0$
$\theta = \frac{\pi}{2} = 90^\circ$