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Physics Question on Motion in a plane

If A⃗\vec{A} and B⃗\vec{B} are non-zero vectors which obey the relation ∣A⃗+B⃗∣=∣A⃗−B⃗∣\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right|, then the angle between them is

A

0∘^{\circ}

B

60∘^{\circ}

C

90∘^{\circ}

D

120∘^{\circ}

Answer

90∘^{\circ}

Explanation

Solution

∣A⃗+B⃗∣=∣A⃗−B⃗∣\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right| ∣A⃗∣2+∣B⃗∣2+2∣A⃗∣∣B⃗∣cosθ\left|\vec{A}\right|^{2}+\left|\vec{B}\right|^{2}+2\left|\vec{A}\right|\left|\vec{B}\right|cos \theta =∣A⃗∣2+∣B⃗∣2−2∣A⃗∣∣B⃗∣cosθ=\left|\vec{A}\right|^{2}+\left|\vec{B}\right|^{2}-2\left|\vec{A}\right|\left|\vec{B}\right| cos \theta ∴cosθ=0⇒θ=90∘\therefore\quad cos\theta=0 \Rightarrow \theta= 90^{\circ}