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Physics Question on Motion in a plane

If ∣A⃗+B⃗∣=∣A⃗−B⃗∣\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right|, then the angle between A⃗\vec{A} and B⃗\vec{B} will be

A

30?30^?

B

45?45^?

C

60?60^?

D

90?90^?

Answer

90?90^?

Explanation

Solution

Let θ\theta be the angle between the vectors A⃗\vec{A} and B⃗\vec{B}. Then ∣A⃗+B⃗∣=A2+B2+2AB cos θ\left|\vec{A}+\vec{B}\right|=\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta} ∣A⃗−B⃗∣=A2+B2−2AB cos θ\left|\vec{A}-\vec{B}\right|=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta} According to given problem ∣A⃗+B⃗∣=∣A⃗−B⃗∣\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right| ∴A2+B2+2AB cos θ\therefore\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta} =A2+B2−2AB cos θ=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta} Squaring both sides, we get A2+B2+2ABcosθ=A2+B2−2ABcosθA^{2}+B^{2}+2ABcos\theta=A^{2}+B^{2}-2ABcos\theta ∴4ABcosθ=0\therefore 4ABcos\theta=0 As A≠0A\ne0, B≠0B\ne0 ∴cosθ=0\therefore cos\theta=0 or θ=90∘\theta=90^{\circ}