Previous Year NEET Question- Optics L-5

Question: 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) $45^{\circ}$

B) $180^{\circ}$

C) $0^{\circ}$

D) $90^{\circ}$

Answer: $90^{\circ}$#

Sol:#

Let the two vectors are $\vec{A}$ and $\vec{B}$.

Given that, $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$ $$ \begin{aligned} & \therefore \sqrt{A^2+B^2+2 A B \cos \theta}=\sqrt{A^2+B^2-2 A B \cos \theta} \ & \Rightarrow 4 A B \cos \theta=0 \ & \because 4 A B \neq 0 \ & \therefore \cos \theta=0 \ & \Rightarrow \theta=90^{\circ} \end{aligned} $$