Question

If the magnitude of sum of the two vectors is equal to the magnitude of difference of the two vectors, then the angle between these vectors is

• A. ${45}^o$
• B. $0^o$
• C. ${90}^{\circ }$
• D. ${180}^o$

Answer 1

The correct option is C ${90}^{\circ }$

Let the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ having magnitude $A$ and $B$ respectively.

Then the magnitude of their sum is given by:
$$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+B^2+2AB\cos \theta }$$
Where, $\theta =$ angle between the vectors

The magnitude of their difference is given by:
$$|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{A^2+B^2-2AB\cos \theta }$$
Given, $$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$$
$$\sqrt{A^2+B^2+2AB\cos \theta }=\sqrt{A^2+B^2-2AB\cos \theta }$$
On squaring both side, we get
$$A^2+B^2+2AB\cos \theta =A^2+B^2-2AB\cos \theta$$
$$4AB\cos \theta =0\Rightarrow \cos \theta =0$$
$$\therefore \theta ={90}^{\circ }$$
Therefore, option $\left(\text{C}\right)$ is the correct answer.