Question

The magnitude of the sum of the two vectors is equal to the difference of their magnitudes, the angle between the vectors is,

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

Answer 1

Answer: (C)

Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively. Then magnitude of their sum is given by:
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+B^2+2AB\cos \theta }\to (1)(\theta =$ angle between the vectors $)$
Magnitude of their difference is given by:
$|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{A^2+B^2-2AB\cos \theta }\to (2)$
As $|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$
$\Rightarrow A^2+B^2+2AB\cos \theta =A^2+B^2-2AB\cos \theta$
$\Rightarrow 4AB\cos \theta =0$ or $\cos \theta =0$
$\Rightarrow \theta =90^{\circ }.$