Question

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

• A. $30^?$
• B. $45^?$
• C. $60^?$
• D. $90^?$

Answer 1

The Correct Option is D

Let $\theta$ be the angle between the vectors $\vec{A}$ and $\vec{B}$. Then $\left|\vec{A}+\vec{B}\right|=\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta}$ $\left|\vec{A}-\vec{B}\right|=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta}$ According to given problem $\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right|$ $\therefore\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta}$ $=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta}$ Squaring both sides, we get $A^{2}+B^{2}+2ABcos\theta=A^{2}+B^{2}-2ABcos\theta$ $\therefore 4ABcos\theta=0$ As $A\ne0$, $B\ne0$ $\therefore cos\theta=0$ or $\theta=90^{\circ}$