Question

if mod a vector plus b vector = mod A vector minus b vector, show that a vector is perpendicular to b vector.

Please bhai log answer Karo. ​

Answer 1

We're given,

$\longrightarrow\left|\vec{\sf{A}}+\vec{\sf{B}}\right|=\left|\vec{\sf{A}}-\vec{\sf{B}}\right|$

If $\sf{\theta}$ is the angle between $\vec{\sf{A}}$ and $\vec{\sf{B}},$

$\longrightarrow\sf{\sqrt{A^2+B^2+2AB\cos\theta}=\sqrt{A^2+B^2-2AB\cos\theta}}$

Squaring both sides,

$\longrightarrow\sf{A^2+B^2+2AB\cos\theta=A^2+B^2-2AB\cos\theta}$

Cancelling like terms,

$\longrightarrow\sf{2AB\cos\theta=-2AB\cos\theta}$

Dividing both sides by $\sf{2AB}$ since $\sf{A,\ B\neq 0,}$

$\longrightarrow\sf{\cos\theta=-\cos\theta}$

$\longrightarrow\sf{\cos\theta+\cos\theta=0}$

$\longrightarrow\sf{2\cos\theta=0}$

$\longrightarrow\sf{\cos\theta=0}$

This implies, since $\sf{0^o\leq\theta\leq180^o,}$

$\longrightarrow\sf{\underline{\underline{\theta=90^o}}}$

This implies the vectors $\vec{\sf{A}}$ and $\vec{\sf{B}}$ are perpendicular to each other.

Answer 2

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