Q126 / 155 · WBJEE 2016 · Physics

The vectors $\mathrm{A}$ and $\mathrm{B}$ are such that $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}| .$ The angle between the two vectors will be

  • A
    $0^{\circ}$
  • B
    $60^{\circ}$
  • Correct answer
    C
    $90^{\circ}$
  • D
    $45^{\circ}$

Solution

Given, $|A+B|=|A-B|$ $\therefore \quad(A+B)^{2}=(A-B)^{2}$ $A^{2}+B^{2}+2 A B \cos \theta=A^{2}+B^{2}-2 A B \cos \theta$ $2 A B \cos \theta+2$ AB $\cos \theta=0$ $4 A B \cos \theta=0$ Here, $\quad A \neq 0, B \neq 0$ $\therefore \quad \cos \theta=0$ $\theta=90^{\circ}$

Hence, the correct answer is option: 'C'.

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