Explanation:
According to the problem, $|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}-\vec{\text{B}}|$
$\Rightarrow\ \sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}\\=\sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}$
$\Rightarrow\ |\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta\\=|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$
$\Rightarrow\ 4|\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$\Rightarrow\ |\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$|\vec{\text{A}}|=0\text{ or }|\vec{\text{B}}|=0\text{ or }\cos\theta=0$
i.e. $\theta=90^\circ$
When $\theta=90^\circ,$ we can say that $\vec{\text{A}}\bot\vec{\text{B}}.$
Hence options (b) and (d) are correct.
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