For the sum $(\vec{A} + \vec{B})$ , difference $(\vec{A} - \vec{B})$ and cross product $(\vec{A} \times \vec{B})$ of two non zero and non linear vectors $\vec{A}$ and $\vec{B}$ to be mutually perpendicular, the compulsory condition is

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$\vec{A}$ and $\vec{B}$ are perpendicular to each other and $| \vec{A} | = | \vec{B} |$

$\vec{A}$ and $\vec{B}$ perpendicular but their magnitudes can be of any non zero value

$| \vec{A} | = | \vec{B} |$ and their directions can be in any non collinear directions

Any one of the above

answer is C.

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Detailed Solution

Let $\vec{A} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{B} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

Given that $\vec{A} + \vec{B}$ is perpendicular to $\vec{A} - \vec{B}$ i.e., $(\vec{A} + \vec{B}) . (\vec{A} - \vec{B}) = 0$ or $(a_{1} + b_{1}) (a_{1} - b_{1}) + (a_{2} + b_{2}) (a_{2} - b_{2}) + (a_{3} + b_{3}) (a_{3} - b_{3}) = 0$ or $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ or $| \vec{A} | = | \vec{B} |$

Cross product of $\vec{A}$ and $\vec{B}$ is perpendicular to the plane formed by $\vec{A}$ and $\vec{B}$ or $\vec{A} + \vec{B}$ and $\vec{A} - \vec{B}$