If |{overrightarrow V _1} + {overrightarrow V | vedclass

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(C) Given: $|{\vec V_1} + {\vec V_2}| = |{\vec V_1} - {\vec V_2}|$Squaring both sides:$|{\vec V_1} + {\vec V_2}|^2 = |{\vec V_1} - {\vec V_2}|^2$Using

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${V_1}$ and ${V_2}$ are mutually perpendicular

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(C) Given: $|{\vec V_1} + {\vec V_2}| = |{\vec V_1} - {\vec V_2}|$Squaring both sides:$|{\vec V_1} + {\vec V_2}|^2 = |{\vec V_1} - {\vec V_2}|^2$Using the property $|\vec A \pm \vec B|^2 = |\vec A|^2 + |\vec B|^2 \pm 2(\vec A \cdot \vec B)$:$|{\vec V_1}|^2 + |{\vec V_2}|^2 + 2(\vec V_1 \cdot \vec V_2) = |{\vec V_1}|^2 + |{\vec V_2}|^2 - 2(\vec V_1 \cdot \vec V_2)$$2(\vec V_1 \cdot \vec V_2) = -2(\vec V_1 \cdot \vec V_2)$$4(\vec V_1 \cdot \vec V_2) = 0$$\vec V_1 \cdot \vec V_2 = 0$This implies that the vectors ${\vec V_1}$ and ${\vec V_2}$ are mutually perpendicular.

If $|{\overrightarrow V _1} + {\overrightarrow V _2}| = |{\overrightarrow V _1} - {\overrightarrow V _2}|$ and ${V_2}$ is finite,then

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