If |vec{v}_1 + vec{v}_2| = |vec{v}_1 - vec{v} | vedclass

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(D) Given the equation $|\vec{v}_1 + \vec{v}_2| = |\vec{v}_1 - \vec{v}_2|$.Squaring both sides,we get $|\vec{v}_1 + \vec{v}_2|^2 = |\vec{v}_1 - \vec{v

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$\vec{v}_1$ and $\vec{v}_2$ are mutually perpendicular

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(D) Given the equation $|\vec{v}_1 + \vec{v}_2| = |\vec{v}_1 - \vec{v}_2|$.Squaring both sides,we get $|\vec{v}_1 + \vec{v}_2|^2 = |\vec{v}_1 - \vec{v}_2|^2$.Using the property $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$,we expand the terms:$(\vec{v}_1 + \vec{v}_2) \cdot (\vec{v}_1 + \vec{v}_2) = (\vec{v}_1 - \vec{v}_2) \cdot (\vec{v}_1 - \vec{v}_2)$.$|\vec{v}_1|^2 + |\vec{v}_2|^2 + 2(\vec{v}_1 \cdot \vec{v}_2) = |\vec{v}_1|^2 + |\vec{v}_2|^2 - 2(\vec{v}_1 \cdot \vec{v}_2)$.Subtracting $|\vec{v}_1|^2 + |\vec{v}_2|^2$ from both sides,we get $2(\vec{v}_1 \cdot \vec{v}_2) = -2(\vec{v}_1 \cdot \vec{v}_2)$.This implies $4(\vec{v}_1 \cdot \vec{v}_2) = 0$,which means $\vec{v}_1 \cdot \vec{v}_2 = 0$.Since the dot product of two non-zero vectors is zero,they must be mutually perpendicular.

Column $I$	Column $II$
$(A)$ $\theta=60^{\circ}$	$(p)$ $n=\sqrt{3}$
$(B)$ $\theta=90^{\circ}$	$(q)$ $n=1$
$(C)$ $\theta=120^{\circ}$	$(r)$ $n=\sqrt{2}$
$(D)$ $\theta=180^{\circ}$	$(s)$ $n=2$

If $|\vec{v}_1 + \vec{v}_2| = |\vec{v}_1 - \vec{v}_2|$ and $\vec{v}_1$ and $\vec{v}_2$ are finite,then:

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