The vector sum of two forces vec{A} and vec{B | vedclass

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(D) Let the two forces be $\vec{A}$ and $\vec{B}$.The vector sum is $(\vec{A} + \vec{B})$ and the vector difference is $(\vec{A} - \vec{B})$.Given tha

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equal in magnitude.

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(D) Let the two forces be $\vec{A}$ and $\vec{B}$.The vector sum is $(\vec{A} + \vec{B})$ and the vector difference is $(\vec{A} - \vec{B})$.Given that the sum is perpendicular to the difference,their dot product must be zero:$(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = 0$Expanding the dot product:$\vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} - \vec{B} \cdot \vec{B} = 0$Since the dot product is commutative,$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$,so the terms cancel out:$|\vec{A}|^2 - |\vec{B}|^2 = 0$$|\vec{A}|^2 = |\vec{B}|^2$$|\vec{A}| = |\vec{B}|$Therefore,the forces are equal in magnitude.

Column $I$	Column $II$
$(A)$ Component along $x$-axis	$(p)$ $5 \text{ unit}$
$(B)$ Component along vector $(2 \hat{i} + \hat{j} + 2 \hat{k})$	$(q)$ $4 \text{ unit}$
$(C)$ Component along $(6 \hat{i} + 8 \hat{j} - 10 \hat{k})$	$(r)$ $0$
$(D)$ Component along $(-3 \hat{i} - 4 \hat{j} + 5 \hat{k})$	$(s)$ None

The vector sum of two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence,the forces $\vec{A}$ and $\vec{B}$ are

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