The vector sum of two forces is perpendicular | vedclass

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(B) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$.The vector sum is $(\vec{F}_1 + \vec{F}_2)$ and the vector difference is $(\vec{F}_1 - \vec{F}_2

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have equal magnitudes.

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(B) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$.The vector sum is $(\vec{F}_1 + \vec{F}_2)$ and the vector difference is $(\vec{F}_1 - \vec{F}_2)$.Given that the sum is perpendicular to the difference,their dot product must be zero:$(\vec{F}_1 + \vec{F}_2) \cdot (\vec{F}_1 - \vec{F}_2) = 0$Expanding the dot product:$\vec{F}_1 \cdot \vec{F}_1 - \vec{F}_1 \cdot \vec{F}_2 + \vec{F}_2 \cdot \vec{F}_1 - \vec{F}_2 \cdot \vec{F}_2 = 0$Since the dot product is commutative $(\vec{F}_1 \cdot \vec{F}_2 = \vec{F}_2 \cdot \vec{F}_1)$,the middle terms cancel out:$|\vec{F}_1|^2 - |\vec{F}_2|^2 = 0$Therefore,$|\vec{F}_1|^2 = |\vec{F}_2|^2$,which implies $|\vec{F}_1| = |\vec{F}_2|$.Thus,the forces have equal magnitudes.

The vector sum of two forces is perpendicular to their vector difference. In this case,the forces:

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