The vector sum of two forces is perpendicular | vedclass

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Question

(A) Let the two force vectors be $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$.Given that their vector sum $(\overrightarrow{F_1} + \overrightarro

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Are equal to each other in magnitude

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(A) Let the two force vectors be $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$.Given that their vector sum $(\overrightarrow{F_1} + \overrightarrow{F_2})$ is perpendicular to their vector difference $(\overrightarrow{F_1} - \overrightarrow{F_2})$.Two vectors are perpendicular if their dot product is zero.Therefore,$(\overrightarrow{F_1} + \overrightarrow{F_2}) \cdot (\overrightarrow{F_1} - \overrightarrow{F_2}) = 0$.Expanding the dot product:$\overrightarrow{F_1} \cdot \overrightarrow{F_1} - \overrightarrow{F_1} \cdot \overrightarrow{F_2} + \overrightarrow{F_2} \cdot \overrightarrow{F_1} - \overrightarrow{F_2} \cdot \overrightarrow{F_2} = 0$.Since the dot product is commutative,$\overrightarrow{F_1} \cdot \overrightarrow{F_2} = \overrightarrow{F_2} \cdot \overrightarrow{F_1}$,so these terms cancel out.We are left with $|\overrightarrow{F_1}|^2 - |\overrightarrow{F_2}|^2 = 0$.This implies $|\overrightarrow{F_1}|^2 = |\overrightarrow{F_2}|^2$,or $|\overrightarrow{F_1}| = |\overrightarrow{F_2}|$.Thus,the forces are equal in magnitude.

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

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