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The vector sum of two forces is perpendicular to their vector differences. In that case, the force:
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The correct answer is:
are equal to each other in magnitude
$(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0$
$\begin{aligned}
& \Rightarrow A^2-\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}+B^2=0 \\
& \Rightarrow \mathrm{A}=\mathrm{B} \quad(\because \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A})
\end{aligned}$
$\begin{aligned}
& \Rightarrow A^2-\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}+B^2=0 \\
& \Rightarrow \mathrm{A}=\mathrm{B} \quad(\because \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A})
\end{aligned}$
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