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The magnitude of the sum of the two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is equal to the magnitude of
the difference of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$. The angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is
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the difference of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$. The angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is
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The correct answer is:
$90^{\circ}$
Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively. Then magnitude of their sum is given by:
$|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos \theta} \rightarrow(1)(\theta=$ angle between the vectors $)$
Magnitude of their difference is given by:
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta} \rightarrow(2)$
As $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
$\Rightarrow \mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta$
$\Rightarrow 4 \mathrm{AB} \cos \theta=0$ or $\cos \theta=0$
$\Rightarrow \theta=90^{\circ}$.
$|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos \theta} \rightarrow(1)(\theta=$ angle between the vectors $)$
Magnitude of their difference is given by:
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta} \rightarrow(2)$
As $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
$\Rightarrow \mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta$
$\Rightarrow 4 \mathrm{AB} \cos \theta=0$ or $\cos \theta=0$
$\Rightarrow \theta=90^{\circ}$.
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