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If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$\vec{a}$ is perpendicular to $\vec{b}$
Given, $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|$
$\Rightarrow|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}-2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|$
$\Rightarrow 4|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=0$
$\Rightarrow \overrightarrow{\mathbf{a}}$ is perpendicular to $\overrightarrow{\mathbf{b}}$
$\Rightarrow|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}-2|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|$
$\Rightarrow 4|\overrightarrow{\mathbf{a}}| \cdot|\overrightarrow{\mathbf{b}}|=0$
$\Rightarrow \overrightarrow{\mathbf{a}}$ is perpendicular to $\overrightarrow{\mathbf{b}}$
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