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If $|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|$, then
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The correct answer is:
$\mathbf{a}$ and $\mathbf{b}$ are perpendicular.
Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors $|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|$
On squaring on both sides, we get $a^2+b^2+2 a b \cos \theta=a^2+b^2-2 a b \cos \theta$
$4 a b \cos \theta=0$
If $a=0, b=0$, then $\cos \theta=0$
$\theta=\left(\frac{2 n-1}{2}\right) \pi$, where $n \in N$
Hence, $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.
On squaring on both sides, we get $a^2+b^2+2 a b \cos \theta=a^2+b^2-2 a b \cos \theta$
$4 a b \cos \theta=0$
If $a=0, b=0$, then $\cos \theta=0$
$\theta=\left(\frac{2 n-1}{2}\right) \pi$, where $n \in N$
Hence, $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.
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