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# If $|\overrightarrow{a+b}|=|\overrightarrow{a-b}|$ then

$\begin{array} {1 1}(1)\overrightarrow{a} \; is\; parallel \;to\; \overrightarrow{b}&(2)\overrightarrow{a} \; is\; perpendicular\; to\; \overrightarrow{b}\\(3)|\overrightarrow{a}|=|\overrightarrow{b}|&(4)\overrightarrow{a}\; and \; \overrightarrow{b}\; are \;unit \;vectors\end{array}$

$|\overrightarrow{a+b}|=|\overrightarrow{a-b}|$
$|\overrightarrow{a+b}|^2=|\overrightarrow{a-b}|^2$
$(\overrightarrow{a+b})^2=(\overrightarrow{a-b})^2$
$\overrightarrow a^2 +\overrightarrow{b}^2+ 2 \overrightarrow{a} .\overrightarrow{b}=\overrightarrow a^2 +\overrightarrow{b}^2- 2\overrightarrow{a} .\overrightarrow{b}$
$4\overrightarrow{a}.\overrightarrow{b} =0$
$\overrightarrow{a}.\overrightarrow{b} =0$
$\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular.
Hence 2 is the correct answer.