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If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is ........ $^o$
$90$
$120$
$45$
$60$
Solution
$\begin{array}{l}
Let\,the\,two\,vectors\,be\,\vec A\,and\,\vec B\,.\\
Then,\,magnitude\,of\,sum\,of\,\vec A\,and\,\vec B\,,\\
\,\,\,\,\,\,\,\,\left| {\vec A + \vec B} \right| = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \\
and\,magnitude\,od\,difference\,of\,\vec A\,and\,\vec B\,,\\
\,\,\,\,\,\,\,\left| {\vec A – \vec B} \right| = \sqrt {{A^2} + {B^2} – 2AB\cos \theta } \,,\\
\,\,\,\,\,\,\,\left| {\vec A + \vec B} \right|\, = \,\,\left| {\vec A – \vec B} \right|\,\,\left( {given} \right)
\end{array}$
$\begin{array}{l}
or\,\,\sqrt {{A^2} + {B^2} + 2AB\cos \theta } \\
\,\,\,\, = \sqrt {{A^2} + {B^2} – 2AB\cos \theta } \\
\Rightarrow \,\,4\,AB\cos \theta = 0\\
\,\,\,\,4\,AB \ne 0,\,\,\therefore \cos \theta = 0\,or\,\theta = {90^ \circ }
\end{array}$
