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Two vectors having equal magnitudes of $x\, units$ acting at an angle of $45^o$ have resultant $\sqrt {\left( {2 + \sqrt 2 } \right)} $ $units$. The value of $x$ is
$0$
$1$
$\sqrt 2 $
$2\sqrt 2 $
Solution
$\begin{array}{l}
Here,\,P = x\,units,\,Q = x\,units,\,\theta = {45^ \circ }\\
R = \sqrt {\left( {2 + \sqrt 2 } \right)} \,units\\
We\,have,\,R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \\
or,\,R = \sqrt {{x^2} + {x^2} + 2.x.x\cos {{45}^ \circ }} \\
or,\,\sqrt {\left( {2 + \sqrt 2 } \right)} = \sqrt {2{x^2} + 2{x^2}\frac{1}{{\sqrt 2 }}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {2{x^2} + \sqrt 2 {x^2}} \\
or,\,\sqrt {\left( {2 + \sqrt 2 } \right)} = \sqrt {{x^2}\left( {2 + \sqrt 2 } \right)} \\
or,\,\sqrt {\left( {2 + \sqrt 2 } \right)} = x\sqrt {\left( {2 + \sqrt 2 } \right)} \Rightarrow x = 1
\end{array}$
