Two vectors having equal magnitudes of x, uni | vedclass

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Question

$\begin{array}{l}
Here,\,P = x\,units,\,Q = x\,units,\,	heta  = {45^ \circ }\
R = \sqrt {\left( {2 + \sqrt 2 } ight)} \,units\
We\,have,\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$\begin{array}{l}
Here,\,P = x\,units,\,Q = x\,units,\,	heta  = {45^ \circ }\
R = \sqrt {\left( {2 + \sqrt 2 } ight)} \,units\
We\,have,\,R = \sqrt {{P^2} + {Q^2} + 2PQ\cos 	heta } \
or,\,R = \sqrt {{x^2} + {x^2} + 2.x.x\cos {{45}^ \circ }} \
or,\,\sqrt {\left( {2 + \sqrt 2 } ight)}  = \sqrt {2{x^2} + 2{x^2}\frac{1}{{\sqrt 2 }}} \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {2{x^2} + \sqrt 2 {x^2}} \
or,\,\sqrt {\left( {2 + \sqrt 2 } ight)}  = \sqrt {{x^2}\left( {2 + \sqrt 2 } ight)} \
or,\,\sqrt {\left( {2 + \sqrt 2 } ight)}  = x\sqrt {\left( {2 + \sqrt 2 } ight)}  \Rightarrow x = 1
\end{array}$

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

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