કયા ખૂણે બે બળો $(x + y)$ અને $(x - y) $ એ પ્રક્રિયા કરે છે. તેથી તેમનું પરિણામી લગભગ $\sqrt {\left( {{x^2}\,\, + \;\,{y^2}} \right)} $ મળે ?

Question

A${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{2({x^2} - {y^2})}}} \right)$
B${\cos ^{ - 1}}\left( { - \frac{{2({x^2} - {y^2})}}{{{x^2} + {y^2}}}} \right)$
C${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}} \right)$
D${\cos ^{ - 1}}\left( { - \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right)$

Medium

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Solution

a
$ |\mathop { {\rm{R}}}\limits^ \to | = \sqrt {{A^2} + \; {B^2} + \; 2AB\cos \theta }$

$\sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {x + y} \right)}^2} + \; {{\left( {x - y} \right)}^2} + \; 2 \left( {x + \; y} \right)\left( {x - y} \right)\cos \theta }$

$ \Rightarrow {x^2} + \; {y^2} = 2{x^2} + \; 2{y^2} + \; 2\left( {{x^2} - {y^2}} \right) \cos \theta$

$ \Rightarrow - \left( {{x^2} + \; {y^2}} \right) = 2\left( {{x^2} - {y^2}} \right) \cos \theta$

$ \Rightarrow \cos \theta = \frac{{ - \left( {{x^2} + \; {y^2}} \right)}}{{2 \left( {{x^2} - {y^2}} \right)}}$

$\Rightarrow \theta = {\cos ^{ - 1}} \left( {\frac{{ - \left( {{x^2} + {y^2}} \right)}}{{2 \left( {{x^2} - {y^2}} \right)}}} \right)$

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