કયા ખૂણે બે બળો (x + y) અને (x - y) એ પ્રક્રિ | vedclass

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Question

$|\mathop {\,{	ext{R}}}\limits^ 	o  |\, = \,\sqrt {{A^2}\, + \,{B^2}\, + \,2AB\cos 	heta } $

$\sqrt {{x^2}\, + \,{y^2}} \, = \sqrt {{{\left

Vedclass · Accepted Answer

${\cos ^{ - 1}}\,\,\frac{{ - \left( {{x^2}\,\, + \;\,{y^2}\,\,} \right)}}{{2\,\,\left( {{x^2}\,\, - \,\,{y^2}} \right)}}$

Vedclass · Answer

$|\mathop {\,{	ext{R}}}\limits^ 	o  |\, = \,\sqrt {{A^2}\, + \,{B^2}\, + \,2AB\cos 	heta } $

$\sqrt {{x^2}\, + \,{y^2}} \, = \sqrt {{{\left( {x\, + \,y} ight)}^2}\, + \,{{\left( {x\, - \,y} ight)}^2}\,+ \,2\,\left( {x\, +\,y} ight)\left( {x\, - \,y} ight)\cos \,	heta } $

$ \Rightarrow \,{x^2}\, + \;{y^2}\, = \,2{x^2}\, + \;2{y^2}\, + \;2\left( {{x^2}\, - \,{y^2}} ight)\cos 	heta$

$\Rightarrow \, - \left( {{x^2}\, + \;{y^2}} ight)\, = \,2\left( {{x^2}\, - \,{y^2}} ight)\cos 	heta $

$\Rightarrow \,\cos \,	heta \, =\,\frac{{ - \left( {{x^2}\, + \;{y^2}} ight)}}{{2\,\left( {{x^2}\, - \,{y^2}} ight)}}\,$

$\Rightarrow \,	heta \, = \,{\cos ^{ - 1}}\,\left( {\frac{{ - \left( {{x^2}\,\, + \,{y^2}} ight)}}{{2\,\left( {{x^2}\, - \,{y^2}} ight)}}} ight) $

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 ?

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