At what angle must the two forces (x + y) and | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$ |\mathop { {m{R}}}\limits^ 	o | = \sqrt {{A^2} + \; {B^2} + \; 2AB\cos 	heta }$

$\sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {x + y} ight)

Vedclass · Accepted Answer

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

Vedclass · Answer

$ |\mathop { {m{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)}}$

$\Rightarrow 	heta = {\cos ^{ - 1}} \left( {\frac{{ - \left( {{x^2} +  {y^2}} ight)}}{{2 \left( {{x^2} - {y^2}} ight)}}} ight)$

At what angle must the two forces $(x + y)$ and $(x -y)$ act so that the resultant may be $\sqrt {({x^2} + {y^2})} $

Similar Questions

While travelling from one station to another, a car travels $75 \,km$ North, $60\, km$ North-east and $20 \,km $ East. The minimum distance between the two stations is.......$km$

A particle is simultaneously acted by two forces equal to $4\, N$ and $3 \,N$. The net force on the particle is

What displacement must be added to the displacement $25\hat i - 6\hat j\,\,m$ to give a displacement of $7.0\, m$ pointing in the $X- $direction

In an octagon $ABCDEFGH$ of equal side, what is the sum of $\overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }+\overrightarrow{ AG }+\overrightarrow{ AH }$ if, $\overrightarrow{ AO }=2 \hat{ i }+3 \hat{ j }-4 \hat{ k }$

Prove the associative law of vector addition.