Question

please with explaination​

Answer 1

$\huge {\orange {\bf {\underline {\underline {Questíon}}}:-}}$

The Vector sum of two vectors $\sf \vec{P}$ and $\sf \vec{Q}$ is $\sf \vec{R}$ .If vector $\sf \vec{Q}$ is reversed, the resultant becomes $\sf \vec{S}$ .then prove that R² + S² = 2(P² + Q²)

$\Huge {\orange {\bf {\underline {\underline {Solutíon}}}:-}}$

According to the question...

Let $\theta$be the angle between vectors $\sf \vec{P}$ and $\sf \vec{Q}$

Then,

$\sf {R}^{2} = {P}^{2} + {Q}^{2} + 2PQcos \theta \: \: \: \: \: \: \: \: .......(1)$

When Vector $\sf \vec{Q}$ is reversed, angle between the vector $\sf \vec{P}$ and $\sf -\vec{Q}$ will become 180° - $\theta$

Thus,

$\sf {S}^{2} = {P}^{2} + {Q}^{2} + 2PQ(180° - \theta)$

$\sf {S}^{2} = {P}^{2} + {Q}^{2} + 2PQ×( - cos \theta)$

$\sf {S}^{2} = {P}^{2} + {Q}^{2} - 2PQ cos \theta \: \: \: \: \: \: \: \: .......(2)$

Adding the equation (1) and (2)

$\sf {R}^{2} + {S}^{2} = {P}^{2} + {Q}^{2} \cancel{+ 2PQcos \theta} + {P}^{2} + {Q}^{2} \cancel{- 2PQcos \theta}$

$\sf {R}^{2} + {S}^{2} = 2({P}^{2} + {Q}^{2} )$

$\large \sf hence \: proved....$