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Question

(c) Let the angle between two vectors $P$ and $Q$ be $\alpha$ and their resultant is $R$

So we can write

$R^{2}=P^{2}+Q^{2}+2 P Q \cos \alpha \l

Vedclass · Accepted Answer

$Q$

Vedclass · Answer

(c) Let the angle between two vectors $P$ and $Q$ be $\alpha$ and their resultant is $R$

So we can write

$R^{2}=P^{2}+Q^{2}+2 P Q \cos \alpha \ldots \ldots[1]$

When $Q$ is doubled then let the resultant vector be $R_{1},$ So we can write

$R_{1}^{2}=P^{2}+4 Q^{2}+4 P Q \cos \alpha \ldots \ldots .[2]$

Again by the given condition $R_{1}$ is perpendicular to $P$

So $4 Q^{2}=P^{2}+R_{1}^{2} \dots \ldots .[3]$

Combining $[ 2]$ and $[ 3]$ we get

$R_{1}^{2}=P^{2}+P^{2}+R_{1}^{2}+4 P Q \cos \alpha$

$\Rightarrow 2 P Q \cos \alpha=-P^{2} \dots \dots[4]$

combining $[ 1]$ and $[ 4]$ we get

$R^{2}=P^{2}+Q^{2}-P^{2}$

$\Rightarrow R=Q$

The resultant of two vectors $\overrightarrow P $ and $\overrightarrow Q $ is $\overrightarrow R .$ If $Q$ is doubled, the new resultant is perpendicular to $P$. Then $R $ equals

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