Physics - Vectors Question with Solution | TestHub

$P+Q=16$(i)

$P^2+Q^2+2PQ\cos \text{θ}=64$(ii)

tan${\text{90}}^{\text{ο}}=\frac{Q\; sin\text{θ}}{P+Q\; cos\text{θ}}$



$\Rightarrow P+Qcos\text{θ}=0$

$cos\text{θ}=-\frac{\text{P}}{\text{Q}}$

From Eq. (ii)

$P^2+Q^2+2\mathit{PQ}\cos \text{θ}=\text{64}$

$P^2+Q^2+\mathit{\text{2.P.Q}}\left(-\frac{P}{Q}\right)=\text{64}$

$-P^2+Q^2=\text{64}$

$\left(Q+P\right)\left(Q-P\right)=\text{64}$

$asP+Q=16$

$Q-P=4$

Solving we get

$2Q=\text{20}$

$Q=\text{10}units$

$P=6units$