Question

If the vectors $3\overrightarrow{p}+\overrightarrow{q};5\overrightarrow{p}-3\overrightarrow{q}$ and $2\overrightarrow{p}+\overrightarrow{q};4\overrightarrow{p}-2\overrightarrow{q}$ are pairs of mutually perpendicular vectors.

Answer 1

If two vectors are perpendicular, then their dot product is zero

$\Rightarrow (3\overrightarrow{p}+\overrightarrow{q}).(5\overrightarrow{p}-3\overrightarrow{q})=0$ (given)

$\Rightarrow 15p^2-9\overrightarrow{p}.\overrightarrow{q}+5\overrightarrow{p}.\overrightarrow{q}-3q^2=0$

$\Rightarrow 15p^2-4\overrightarrow{p}.\overrightarrow{q}-3q^2=0$ ....$\left(1\right)$

$\Rightarrow (2\overrightarrow{p}+\overrightarrow{q}).(4\overrightarrow{p}-2\overrightarrow{q})=0$ (given)

$\Rightarrow 8p^2-4\overrightarrow{p}.\overrightarrow{q}+4\overrightarrow{p}.\overrightarrow{q}-2q^2=0$

$\Rightarrow 8p^2=2q^2$

$\Rightarrow 4p^2=q^2$ ...$\left(2\right)$

$\Rightarrow 2|p|=|q|$ ...$\left(3\right)$

Substituting $\left(2\right)$ in $\left(1\right)$, we get

$\Rightarrow 15p^2-4\overrightarrow{p}.\overrightarrow{q}-3\left(4p^2\right)=0$

$\Rightarrow 15p^2-4\overrightarrow{p}.\overrightarrow{q}-12p^2=0$

$\Rightarrow 3p^2-4\overrightarrow{p}.\overrightarrow{q}=0$

$\Rightarrow \overrightarrow{p}.\overrightarrow{q}=\frac{3p^2}{4}$ ...$\left(4\right)$

We know that ,$\overrightarrow{p}.\overrightarrow{q}=|p||q|cos(p,q)$

$\Rightarrow \cos (p,q)=\frac{\overrightarrow{p}.\overrightarrow{q}}{|p||q|}$ ...$\left(5\right)$

Substituting $\left(3\right)$ and $\left(4\right)$ in $\left(5\right)$ we get,

$\Rightarrow \cos (p,q)=\frac{\frac{3p^2}{4}}{2|p||p|}$

$\Rightarrow \cos (p,q)=\frac{3p^2}{8p^2}$

$\Rightarrow \cos (p,q)=\frac{3}{8}$

Therefore angle between $\overrightarrow{p}and\overrightarrow{q}is{\cos }^{-1}\frac{3}{8}$