Question

If the direction ratio of two vectors are connected by the relations $p+q+r=0$ and $p^2+q^2-r^2=0$, find the angle between them.

Answer 1

$p+q+r=0........1$

let $p=-(q+r)$

$p^2+q^2+r^2=0............2$

put value of $p$ from eq. 1 in eq. 2

$[-(q+r){]}^2+q^2+r^2=0$

$q^2+r^2+2qr+q^2-r^2=0$

$2q^2+2qr=0$

$2q(q+r)=0$

$\therefore q=0orq=r$

now, $p=-(q+r)$

therefore, if $q=0;p=-r;q=0$

$\therefore \frac{p}{1}=\frac{r}{-1};q=0$

$\therefore$ direction ratio of first vector are

$a_1=1;b_1=0,c_1=-1$

if $q=-r,p=-(-r+r)=0$

$p=0$ and $q=-r$

$\therefore$ direction ration of the second vector is

$a_2=0,b_2=1,c_2=1$

$Q=$ angle between two vectors

them, cos $Q=$ $\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a^2+1^2(-1{)}^2}}$

$\left|\frac{1\left(0\right)+\left(0\right)1+(-1)\left(1\right)}{\sqrt{1^2+0^2+(-1{)}^2}\sqrt{0^2+1^2+(-1{)}^2}}\right|$

$\frac{1}{|\sqrt{2}|\sqrt{2}|}=\frac{1}{2}$

$cos\theta =\frac{cos\pi }{3}$

$\therefore \theta =\frac{\pi }{3}$