Lining a Mathematical Heart #1

$p+q+r=0$ and $p^2+q^2+r^2=0$ .....(given)

Therefore , $r=-(p+q).....(1)$ . $\implies p^2+q^2-[-(p+q)]^2=0$ .....(from 1), Therefore , $p^2+q^2-(p^2+2pq+q^2)=0$ $\implies -2pq=0$ $\implies p=0$ or $q=0$ and if $p=0$ then from $(1)$ we get $r=-q$ . Direction ratios of the first vector are $0,q,-q$ i.e. $0,1,-1$ .

If $q=0$ ,then from $(1)$ ,we get $r=-p$ ,so direction ratios of the second vector are $p,0,-p$ i.e. $1,0,-1$ .

Let $\theta$ be the angle between the 2 vectors ...

Therefore , $cos\theta=\bigg|\dfrac{0(1)+1(0)+(-1)(-1)}{\sqrt{0^2+1^2+(-1)^2}.\sqrt{1^2+0^2+(-1)^2}}\bigg|$ $=\bigg|\dfrac{1}{\sqrt{2}\sqrt{2}}\bigg|$ So , $cos\theta=\dfrac{1}{2}$ or $cos\theta=cos\dfrac{\pi}{3}$ $\implies \boxed{\theta=\dfrac{\pi}{3}}$