JEE Main 2020 — Vector Algebra Question with Solution

Let $a\cos \theta =b\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=c\cos \left(\theta +\frac{4\mathrm{\pi }}{3}\right)=k$

$\Rightarrow a=\frac{k}{\cos \theta }, b=\frac{k}{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}, c=\frac{k}{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}$

$\Rightarrow ab+bc+ca=k^2\frac{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)+\cos \theta +\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)\cos \theta \cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}$

$=k^2\left[\frac{\cos \theta +2\cos \left(\mathrm{\theta }+\mathrm{\pi }\right).\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{\cos \theta .\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right).\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}\right]$

$=k^2\left[\frac{\cos \theta -2\cos \theta .{\displaystyle \frac{1}{2}}}{\cos \theta .\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right).\cos \left(\theta +{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}\right]=0$

Let $\varphi$ be the angle between two given vectors.

$\therefore \cos \varphi =\frac{\left(a\hat{i}+b\hat{j}+c\hat{k}\right).\left(b\hat{i}+c\hat{j}+a\hat{k}\right)}{\sqrt{a^2+b^2+c^2 }·\sqrt{b^2+c^2+a^2 }}=\frac{ab+bc+ca}{a^2+b^2+c^2}=0$

$\Rightarrow \varphi =\frac{\pi }{2}$