Let $a,b,c\in R$ be such that $a^2+b^2+c^2=1$. If $a\cos \theta =b\cos \left(\theta +\frac{2\pi }{3}\right)=c\cos \left(\theta +\frac{4\pi }{3}\right)$, where $\theta =\frac{\pi }{9}$, then the angle between the vectors $a\widehat{i}+b\widehat{j}+c\widehat{k}$ and $b\widehat{i}+c\widehat{j}+a\widehat{k}$ is

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

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

Now, $\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)\left(b\widehat{i}+c\widehat{j}+a\widehat{k}\right)=ab+bc+ac$

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

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

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

$\therefore$Angle between $a\widehat{i}+b\widehat{j}+c\widehat{k}$  and $b\widehat{i}+c\widehat{j}+a\widehat{k}$  is $\frac{\pi }{2}$