Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{9}$
• D. $0$

Answer 1

The correct option is A $\frac{\pi }{2}$

Let the angle between the vectors $a\hat{i}+b\hat{j}+c\hat{k}$ and $b\hat{i}+c\hat{j}+a\hat{k}$ be $\alpha$
$\cos \alpha =\frac{\overrightarrow{p}\cdot \overrightarrow{q}}{\left|p\right|\left|q\right|}=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{ab+bc+ca}{1}$
$\cos \alpha =ab+bc+ca$
$\cos \alpha =abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\cdots \left(1\right)$

$a\cos {20}^{\circ }=b\cos \left({140}^{\circ }\right)=c\cos \left({260}^{\circ }\right)=\lambda$
$\Rightarrow \frac{1}{a}=\frac{\cos {20}^{\circ }}{\lambda },\frac{1}{b}=\frac{\cos \left({140}^{\circ }\right)}{\lambda },\frac{1}{c}=\frac{\cos \left({260}^{\circ }\right)}{\lambda }$
Putting in equation $\left(1\right)$, we get
$\Rightarrow \cos \alpha =\frac{abc}{\lambda }(\cos {20}^{\circ }+\cos {140}^{\circ }+\cos {260}^{\circ })$
$\Rightarrow \cos \alpha =\frac{abc}{\lambda }(\cos {20}^{\circ }+2\cos {200}^{\circ }\cdot \cos {60}^{\circ })$
$\Rightarrow \cos \alpha =\frac{abc}{\lambda }(\cos {20}^{\circ }-\cos {20}^{\circ })$
$\Rightarrow \cos \alpha =0$
$\therefore \alpha =\frac{\pi }{2}$