Question

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 $a \hat{i}+b \hat{j}+c \hat{k}$ and $b \hat{i}+c \hat{j}+a \hat{k}$ is :

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

Answer 1

Answer: (A)

$\cos \phi=\frac{\overline{ p } \cdot \overline{ q }}{|\overline{ p }||\overline{ q }|}=\frac{ ab + bc + ca }{ a ^{2}+ b ^{2}+ c ^{2}}=\frac{\Sigma ab }{1}$
$=abc \left(\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }\right)$
$=\frac{abc}{\lambda}\left(\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)\right)$
$=\frac{abc}{\lambda}\left(\cos +2 \cos (\theta+\pi) \cos \frac{\pi}{3}\right)$
$=\frac{a b c}{\lambda}(\cos \theta-\cos \theta)=0$
$\phi=\frac{\pi}{2}$