Let the function


Let $a, b, c \in R$ be such that $a^{2}+b^{2}+c^{2}=1$.

If $a \cos \theta=b \cos \left(\theta+\frac{2 \pi}{3}\right)=\cos \left(\theta+\frac{4 \pi}{3}\right)$

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 :

  1. $\frac{\pi}{2}$

  2. 0

  3. $\frac{\pi}{9}$

  4. $\frac{2 \pi}{3}$

Correct Option: 1


$\cos \phi=\frac{\bar{p} \cdot \bar{q}}{|\bar{p}||\bar{q}|}=\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=\frac{\Sigma a b}{1}$

$=a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$=\frac{\mathrm{abc}}{\lambda}\left(\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)\right)$

$=\frac{\mathrm{abc}}{\lambda}\left(\cos +2 \cos (\theta+\pi) \cos \frac{\pi}{3}\right)$

$=\frac{\mathrm{abc}}{\lambda}(\cos \theta-\cos \theta)=0$


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