Prove the following

Question:

Let $a, b, c \in \mathbf{R}$ 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 :

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

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

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

  4. 0


Correct Option: 1

Solution:

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

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

$a b+b c+c a=k^{2} \frac{\left[\cos \left(\theta+\frac{4 \pi}{3}\right)+\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)\right]}{\cos \left(\theta+\frac{4 \pi}{3}\right) \cdot \cos \theta \cdot \cos \left(\theta+\frac{2 \pi}{3}\right)}$

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

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

$\cos \phi=\frac{(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(b \hat{i}+c \hat{j}+a \hat{k})}{\sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{b^{2}+c^{2}+a^{2}}}$

$=a b+b c+c a=0$

$\phi=\frac{\pi}{2}$

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