Question

Let $\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R}$ be such that ${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=1$, if $\mathrm{acos\theta }=\mathrm{bcos}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=\mathrm{ccos}\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)$, where $\mathrm{\theta }=\frac{\mathrm{\pi }}{9}$,

Then the angle between the vectors $\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}$ and $\mathrm{b}\hat{\mathrm{i}}+\mathrm{c}\hat{\mathrm{j}}+\mathrm{a}\hat{\mathrm{k}}$ is:

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

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct answer:

Step 1: Finding the dot product of vectors $\overrightarrow{p}$and $\overrightarrow{q}$

Given: The vectors are $\overrightarrow{\mathrm{p}}=\mathrm{a}\hat{\mathrm{ı}}+\mathrm{b}\hat{\mathrm{ȷ}}+\mathrm{c}\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{q}}=\mathrm{b}\hat{\mathrm{ı}}+\mathrm{c}\hat{\mathrm{ȷ}}+\mathrm{a}\hat{\mathrm{k}}$. Let $\alpha$ is angle between vectors.

$$\begin{gathered} \Rightarrow \overrightarrow{\mathrm{p}}·\overrightarrow{\mathrm{q}}=\left|\overrightarrow{\mathrm{p}}\right|\left|\overrightarrow{\mathrm{q}}\right|\cos \alpha \left[Formulafordotproductofvectors\right] \\ \Rightarrow \mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\left(\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\right)\left(\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\right)\cos \alpha \end{gathered}$$

$\because {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=1$(given)

$$\begin{gathered} \Rightarrow \mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\left(\sqrt{1}\right)\left(\sqrt{1}\right)\cos \alpha \\ \Rightarrow \mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\cos \alpha \dots \left(1\right) \end{gathered}$$

Step 2: Finding the value of $\cos \alpha$

We know that

$$\begin{gathered} \Rightarrow \mathrm{acos}\theta =\mathrm{bcos}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)\left[Given\right] \\ \Rightarrow \mathrm{acos}\theta =\mathrm{bcos}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)=\lambda \left[Say\right] \\ \Rightarrow \mathrm{a}=\frac{\mathrm{\lambda }}{\mathrm{cos\theta }},\mathrm{b}=\frac{\mathrm{\lambda }}{\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)},\mathrm{c}=\frac{\mathrm{\lambda }}{\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)} \end{gathered}$$

Now equation $\left(1\right)$ becomes,

$\Rightarrow \cos \alpha =\left\{\frac{\mathrm{\lambda }}{\mathrm{cos\theta }}·\frac{\mathrm{\lambda }}{\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)}+\frac{\mathrm{\lambda }}{\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\cos \left(\mathrm{\theta }+{\displaystyle \frac{4\mathrm{\pi }}{3}}\right)}+\frac{\lambda }{\cos \left(\theta +\frac{4\pi }{3}\right)\left(\cos \theta \right)}\right\}$

$={\lambda }^2\left\{\frac{\cos \left(\theta +\frac{4\pi }{3}\right)+\cos \theta +\cos \left(\theta +\frac{2\pi }{3}\right)}{\cos \theta ·\cos \left(\theta +\frac{2\pi }{3}\right)·\cos \left(\theta +\frac{4\pi }{3}\right)}\right\}\left[\mathrm{Taking}\mathrm{LCM}\right]$

Step 3: Finding the value of $\alpha$:

We know that

$\mathrm{cosA}+\mathrm{cosB}=2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$
Consider $\mathrm{A}=\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}$and $\mathrm{B}=\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}$
$$\begin{gathered} \Rightarrow \cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)+\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=2\cos \left(\mathrm{\theta }+\mathrm{\pi }\right)\cos \left(\frac{\mathrm{\pi }}{3}\right) \\ =2\cos \left(\mathrm{\theta }+\mathrm{\pi }\right)\frac{1}{2} \\ =\cos \left(\mathrm{\theta }+\mathrm{\pi }\right) \\ =\cos \theta \end{gathered}$$

$\Rightarrow \cos \alpha ={\mathrm{\lambda }}^2\left\{\frac{\mathrm{cos\theta }-\mathrm{cos\theta }}{\mathrm{cos\theta }·\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)}\right\}$

$$\begin{gathered} \Rightarrow \cos \alpha ={\mathrm{\lambda }}^2\left(0\right) \\ \Rightarrow \cos \alpha =0 \\ \Rightarrow \alpha ={\cos }^{-1}\left(0\right) \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Therefore, the angle between $\overrightarrow{\mathrm{p}}$and $\overrightarrow{\mathrm{q}}$is $\frac{\mathrm{\pi }}{2}$.

Therefore, option (A) is the correct answer.