Question

Find the angle between the lines vector $\overrightarrow{r}=\left(2i-3j+k\right)+l\left(i+j+3k\right)$ and vector $=\left(i-j-k\right)+\mu \left(2i-3j+k\right)$

• A. $\frac{\pi }{2}$
• B. ${\cos }^{-1}\left(\frac{9}{\sqrt{91}}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{\sqrt{154}}\right)$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is C

${\cos }^{-1}\left(\frac{2}{\sqrt{154}}\right)$

Explanation for correct answer:

Find the angle:

Given, vectors $\overrightarrow{r}=\left(2i-3j+k\right)+l\left(i+j+3k\right)$ and $\left(i-j-k\right)+\mu \left(2i-3j+k\right)$

We know that the angle between two vectors $\overrightarrow{a}\&\overrightarrow{b}$ is $\mathit{c}\mathit{o}\mathit{s}\left(\theta \right)\mathbf{=}\frac{\overrightarrow{\mathbf{a}}\mathbf{·}\overrightarrow{\mathbf{b}}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$

Required angle is $\cos \left(\theta \right)=\frac{\left(i+j+3k\right)·\left(2i-3j+k\right)}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(3\right)}^2}\times \sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(1\right)}^2}}$

$\Rightarrow$$\cos \left(\theta \right)=\frac{2-3+3}{\sqrt{11}\times \sqrt{14}}$

$\Rightarrow$$\cos \left(\theta \right)=\frac{2}{\sqrt{154}}$

$\therefore$ $\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{154}}\right)$

Hence, correct answer is option $\left(C\right)$.