Question

Find the angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$.

Answer 1

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be the vectors in the direction of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$

$\therefore \overrightarrow{a}=4\hat{i}+\hat{j}+8\hat{k}and\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=4\left(2\right)+1\left(2\right)+8\left(1\right)=8+2+8=18$

Also $|\overrightarrow{a}|=\sqrt{4^2+1^2+8^2}=\sqrt{16+1+64}=\sqrt{81}=9$ and $|\overrightarrow{b}|=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$

Let $\theta$ be the angle between the two lines.

$\therefore cos\theta =\left|\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}\right|$

$=\left|\frac{18}{9\times 3}\right|$

$\therefore cos\theta =\frac{2}{3}$

$\therefore \theta =co{s}^{-1}\left(\frac{2}{3}\right)$