Question

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

Answer 1

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

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

${\displaystyle \therefore \overline{a}.\overline{b}=4\times 2+1\times 2+8\times 1=8+2+8=18}$

and

${\displaystyle \left|\overline{a}\right|=\sqrt{16+1+64}=\sqrt{81}=9}$

${\displaystyle \overline{b}=\sqrt{4+1+4}=\sqrt{9}=3}$ 

Let Θ be the acute angle between the two given lines.

${\displaystyle \therefore \cos \theta =\mid \frac{\overline{a}.\overline{b}}{\left|\overline{a}\right|.\left|\overline{b}\right|}=\frac{18}{9\times 3}=\frac{2}{3}}$

${\displaystyle \theta ={\cos }^{-1}\left(\frac{2}{3}\right)}$