The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and…
The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is
$\cos ^{-1}\left(\frac{3}{4}\right)$
$\cos ^{-1}\left(\frac{1}{3}\right)$
$\cos ^{-1}\left(\frac{1}{2}\right)$
$\cos ^{-1}\left(\frac{2}{3}\right)$
Solution
Let $\bar{a}$ and $\bar{b}$ be the vectors in the direction of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ respectively.
$\therefore \quad \overline{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+8 \hat{\mathrm{k}} \quad$ and $\quad \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\bar{a} \cdot \bar{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18$
Also $|\bar{a}| \quad=\sqrt{16+1+64}=\sqrt{81}=9$ and $|\bar{b}|=\sqrt{4+1+4}=\sqrt{9}=3$
Let $\theta$ be the acute angle between the two given lines.
$\cos \theta=\frac{\bar{a} \cdot \bar{b}}{|\bar{a}| \cdot|\bar{b}|}=\frac{18}{9 \times 3}=\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)$