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

(D)
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.
\(\cos \theta=\frac{\bar{a} \cdot \bar{b}}{|\bar{a}| \cdot|\bar{b}|}=\frac{18}{9 \times 3}=\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\)