Question

The angle between the lines $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{2}\mathrm{and},\frac{x-1}{-\sqrt{3}-1}=\frac{y-1}{\sqrt{3}-1}=\frac{z-1}{4}$ is
(a) ${\cos }^{-1}\left(\frac{1}{65}\right)$

(b) $\frac{\mathrm{\pi }}{6}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{4}$

Answer 1

(c) $\frac{\mathrm{\pi }}{3}$

We have

$$\begin{gathered} \frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{2} \\ \frac{x-1}{-\sqrt{3}-1}=\frac{y-1}{\sqrt{3}-1}=\frac{z-1}{4} \end{gathered}$$


The direction ratios of the given lines are proportional to 1, 1, 2 and $-\sqrt{3}-1,\sqrt{3}-1,4$.

The given lines are parallel to vectors $\overrightarrow{b_1}=\hat{i}+\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b_2}=\left(-\sqrt{3}-1\right)\hat{i}+\left(\sqrt{3}-1\right)\hat{j}+4\hat{k}$.

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

Now,

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(\hat{i}+\hat{j}+2\hat{k}\right).\left\{\left(-\sqrt{3}-1\right)\hat{i}+\left(\sqrt{3}-1\right)\hat{j}+4\hat{k}\right\}}{\sqrt{1^2+1^2+1^2}\sqrt{{\left(-\sqrt{3}-1\right)}^2+{\left(\sqrt{3}-1\right)}^2+4^2}} \\ =\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}} \\ =\frac{6}{6\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$