Question

Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.

Answer 1

We have
2x = 3y = −z
6x = −y = −4z

The given lines can be re-written as

$$\begin{gathered} \frac{x}{{\displaystyle \frac{1}{2}}}=\frac{y}{{\displaystyle \frac{1}{3}}}=\frac{z}{-1}\mathrm{and}\frac{x}{{\displaystyle \frac{1}{6}}}=\frac{y}{-1}=\frac{z}{-{\displaystyle \frac{1}{4}}} \\ \Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{z}{-6}\mathrm{and}\frac{x}{2}=\frac{y}{-12}=\frac{z}{-{\displaystyle 3}} \end{gathered}$$

These lines are parallel to vectors $\overrightarrow{b_1}=3\hat{i}+2\hat{j}-6\hat{k}\mathrm{and}\overrightarrow{b_2}=2\hat{i}-12\hat{j}-3\hat{k}$.

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

Now,

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(3\hat{i}+2\hat{j}-6\hat{k}\right).\left(2\hat{i}-12\hat{j}-3\hat{k}\right)}{\sqrt{3^2+2^2+{\left(-6\right)}^2}\sqrt{2^2+{\left(-12\right)}^2+{\left(-3\right)}^2}} \\ =\frac{6-24+18}{\sqrt{9+4+36}\sqrt{4+144+9}} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$