Question

Find the angle between the lines $2x=3y=-z$ and $6x=-y=-4z$. [CBSE 2015]

Answer 1

The equations of the given lines can be re-written as

$\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ and $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$

We know that angle between the lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.

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

$$\begin{gathered} \therefore \cos \theta =\frac{3\times 2+2\times \left(-12\right)+\left(-6\right)\times \left(-3\right)}{\sqrt{3^2+2^2+{\left(-6\right)}^2}\sqrt{2^2+{\left(-12\right)}^2+{\left(-3\right)}^2}} \\ =\frac{6-24+18}{\sqrt{49}\sqrt{157}} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Thus, the angle between the given lines is $\frac{\mathrm{\pi }}{2}$.