Q. The angle between the lines $2x = 3y = - z$ and $6x = - y = - 4z$ is

Solution:

Given $\ce{L_1 : 2x = 3y = - z}$ and $L_2 : 6x = - y = -4x $
$\Rightarrow \; L_1 : \frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{-1}$ and $L_2 : \frac{x}{\frac{1}{16}} = \frac{y}{-1} = \frac{z}{ - \frac{1}{4}}$
Now, $a_1 a_2 + b_1 b_2 + c_1 c_2 = \left(\frac{1}{2} \times\frac{1}{6}\right) + \left(\frac{1}{3} \times-1\right) + \left(-1 \times- \frac{1}{4}\right) = \frac{1}{12} - \frac{1}{3} + \frac{1}{4}= 0 $
$\therefore$ Angle between $L_1$ and $L_2$ is $90^{\circ}$