Question

Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.

 

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 θ  be the angle between the given lines. 

\[\therefore \cos\theta = \frac{3 \times 2 + 2 \times \left( - 12 \right) + \left( - 6 \right) \times \left( - 3 \right)}{\sqrt{3^2 + 2^2 + \left( - 6 \right)^2}\sqrt{2^2 + \left( - 12 \right)^2 + \left( - 3 \right)^2}}\]

\[ = \frac{6 - 24 + 18}{\sqrt{49}\sqrt{157}}\]

\[ = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

Thus, the angle between the given lines is  \[\frac{\pi}{2}\] .