Question

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

Answer 1

Given lines are 2x=3y=-z i.e., $\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ and $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$ And their respective d.r.'s are 3, 2, -6 and 2,-12, -3 $\therefore$ required angle between lines is $\theta =co{s}^{-1}\left(\frac{|3\times 2+2\times (-12)+(-6)(-3)|}{\sqrt{9+4+36}\sqrt{4+144+9}}\right)$ $\Rightarrow \theta =co{s}^{-1}\left(\frac{|6-24+18|}{\sqrt{9+4+36}\sqrt{4+144+9}}\right)=co{s}^{-1}\left(0\right)=\frac{\pi }{2}$.