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

Question

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

Accepted Answer

The given lines are $2x = 3y =  - z$

or $$\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$$ $$\,\,\,\left[ {} \right.$$ Dividing by $6$ $$\left. {} \right]$$

and $6x =  - y =  - 4z$

or $$\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$$ $$\,\,\,\,\left[ {} \right.$$ Dividing by $12$ $$\left. {} \right]$$

$$\therefore$$ Angle between two lines is

$$\cos \theta  = {{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)} \over {\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }}$$

$$ = {{6 - 24 + 18} \over {\sqrt {49} \sqrt {157} }} = 0 \Rightarrow \theta  = {90^ \circ }$$

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