Question
The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is :
A.
$${0^ \circ }$$
B.
$${90^ \circ }$$
C.
$${45^ \circ }$$
D.
$${30^ \circ }$$
Answer :
$${90^ \circ }$$
Solution :
The given lines are $$2x=3y=-z$$
or $$\frac{x}{3} = \frac{y}{2} = \frac{z}{{ - 6}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Dividing by 6}}} \right]$$
and $$6x=-y=-4z$$
or $$\frac{x}{2} = \frac{y}{{ - 12}} = \frac{z}{{ - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Dividing by 12}}} \right]$$
$$\therefore $$ Angle between two lines is
$$\eqalign{
& \cos \,\theta = \frac{{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)}}{{\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }} \cr
& \Rightarrow \cos \,\theta = \frac{{6 - 24 + 18}}{{\sqrt {49} \sqrt {157} }} = 0 \cr
& \Rightarrow \theta = {90^ \circ } \cr} $$