Question

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

• A. $0^{\circ }$
• B. ${90}^{\circ }$
• C. ${45}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

The given lines are $2x=3y=-z\text{and}6x=-y=-4z$
or $L_1:\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}\text{and}{L}_2:\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$
$\Rightarrow D.R^{\prime }s$ of line $L_1:(a_1,b_1,c_1)=(3,2,-6)$
$D.R^{\prime }s$ of line $L_2:(a_2,b_2,c_2)=(2,-12,-3)$
Now $\left(3\right)\left(2\right)+\left(2\right)(-12)+(-6)(-3)=0$
we know, angle between two lines $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\because a_1{a}_2+b_1{b}_2+c_1{c}_2=\left(3\right)\left(2\right)+\left(2\right)(-12)+(-6)(-3)=0$
$\Rightarrow \theta ={90}^{\circ }$