Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equations of the given lines can be written in standard form by dividing by the LCM of the coefficients.

For the first line, $3x=2y=-z$, dividing by $6$ gives:

$\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

The direction ratios of the first line are $a_1=2$, $b_1=3$, $c_1=-6$.

For the second line, $-x=6y=-4z$, dividing by $12$ gives:

$\frac{x}{-12}=\frac{y}{2}=\frac{z}{-3}$

The direction ratios of the second line are $a_2=-12$, $b_2=2$, $c_2=-3$.

Let $\theta$ be the angle between the lines. Then,

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

$\cos \theta =\frac{|(2)(-12)+(3)(2)+(-6)(-3)|}{\sqrt{2^2+3^2+(-6{)}^2}\sqrt{(-12{)}^2+2^2+(-3{)}^2}}$

$\cos \theta =\frac{|-24+6+18|}{\sqrt{49}\sqrt{157}}$

$\cos \theta =0$

$\theta =\frac{\pi }{2}$

Answer: $\frac{\pi }{2}$