Question

Angle between the lines $3x=6y=2z$ and $3x+2y+z-5=0=x+y-2z-3$ is?

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

The given two planes have a line of intersection,

whose equation can be found by,

$\begin{array}{c}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 1\\ 1& 1& -2\end{array}\right|\\ =\hat{i}(-4-1)-\hat{j}(-6-1)+\hat{k}(3-2)\\ =-5\hat{i}+7\hat{j}+\hat{k}\end{array}$

Given line is,

$\begin{array}{c}3x=6y=2z\\ \frac{x}{2}=y=\frac{z}{3}\end{array}$

so, its vector equation is $2\hat{i}+\hat{j}+3\hat{k}$.

Angle between line is cos$\theta$=$\frac{(-5)\left(2\right)+7\left(1\right)+1\left(3\right)}{\sqrt{25+49+1}\sqrt{4+1+9}}$

= $\frac{0}{\sqrt{25+49+1}\sqrt{4+1+9}}$

cos$\theta$ = 0

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

So that the correct Option is D