Question

The angle between the lines $x-3y-4=0,$ $4y-z+5=0$ and $x+3y-11=0,$ $2y-z+6=0$ is

• A. ${\cos }^{-1}\left(\sqrt{\frac{2}{13}}\right)$
• B. ${\cos }^{-1}\left(\sqrt{\frac{1}{14}}\right)$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

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

The lines $x-3y-4=0=4y-z+5$ is the line of intersection of planes $x-3y-4=0$ and $4y-z+5=0.$
Vector along the line of intersection is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 0\\ 0& 4& -1\end{array}\right|=3\hat{i}+\hat{j}+4\hat{k}=\overrightarrow{a}$

Similarly, the lines $x+3y-11=0=2y-z+6$ is the line of intersection of planes $x+3y-11=0$ and $2y-z+6=0.$
Vector along the line of intersection is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& 0\\ 0& 2& -1\end{array}\right|=-3\hat{i}+\hat{j}+2\hat{k}=\overrightarrow{b}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=\left(3\right)(-3)+\left(1\right)\left(1\right)+\left(4\right)\left(2\right)=0$
Therefore, lines are perpendicular.