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. 60${}^0$
• B. 30${}^0$
• C. 90${}^0$
• D. 45${}^0$

Answer 1

Answer: (C)

90${}^0$

Given,

$x-3y-4=0$ and $4y-z+5=0$ is

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 0\\ 0& 4& -1\end{array}\right|$

$=3i+j+4\hat{k}=$

Vector along the line of intersection of planes

$x+3y-11=0x+3y-11=0$ and $2y-z+6=02y-z+6=0$ is

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& 0\\ 0& 2& -1\end{array}\right|$

$=-3i+j+2\hat{k}$

$\therefore$ Angle between lines be $\theta$

$\therefore \cos \theta =\frac{-9+1+8}{\sqrt{9+1+16}\sqrt{9+1+4}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta ={90}^{\circ }$