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

Hint:

A line represented by two planes has a direction vector equal to the cross product of the planes' normals.

Answer 1

The Correct Option is A

Step 1: Find Direction Ratios of Line 1
Normals to planes: $\vec{n_1} = (1, -3, 0), \vec{n_2} = (0, 4, -1)$.
Direction $\vec{b_1} = \vec{n_1} \times \vec{n_2} = (3, 1, 4)$.
Step 2: Find Direction Ratios of Line 2
Normals: $\vec{n_3} = (1, 3, 0), \vec{n_4} = (0, 2, -1)$.
Direction $\vec{b_2} = \vec{n_3} \times \vec{n_4} = (-3, 1, 2)$.
Step 3: Calculate Dot Product
$\vec{b_1} \cdot \vec{b_2} = 3(-3) + 1(1) + 4(2) = -9 + 1 + 8 = 0$.
Since the dot product is zero, the lines are perpendicular ($\pi/2$).
Final Answer: (A)