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: Understanding the Concept:
Each line is the intersection of two planes.
Direction vector of a line is the cross product of normals of the two planes.
Step 2: Key Formula or Approach:
$d_1 = (1, -3, 0) \times (0, 4, -1) = (3, 1, 4)$.
$d_2 = (1, 3, 0) \times (0, 2, -1) = (-3, 1, 2)$.
Step 3: Detailed Explanation:
Check dot product: $d_1 \cdot d_2 = (3)(-3) + (1)(1) + (4)(2) = -9 + 1 + 8 = 0$.
The angle is $90^\circ$ since the dot product is zero.
Step 4: Final Answer:
The angle is $\frac{\pi}{2}$.