Question

The angle between the lines \(3x = 2y = -z\) and \(-x = 6y = -4z\) is

• A. \(\pi/3\)
• B. \(\pi/4\)
• C. \(\pi/2\)
• D. \(\pi/6\)

Hint:

If the dot product of direction vectors is zero, then the lines are perpendicular and the angle is \(\pi/2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The angle between lines depends on their direction ratios (DRs).
Step 2: Key Formula or Approach:
For lines \(\frac{x}{a_1} = \frac{y}{b_1} = \frac{z}{c_1}\) and \(\frac{x}{a_2} = \frac{y}{b_2} = \frac{z}{c_2}\), \(\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{\sum a_1^2} \sqrt{\sum a_2^2}}\).
Step 3: Detailed Explanation:
Line 1: \(3x=2y=-z \implies \frac{x}{1/3} = \frac{y}{1/2} = \frac{z}{-1} \implies \text{DRs} = (2, 3, -6)\).
Line 2: \(-x=6y=-4z \implies \frac{x}{-1} = \frac{y}{1/6} = \frac{z}{-1/4} \implies \text{DRs} = (-12, 2, -3)\).
Dot product \(= (2)(-12) + (3)(2) + (-6)(-3) = -24 + 6 + 18 = 0\).
Since dot product is \(0\), lines are perpendicular.
Step 4: Final Answer:
Angle is \(\pi/2\).