Question

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

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{2}\)
• D. \(\frac{\pi}{6}\)

Hint:

Convert each line to symmetric form to get its direction ratios, then take the dot product.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The angle between lines with direction ratios \(\vec b_1\) and \(\vec b_2\) satisfies \(\cos\theta=\dfrac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}\).

Step 2: Direction Ratios
\(3x=2y=-z\Rightarrow\dfrac{x}{1/3}=\dfrac{y}{1/2}=\dfrac{z}{-1}\), so \(\vec b_1=(2,3,-6)\).
\(-x=6y=-4z\Rightarrow\dfrac{x}{-1}=\dfrac{y}{1/6}=\dfrac{z}{-1/4}\), so \(\vec b_2=(-12,2,-3)\).

Step 3: Calculation
\(\vec b_1\cdot\vec b_2=-24+6+18=0\)

Step 4: Result
The lines are perpendicular, so \(\theta=\dfrac{\pi}{2}\).

Final Answer:
The lines are perpendicular, so the angle is pi by two. \[ \boxed{\text{(C) }\dfrac{\pi}{2}} \]