Question

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

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

Hint:

Convert each line to symmetric form to read its direction ratios, then test the dot product.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The angle between two lines in space comes from the dot product of their direction ratios. First we need these ratios from the given equations.

Step 2: Key Formula or Approach:
For a line \(\dfrac{x}{l} = \dfrac{y}{m} = \dfrac{z}{n}\), the direction ratios are \((l, m, n)\). Rewrite \(2x = 3y = -z = k\) as \(x = \dfrac k2\), \(y = \dfrac k3\), \(z = -k\).

Step 3: Detailed Explanation:
First line: \(\left(\dfrac12, \dfrac13, -1\right)\). Multiply by 6 to clear fractions: \((3, 2, -6)\).
Second line: \(6x = -y = -4z = k\) gives \(x = \dfrac k6\), \(y = -k\), \(z = -\dfrac k4\). Direction \(\left(\dfrac16, -1, -\dfrac14\right)\). Multiply by 12: \((2, -12, -3)\).
Dot product:
\[ 3(2) + 2(-12) + (-6)(-3) = 6 - 24 + 18 = 0 \]
So \(\cos\theta = 0\) and the lines are perpendicular.
\[ \theta = \frac\pi2 \]
The other options give \(\cos\theta\) values of \(\tfrac1{\sqrt2}\), \(\tfrac12\) or \(\tfrac{\sqrt3}{2}\), none of which matches 0.

Final Answer:
The angle between the lines is \(\dfrac\pi2\), option (D). \[ \boxed{\frac{\pi}{2} \text{ (D)}} \]