<p>Given lines are \(2x = 3y = -z\) and \(6x = -y = -4z\). The angle between the two lines is:</p>
Step-by-Step Solution
Key Concept: Convert both lines to parametric form to extract direction vectors, then use the dot product formula: cos θ = |d₁·d₂|/(|d₁||d₂|) where d₁ and d₂ are direction vectors. The absolute value ensures we get the acute angle between lines.
Step 1: Extract direction vector from first line 2x = 3y = -z Let 2x = 3y = -z = t, then x = t/2, y = t/3, z = -t Direction vector: d_1 = (1/2, 1/3, -1) or proportionally (3, 2, -6) Step 2: Extract direction vector from second line 6x = -y = -4z Let 6x = -y = -4z = s, then x = s/6, y = -s, z = -s/4 Direction vector: d_2 = (1/6, -1, -1/4) or proportionally (2, -12, -3) Step 3: Calculate dot product d_1 · d_2 = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0 Step 4: Apply angle formula cos θ = |d_1·d_2|/(|d_1||d_2|) = |0|/(√(9+4+36)√(4+144+9)) = 0 ∴ θ = 90° (or π/2 radians) The lines are perpendicular . ∴ Answer: B
Correct Answer: B