<p>The angle between the lines \(2x = 3y = -z\) and \(6x = -y = -4z\) is</p>
Step-by-Step Solution
Key Concept: Convert both lines to standard direction vector form by expressing them as ratios, then use the dot product formula cos θ = |d₁·d₂|/(|d₁||d₂|) to find the angle between them.
Step 1: Convert first line 2x = 3y = -z to direction ratios. Let 2x = 3y = -z = k, then x = k/2, y = k/3, z = -k Direction ratios: d_1 = (1/2, 1/3, -1) or multiplying by 6: d_1 = (3, 2, -6) Step 2: Convert second line 6x = -y = -4z to direction ratios. Let 6x = -y = -4z = k, then x = k/6, y = -k, z = -k/4 Direction ratios: d_2 = (1/6, -1, -1/4) or multiplying by 12: d_2 = (2, -12, -3) Step 3: Calculate dot product and magnitudes. d_1 · d_2 = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0 |d_1| = √(9 + 4 + 36) = √49 = 7 |d_2| = √(4 + 144 + 9) = √157 Step 4: Find angle using cos θ = |d_1·d_2|/(|d_1||d_2|) cos θ = |0|/(7√157) = 0 ∴ θ = 90° (lines are perpendicular) Answer: B
Correct Answer: B