<p>Given lines can be written in vector forms as \(\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \lambda(\hat{i} + \hat{j} - k\hat{k})\) and \(\vec{l} = (\hat{i} + 4\hat{j} + 5\hat{k}) + u(k\hat{i} + 2\hat{j} + \hat{k})\). The two lines will be coplanar if:</p>
Step-by-Step Solution
Key Concept: Two lines in 3D are coplanar if and only if the scalar triple product of (difference of position vectors) and both direction vectors equals zero: (P₂ - P₁) · (d₁ × d₂) = 0.
Step 1: Identify position and direction vectors. Line 1: P_1 = (2, 3, 4), d_1 = (1, 1, -1) Line 2: P_2 = (1, 4, 5), d_2 = (k, 2, 1) Step 2: For coplanarity, compute (P_2 - P_1) · (d_1 × d_2) = 0. P_2 - P_1 = (-1, 1, 1) Step 3: Calculate d_1 × d_2: d_1 × d_2 = | i j k | = (1+2) i - (1+k) j + (2-k) k |1 1 -1| |k 2 1| = (3, -1-k, 2-k) Step 4: Apply coplanarity condition: (-1, 1, 1) · (3, -1-k, 2-k) = 0 -3 - (1+k) + (2-k) = 0 -3 - 1 - k + 2 - k = 0 -2 - 2k = 0 k = -1 ∴ Answer: B (k = -1)
Correct Answer: B