<p>If the lines \(\dfrac{x-2}{1} = \dfrac{y-3}{1} = \dfrac{z-4}{-k}\) and \(\dfrac{x-1}{k} = \dfrac{y-4}{2} = \dfrac{z-5}{1}\) are coplanar, then \(k\) can have</p>
<p>exactly one value.</p>
<p>exactly two values.</p>
<p>exactly three values.</p>
<p>any value.</p>
Step-by-Step Solution
Key Concept: Two lines are coplanar if and only if the scalar triple product of their direction vectors and the vector connecting any point on one line to any point on the other equals zero. This gives us the condition: (P₂ - P₁) · (d₁ × d₂) = 0.
Step 1: Identify direction vectors and points. Line 1: Point P_1 = (2, 3, 4), Direction d_1 = (1, 1, -k) Line 2: Point P_2 = (1, 4, 5), Direction d_2 = (k, 2, 1) Step 2: Find the vector connecting the two points. P_1P_2 = P_2 - P_1 = (1-2, 4-3, 5-4) = (-1, 1, 1) Step 3: Compute the cross product d_1 × d_2. d_1 × d_2 = |i j k| |1 1 -k| |k 2 1| = i(1·1 - (-k)·2) - j(1·1 - (-k)·k) + k(1·2 - 1·k) = i(1 + 2k) - j(1 + k^2) + k(2 - k) = (1 + 2k, -(1 + k^2), 2 - k) Step 4: Apply coplanarity condition: P_1P_2 · (d_1 × d_2) = 0. (-1, 1, 1) · (1 + 2k, -(1 + k^2), 2 - k) = 0 -1(1 + 2k) + 1(-(1 + k^2)) + 1(2 - k) = 0 -1 - 2k - 1 - k^2 + 2 - k = 0 -k^2 - 3k = 0 k(-k - 3) = 0 k = 0 or k = -3 ∴ Answer: B
Correct Answer: B