<p>If the lines \(\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}\) and \(\dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1}\) intersect, then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Two lines in 3D intersect if and only if they are coplanar, which means the scalar triple product of (point difference, direction₁, direction₂) equals zero. This condition eliminates the parameter k uniquely.
Step 1: Extract line parameters. Line 1: Point A_1(1, -1, 1), direction d _1 = (2, 3, 4) Line 2: Point A_2(3, k, 0), direction d _2 = (1, 2, 1) Step 2: For intersecting lines, use coplanarity condition. Vector A_1A_2 = (3-1, k-(-1), 0-1) = (2, k+1, -1) The scalar triple product must equal zero: $\begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$ Step 3: Expand the determinant along row 1. $2\begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} - (k+1)\begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} - 1\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 0$ $2(3-8) - (k+1)(2-4) - 1(4-3) = 0$ $2(-5) - (k+1)(-2) - 1 = 0$ $-10 + 2(k+1) - 1 = 0$ $-10 + 2k + 2 - 1 = 0$ $2k - 9 = 0$ $k = \frac{9}{2}$ ∴ Answer: C
Correct Answer: C