MHTCET 11 May 2024 Shift 1 Paper

Concept:Two lines intersect when the shortest distance between them is zero, which makes the determinant of their direction vectors and the vector joining their points zero.Explanation:For the first line, a point is $(-1, -k, 4)$ and its direction ratios are $(-10, -1, 1)$.For the second line, a point is $(-10, -1, 1)$ and its direction ratios are $(-1, -3, 4)$.The vector joining the two points is:$$(-10 - (-1),\; -1 - (-k),\; 1 - 4) = (-9,\; k - 1,\; -3)$$Since the lines intersect, the determinant formed by this vector and the two direction vectors must be $0$:$$\begin{vmatrix} -9 & k-1 & -3 \\ -10 & -1 & 1 \\ -1 & -3 & 4 \end{vmatrix} = 0$$Expanding the determinant gives:$$(-9)[(-1)(4) - (1)(-3)] - (k-1)[(-10)(4) - (1)(-1)] + (-3)[(-10)(-3) - (-1)(-1)] = 0$$$$9 + 39(k-1) - 87 = 0$$$$39k - 117 = 0$$$$k = 3$$Answer:The value of $k$ is $3$, which matches option B.