COMEDK 2025 Afternoon Solved Paper

Concept:The shortest distance between two skew lines is the length of the projection of the vector joining a point on each line onto the common perpendicular direction.Explanation:Identify a point and direction vector for each line.Line 1: Point $P_1 = (8, -9, 10)$ and direction $\mathbf{d}_1 = (3, -16, 7)$.Line 2: Point $P_2 = (15, 29, 5)$ and direction $\mathbf{d}_2 = (3, 8, -5)$.Compute the cross product of the direction vectors:$$\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = (24, 36, 72).$$Find its magnitude: $|\mathbf{d}_1 \times \mathbf{d}_2| = \sqrt{24^2 + 36^2 + 72^2} = 84$.Form the vector from $P_1$ to $P_2$:$\overrightarrow{P_1P_2} = (15-8,\; 29-(-9),\; 5-10) = (7, 38, -5)$.Compute the absolute value of the scalar triple product:$|\overrightarrow{P_1P_2} \cdot (\mathbf{d}_1 \times \mathbf{d}_2)| = |(7,38,-5)\cdot(24,36,72)| = |7\cdot24 + 38\cdot36 - 5\cdot72| = |168 + 1368 - 360| = 1176$.Apply the shortest distance formula:$\text{Distance} = \dfrac{|\overrightarrow{P_1P_2} \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|} = \dfrac{1176}{84} = 14$.Answer:14 units (Option C).