MHTCET 15 Apr 2026 Shift 1 Paper

Concept:The direction ratios of a line perpendicular to two given lines are found using the cross product of their direction ratios.The direction cosines are then obtained by dividing each direction ratio by the magnitude of the direction vector.Explanation:From the first line, the direction ratios are $(2, -3, 1)$.From the second line, the direction ratios are $(1, 2, -2)$.A line perpendicular to both lines has direction ratios equal to the cross product of these two vectors:$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{vmatrix}$$This gives $\hat{i}(6 - 2) - \hat{j}(-4 - 1) + \hat{k}(4 + 3)$.Simplifying, the direction ratios are $(4, 5, 7)$.The magnitude of this direction vector is $\sqrt{4^2 + 5^2 + 7^2}$.So, magnitude $= \sqrt{16 + 25 + 49} = \sqrt{90} = 3\sqrt{10}$.Therefore, the direction cosines are $\left(\frac{4}{3\sqrt{10}}, \frac{5}{3\sqrt{10}}, \frac{7}{3\sqrt{10}}\right)$.Since direction cosines can be taken along either direction, both positive and negative signs are valid.Answer:The required direction cosines are $\frac{\pm 4}{3\sqrt{10}}, \frac{\pm 5}{3\sqrt{10}}, \frac{\pm 7}{3\sqrt{10}}$.Hence, the correct option is Option D.