MHTCET 20 Apr 2026 Shift 1 Paper

Concept:The shortest distance between two skew lines is given by the projection of the vector joining any two points on the lines onto the cross product of their direction vectors.Explanation:For the first line, a point is $A(3,8,3)$ and its direction vector is $\vec{b}_{1}=3\hat{i}-\hat{j}+\hat{k}$.For the second line, a point is $B(-3,-7,6)$ and its direction vector is $\vec{b}_{2}=-3\hat{i}+2\hat{j}+4\hat{k}$.The vector joining the points is $\overrightarrow{AB}=(-3-3)\hat{i}+(-7-8)\hat{j}+(6-3)\hat{k}=-6\hat{i}-15\hat{j}+3\hat{k}$.Now compute the cross product:$$\vec{b}_{1}\times\vec{b}_{2}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix} =-6\hat{i}-15\hat{j}+3\hat{k}$$Then,$$\overrightarrow{AB}\cdot(\vec{b}_{1}\times\vec{b}_{2})=(-6)(-6)+(-15)(-15)+(3)(3)=36+225+9=270$$The magnitude of the cross product is:$$|\vec{b}_{1}\times\vec{b}_{2}|=\sqrt{(-6)^{2}+(-15)^{2}+3^{2}}=\sqrt{270}=3\sqrt{30}$$Therefore, the shortest distance is:$$d=\frac{|\overrightarrow{AB}\cdot(\vec{b}_{1}\times\vec{b}_{2})|}{|\vec{b}_{1}\times\vec{b}_{2}|} =\frac{270}{3\sqrt{30}}=3\sqrt{30}$$Answer:The correct option is B, i.e. $3\sqrt{30}$.