MHTCET 16 Apr 2026 Shift 2 Paper

Concept:For a line parallel to $\vec{a} \times \vec{b}$, its direction ratios are proportional to the components of $\vec{a} \times \vec{b}$.Explanation:Given vectors:$\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=3\hat{i}-4\hat{j}$.So, $\vec{a}=(2,1,2)$ and $\vec{b}=(3,-4,0)$.Compute the cross product:$$\vec{a}\times\vec{b} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 2\\ 3 & -4 & 0 \end{vmatrix} =\hat{i}(1\cdot0-2(-4))-\hat{j}(2\cdot0-2\cdot3)+\hat{k}(2(-4)-1\cdot3) =8\hat{i}+6\hat{j}-11\hat{k}$$Thus, direction ratios of the required line are $8,\ 6,\ -11$.The line passes through $(0,0,0)$ and $(4,3,c)$, so its direction ratios are also $4,\ 3,\ c$.Since the line is parallel to $\vec{a}\times\vec{b}$, these direction ratios must be proportional:$$(4,3,c)\parallel(8,6,-11)$$Here, $4=\frac{8}{2}$ and $3=\frac{6}{2}$, so the proportionality factor is $\frac12$.Hence,$$c=\frac{-11}{2}$$So the line passes through $\left(4,3,-\frac{11}{2}\right)$ with direction ratios $8,\ 6,\ -11$.The Cartesian equation is:$$\frac{x-4}{8}=\frac{y-3}{6}=\frac{z+\frac{11}{2}}{-11}$$Answer:Option C: $\frac{x-4}{8}=\frac{y-3}{6}=\frac{z+\frac{11}{2}}{-11}$