MHTCET 20 Apr 2026 Shift 1 Paper

Concept:The plane perpendicular to line $AB$ has its normal vector parallel to $\overrightarrow{AB}$, and it passes through the midpoint of $A$ and $B$.Explanation:Given points are $A(4,5,-10)$ and $B(-1,2,1)$.Direction vector of $AB$ is:$$\overrightarrow{AB} = (-1-4)\hat{i} + (2-5)\hat{j} + (1+10)\hat{k}$$$$\overrightarrow{AB} = -5\hat{i} - 3\hat{j} + 11\hat{k}$$So the normal vector of the required plane is $\vec{n} = -5\hat{i} - 3\hat{j} + 11\hat{k}$.Midpoint $M$ of $A$ and $B$ is:$$M = \left( \frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2} \right)$$$$M = \left( \frac{3}{2}, \frac{7}{2}, -\frac{9}{2} \right)$$Thus position vector of $M$ is $\vec{a} = \frac{3}{2}\hat{i} + \frac{7}{2}\hat{j} - \frac{9}{2}\hat{k}$.Equation of plane with normal $\vec{n}$ passing through $\vec{a}$ is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Compute $\vec{a} \cdot \vec{n}$:$$\vec{a} \cdot \vec{n} = \frac{3}{2}(-5) + \frac{7}{2}(-3) + \left(-\frac{9}{2}\right)(11)$$$$= -\frac{15}{2} - \frac{21}{2} - \frac{99}{2} = -\frac{135}{2}$$Hence the plane equation is:$$\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) = -\frac{135}{2}$$$$\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) + \frac{135}{2} = 0$$Answer:Option A: $\bar{r} \cdot(-5 \hat{i}-3 \hat{j}+11 \hat{k})+\frac{135}{2}=0$