MHTCET 15 May 2024 Shift 2 Paper

Concept:The vector joining the origin to the foot of the perpendicular is normal to the plane.Explanation:Let the foot of the perpendicular be $\mathrm{M}(2,1,-2)$.The normal vector to the plane is$$\vec{n} = \overrightarrow{OM} = 2\hat{i} + \hat{j} - 2\hat{k}$$Since $\mathrm{M}$ lies on the plane, its position vector is $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$.The vector equation of a plane with normal $\vec{n}$ and passing through point with position vector $\vec{a}$ is$$\overline{\mathrm{r}} \cdot \vec{n} = \vec{a} \cdot \vec{n}$$Now compute the right-hand side:$$\vec{a} \cdot \vec{n} = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k})$$$$= 4 + 1 + 4 = 9$$Therefore, the required vector equation is$$\overline{\mathrm{r}} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 9$$Answer:Option A: $\overline{\mathrm{r}} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 9$