The foot of the perpendicular drawn from the | vedclass

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Question

(A) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $M(2, 1, -2)$.Since $OM$ is the normal to the plane,the normal vector $\vec{n}

Vedclass · Accepted Answer

$\bar{r} \cdot (2 \hat{i} + \hat{j} - 2 \hat{k}) = 9$

Vedclass · Answer

(A) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $M(2, 1, -2)$.Since $OM$ is the normal to the plane,the normal vector $\vec{n}$ is given by the vector $\vec{OM} = (2 - 0)\hat{i} + (1 - 0)\hat{j} + (-2 - 0)\hat{k} = 2\hat{i} + \hat{j} - 2\hat{k}$.The vector equation of a plane passing through a point with position vector $\vec{a}$ and having normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Here,$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$.Thus,$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k})$.Calculating the dot product: $(2 	imes 2) + (1 	imes 1) + (-2 	imes -2) = 4 + 1 + 4 = 9$.Therefore,the equation of the plane is $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 9$.

The foot of the perpendicular drawn from the origin to a plane is $M(2, 1, -2)$. Find the vector equation of the plane.

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