Find the equation of a plane,given that the f | vedclass

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(D) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $A(2, 1, 2)$.The vector $\vec{OA}$ is normal to the plane.$\vec{OA} = (2 - 0)\

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$2x + y + 2z = 9$

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(D) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $A(2, 1, 2)$.The vector $\vec{OA}$ is normal to the plane.$\vec{OA} = (2 - 0)\hat{i} + (1 - 0)\hat{j} + (2 - 0)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$.The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Here,$\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$.Substituting these values,we get:$(\vec{r} - (2\hat{i} + \hat{j} + 2\hat{k})) \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 0$$\vec{r} \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = (2\hat{i} + \hat{j} + 2\hat{k}) \cdot (2\hat{i} + \hat{j} + 2\hat{k})$$\vec{r} \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9$.In Cartesian form,where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,the equation is $2x + y + 2z = 9$.

Find the equation of a plane,given that the foot of the perpendicular drawn to the plane from the origin is $(2, 1, 2)$.

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