Foot of perpendicular from point to plane

Verify if point lies on plane

$$ p: \textbf{r} \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) = 3 $$

Any point that lies on the plane must satisfy the equation of the plane.

Checking if the point (-1, 1, 1) lies on the plane: \begin{align} \left( \begin{matrix} -1 \\ 1 \\ 1 \end{matrix} \right) \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) & = 1 + 0 + 2 \\ & = 3 \\ \\ \therefore (-1, 1, 1) & \text{ lies on plane } p \end{align}

Checking if the point (1, 1, 1) lies on the plane: \begin{align} \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) & = -1 + 0 + 2 \\ & = 1 \\ \\ \therefore (1, 1, 1) & \text{ does not lie on plane } p \end{align}

Foot of perpendicular from point to plane

In the diagram above, the foot of perpendicular from point $Q$ to the plane $p$ is denoted by the point $F$.

There are two ways to find the coordinates of $F$.

Given $$ Q(1, 1, 1) \text{ and } p: \textbf{r} \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) = 3 $$

Form the vector equation of line $QF$, which is parallel to the normal vector and passes through Q: \begin{align} l_{QF}: \textbf{r} & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + \lambda \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right), \lambda \in \mathbb{R} \\ \\ \text{Since } F & \text{ lies on } l_{QF}, \\ \overrightarrow{OF} & =\left( \begin{matrix} 1 - \lambda \\ 1 \\ 1 + 2\lambda \end{matrix} \right) \end{align}

Since $F$ is a point on the plane, it must satisfy the equation of the plane: \begin{align} \text{Since } F & \text{ lies on } p, \\ \overrightarrow{OF} \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) & = 3 \\ \left( \begin{matrix} 1 - \lambda \\ 1 \\ 1 + 2\lambda \end{matrix} \right) \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) & = 3 \\ -1 + \lambda + 0 + 2 + 4\lambda & = 3 \\ 1 + 5\lambda & = 3 \\ 5\lambda & = 2 \\ \lambda & = {2 \over 5} \\ \\ \overrightarrow{OF} & = \left( \begin{matrix} 1 - {2 \over 5} \\ 1 \\ 1 + 2\left(2 \over 5\right) \end{matrix} \right) \\ & = \left( \begin{matrix} {3 \over 5} \\ 1 \\ {9 \over 5} \end{matrix} \right) \\ \\ \therefore & \phantom{.} F \left( {3 \over 5}, 1, {9 \over 5} \right) \end{align}

Find a point that satisfies the equation of the plane: \begin{align} \left( \begin{matrix} -3 \\ 0 \\ 0 \end{matrix} \right) \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) & = 3 + 0 + 0 \\ & = 3 \\ \\ \text{Let } A \text{ denote } & \text{the point (-3, 0, 0) on the plane} \end{align}

Find the projection vector, $\overrightarrow{QF}$ \begin{align} \overrightarrow{QF} & = \left( \overrightarrow{QA} \cdot \hat{ \textbf{n} } \right) \hat{ \textbf{n} } \\ \\ \overrightarrow{QA} & = \overrightarrow{OA} - \overrightarrow{OQ} \\ & = \left( \begin{matrix} -3 \\ 0 \\ 0 \end{matrix} \right) - \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} -4 \\ -1 \\ -1 \end{matrix} \right) \\ \\ \hat{ \textbf{n} } & = {1 \over | \textbf{n} | } \textbf{n} \\ & = {1 \over \sqrt{(-1)^2 + 0^2 + 2^2} } \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ & = {1 \over \sqrt{5} } \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ \\ \overrightarrow{QF} & = \left[ \left( \begin{matrix} -4 \\ -1 \\ -1 \end{matrix} \right) \cdot {1 \over \sqrt{5} } \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \right] {1 \over \sqrt{5} } \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ & = {1 \over 5} \left[ \left( \begin{matrix} -4 \\ -1 \\ -1 \end{matrix} \right) \cdot \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \right] \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ & = {1 \over 5} [ 4 + 0 + (-2)] \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ & = {2 \over 5} \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ \\ \overrightarrow{OF} & = \overrightarrow{OQ} + \overrightarrow{QF} \\ & = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) + {2 \over 5} \left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) \\ & = \left( \begin{matrix} {3 \over 5} \\ 1 \\ {9 \over 5} \end{matrix} \right) \\ \\ \therefore & \phantom{.} F \left( {3 \over 5}, 1 , {9 \over 5} \right) \end{align}