### Instruction: Let $P$ be a plane passing through the origin.  When $\begin{pmatrix} 5 \\ 3 \\ 5 \end{pmatrix}$ is projected onto plane $P,$ the result is $\begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix}.$  When $\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}$ is projected onto plane $P,$ what is the result?### Response: To find the projection of a vector onto a plane, I need to know the normal vector of the plane, which is perpendicular to the plane.
I can use the given information to find the normal vector of plane $P$.
If I subtract the projection from the original vector, I get a vector that is orthogonal to the plane.
So, $\begin{pmatrix} 5 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$ is a vector orthogonal to plane $P$.
This means that $\begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$ is a scalar multiple of the normal vector of plane $P$.
I can choose any scalar multiple as the normal vector, so I will pick $\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ as the normal vector of plane $P$ for convenience.
Now that I have the normal vector, I can use the formula for the projection of a vector onto a plane.
The formula is $\text{proj}_{P}(\vec{v}) = \vec{v} - \frac{\vec{n} \cdot \vec{v}}{\vec{n} \cdot \vec{n}} \vec{n}$, where $\vec{v}$ is the vector to be projected, $\vec{n}$ is the normal vector of the plane, and $\cdot$ is the dot product.
I want to find the projection of $\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}$ onto plane $P$, so I plug in the values into the formula.
$\text{proj}_{P}(\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}) = \begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix} - \frac{\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}}{\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$
I simplify the dot products in the numerator and denominator.
$\text{proj}_{P}(\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}) = \begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix} - \frac{4 + 0 + 14}{1 + 1 + 4} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$
I simplify the fraction and multiply it by the normal vector.
$\text{proj}_{P}(\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}) = \begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix} - \frac{9}{3} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix} - \begin{pmatrix} 3 \\ -3 \\ 6 \end{pmatrix}$
I subtract the vectors and get the final answer.