The problem
Find the orthogonal projection of $v=\begin{pmatrix} 4 \cr -6 \cr -6 \end{pmatrix}$ onto the line L through $v=\begin{pmatrix} 8 \cr -2 \cr -5 \end{pmatrix}$ and the origin.
What I've done so far
$$\operatorname{Proj}_L(\mathbf{x}) = \frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v}$$
So, in this case, we have
$$\mathbf{v} = \begin{pmatrix}4\\-6\\-6\end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix}8\\-2\\-5\end{pmatrix},$$
so that
$$\mathbf{v} \cdot \mathbf{x} = 4 \cdot 8 -6 \cdot -2 - 6 \cdot -5 = 74, \quad \mathbf{v} \cdot \mathbf{v} = 4^2 - 6^2 - 6^2 = -56,$$
hence
$$\operatorname{Proj}_L(\mathbf{x}) = \frac{74}{-56}\begin{pmatrix}4\\-6\\-6\end{pmatrix}.$$
Am I right so far with this? Would I need to go further, with the following:
$$\operatorname{Refl}_L(\mathbf{x}) = 2\operatorname{Proj}_L(\mathbf{x}) - \mathbf{x} = 2\frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v} - \mathbf{x},$$
which yields
$$\operatorname{Refl}_L(\mathbf{x}) = 2 \cdot \frac{74}{-56}\begin{pmatrix}4\\-6\\-6\end{pmatrix} - \begin{pmatrix}8\\-2\\-5\end{pmatrix} = \begin{pmatrix}-130\\125\\146\end{pmatrix}.$$