### Instruction: Let $\mathbf{a}$ and $\mathbf{b}$ be orthogonal vectors.  If $\operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 0 \\ 13 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix},$ then find $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 0 \\ 13 \end{pmatrix}.$.### Response: The problem asks me to find the projection of a vector onto another vector, given the projection onto a third vector that is orthogonal to the second one.
I recall that the projection of a vector $\mathbf{u}$ onto a vector $\mathbf{v}$ is given by $\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$
I also know that if two vectors are orthogonal, their dot product is zero.
Therefore, if $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a} \cdot \mathbf{b} = 0.$
This means that the projection of any vector onto $\mathbf{b}$ is orthogonal to the projection of the same vector onto $\mathbf{a}.$
In other words, $\operatorname{proj}_{\mathbf{b}} \mathbf{u}$ is the component of $\mathbf{u}$ that is perpendicular to $\mathbf{a}.$
I can use this fact to find $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 0 \\ 13 \end{pmatrix}$ by subtracting $\operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 0 \\ 13 \end{pmatrix}$ from $\begin{pmatrix} 0 \\ 13 \end{pmatrix}.$
That is, $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 0 \\ 13 \end{pmatrix} = \begin{pmatrix} 0 \\ 13 \end{pmatrix} - \operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 0 \\ 13 \end{pmatrix} = \begin{pmatrix} 0 \\ 13 \end{pmatrix} - \begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 9 \end{pmatrix}.$