### Instruction: Find the projection of $\mathbf{a}$ onto $\mathbf{b} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ if $\mathbf{a} \cdot \mathbf{b} = 2.$.### Response: I recall that the projection of $\mathbf{a}$ onto $\mathbf{b}$ is given by the formula $\operatorname{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}.$
I already know $\mathbf{a} \cdot \mathbf{b} = 2$, so I only need to find $\mathbf{b} \cdot \mathbf{b}$ and multiply by $\mathbf{b}$.
I can compute $\mathbf{b} \cdot \mathbf{b}$ using the dot product formula: $\mathbf{b} \cdot \mathbf{b} = (1)^2 + (-3)^2 = 1 + 9 = 10.$
So, $\operatorname{proj}_{\mathbf{b}} \mathbf{a} = \frac{2}{10} \mathbf{b} = \frac{1}{5} \begin{pmatrix} 1 \\ -3 \end{pmatrix}.$
This is a scalar multiple of $\mathbf{b}$, as expected.