### Instruction: Find the projection of the vector $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ onto the vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}.$.### Response: To find the projection of one vector onto another, I need to use the formula $\text{proj}_\mathbf{u} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|^2} \mathbf{u}$, where $\mathbf{u}$ is the vector I am projecting onto, and $\mathbf{v}$ is the vector I am projecting.
In this case, $\mathbf{u} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$.
I can compute the dot product of $\mathbf{u}$ and $\mathbf{v}$ by multiplying the corresponding components and adding them up: $\mathbf{u} \cdot \mathbf{v} = (2)(4) + (0)(5) = 8$.
I can also compute the norm of $\mathbf{u}$ by using the Pythagorean theorem: $\|\mathbf{u}\| = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2$.
Now I can plug these values into the formula: $\text{proj}_\mathbf{u} \mathbf{v} = \frac{8}{2^2} \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \frac{8}{4} \begin{pmatrix} 2 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 2 \\ 0 \end{pmatrix}$.
This is the vector that points in the same direction as $\mathbf{u}$, but has the same length as the component of $\mathbf{v}$ along $\mathbf{u}$.
I can write the answer as a column vector or as an ordered pair.