### Instruction: When every vector on the line $y = \frac{5}{2} x + 4$ is projected onto a certain vector $\mathbf{w},$ the result is always the vector $\mathbf{p}.$  Find the vector $\mathbf{p}.$.### Response: To project any vector $\mathbf{v}$ onto $\mathbf{w},$ I need to use the formula $\frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w}.$
But the problem says that the projection is always the same vector $\mathbf{p},$ regardless of the choice of $\mathbf{v}$ on the line.
That means that the scalar factor $\frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}}$ must be constant for all $\mathbf{v}$ on the line.
How can I find that constant?
Well, one way is to pick a specific vector $\mathbf{v}$ on the line and find its projection onto $\mathbf{w}.$
Then I can solve for the constant and use it to find $\mathbf{p}.$
What's a convenient vector $\mathbf{v}$ on the line?
I can use the y-intercept, which is $(0,4).$
That's easy to work with.
So I'll plug that into the formula and get $\frac{(0,4) \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w}.$
Now I need to find $\mathbf{w}.$
I don't know its exact components, but I do know something about its direction.
Since the projection is always the same vector, $\mathbf{w}$ must be perpendicular to the line.
That means that $\mathbf{w}$ must have a slope that is the negative reciprocal of the slope of the line.
In other words, $\frac{w_2}{w_1} = - \frac{2}{5},$ where $w_1$ and $w_2$ are the x- and y-components of $\mathbf{w},$ respectively.
This gives me some freedom to choose $\mathbf{w},$ as long as I keep the ratio of its components fixed.
I'll pick a simple one, like $\mathbf{w} = (5,-2).$
This makes the dot product calculations easier.
If I plug this into the formula, I get $\frac{(0,4) \cdot (5,-2)}{(5,-2) \cdot (5,-2)} (5,-2) = \frac{-8}{29} (5,-2).$
So this is the projection of $(0,4)$ onto $\mathbf{w},$ and it must be equal to $\mathbf{p},$ by the problem's assumption.