This is a practice problem in a 400-level linear algebra course. This is supposed to be review of 200-level material. For some reason, I'm having a mental block trying to understand the solution given.
Problem: Let $\bar{a}=\big[\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\big]$, $\bar{b}=\big[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\big]$, and $\bar{v}=\big[\begin{smallmatrix}0 \\ 0 \end{smallmatrix}\big]$. Find the orthogonal projection $\bar{p}$ of the vector $\bar{v}$ on the line $\{t \bar{a}+\bar{b}, -\infty < t < \infty \}$.
Solution: Let $\bar{p} = t\bar{a}+\bar{b}$, then $0=\bar{a}^T(\bar{v}-\bar{p})$, and $t=-\frac{1}{2}$. Finally $\bar{p}=\big[\begin{smallmatrix} 1/2 \\ -1/2 \end{smallmatrix}\big]$.
The part I don't understand is that $\bar{v}=\big[\begin{smallmatrix}0 \\ 0 \end{smallmatrix}\big]$. I keep wanting to apply the formula: $$ \text{proj}_\bar{p}\bar{v}=\frac{\bar{v}\cdot\bar{p}}{\bar{p}\cdot\bar{p}}\bar{p}$$ But $\bar{v}$ just ends up cancelling everything out. Is there another formula I'm missing?
If someone can explain the reasoning behind this, I would greatly appreciate it.
$v$ is the vector (the origin in this case) you are projecting on to the line $t \mapsto ta+b$. The line does not pass through the origin.
Solve $\min_t {1 \over 2} \|ta+b-v\|^2 $. Differentiating and finding where the derivative is zero gives $a^T (ta+b-v) = 0$ or $t = {a^T(v-b) \over \|a\|^2}$, computing gives $t=-{1 \over 2}$ and so the closest point is $ta+b = \binom{{1 \over 2}}{-{1 \over 2}}$.