Problem: Suppose a helicopter is currently at the position $(-1,0, 1)$ and moves in the direction $(1,2,2)$.
a) Find a system of linear equations $Ax = b$, such that its solution set is the set of all points along the plane's tranjectory.
b) Now suppose a second helicopter is at $(0,0,0)$ and travels along the same direction $(1,2,2)$. Let $T$ be the smallest subspace containing this trajectory. Compute a matrix $P \in R^{3×3}$ such that, given a point $x \in R^3$ , the point in T that is closest to x is equal to P x.
Here's what I have so far:
a) I'll first find a matrix A so that $$A\begin{bmatrix} 1 \\ 2 \\ 2\\ \end{bmatrix} = 0$$ This will model what it'd be if we started at the origin instead. Just some basic algebra will give us $$A = \begin{bmatrix} 2 & 1 & -2\\ 2 & -1 & 0\\ -4 & 1 & 1 \end{bmatrix}$$ Now we need to shift by our starting point. $$A \left( \begin{bmatrix} 1 \\ 2 \\ 2\\ \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \\ 1\\ \end{bmatrix}\right) = \begin{bmatrix} -4 \\ -2 \\ 5\\ \end{bmatrix}$$ So the answer should be $$\begin{bmatrix} 2 & 1 & -2\\ 2 & -1 & 0\\ -4 & 1 & 1 \end{bmatrix}x = \begin{bmatrix} -4 \\ -2 \\ 5\\ \end{bmatrix}$$ right?
Now for part b) I'm totally lost on how to go about doing this.
I think I need to find a projection matrix into subspace T, but I don't even know what T is. Any ideas for how to go about this?
Since $T$ is the smallest subspace containing the trajectory from $u = (0,0,0)$ to $v = (1,2,2)$ then $T$ is of course the straight line spanned by the vector $v-u = v$. This is because this subspace has dimension $1$. The only subspace with a smaller dimension is the null subspace (dimension $0$) but this is not the case, because we already know that $T$ contains the vector $v$.