I have a plane, $\Pi$, a line, $D$, and a basis, $C$:
$$ \Pi : x + 2y + 2z = 0\\ D : \begin{cases} x = t \\ y = 2t \\ z = 4t \\ \end{cases},\;\; t \in \mathbb{R}\\ C = (\vec{i}, \vec{j}, \vec{k}) $$
There's also $T$, which is defined as "the projection on the plane $\Pi$ parallel to the line $D$."
Let $B$ be a basis, with $\vec{b_1}$ and $\vec{b_2}$ two vectors from the plane, and $\vec{b_3}$ the direction vector of $D$.
$$ B = (\vec{b_1}, \vec{b_2}, \vec{b_3}) = (2\vec{i} - k, 2\vec{i} - \vec{j}, \vec{i} + 2\vec{j} + 4\vec{k}) $$
This means that:
$$ [T]_B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$
Right?
The final question is to give $A = [T]_C$ . To get that, I need to use the following transition matrices, which I have to find:
$$ [T]_C = \; _CP_B \; [T]_B \; _BP_C $$
I got
$$ _BP_C = \begin{bmatrix} \frac{1}{12} & \frac{1}{6} & \frac{1}{6} \\ \frac{1}{6} & -\frac{2}{3} & \frac{1}{3} \\ -\frac{5}{24} & 0 & -\frac{1}{3} \\ \end{bmatrix}\\ _CP_B = \begin{bmatrix} 2 & 2 & 1 \\ 0 & -1 & 2 \\ -1 & 0 & 3 \\ \end{bmatrix}\\ [T]_C = \begin{bmatrix} \frac{1}{6} & 0 & \frac{5}{12} \\ \frac{1}{3} & 1 & -\frac{7}{6} \\ -\frac{5}{12} & -\frac{5}{12} & -\frac{5}{24} \\ \end{bmatrix} $$
My problem is that the trace of $[T]_C$ is supposed to be equal to $2$ and its determinant to $0$. The determinant is $0$, but the trace isn't 2 (it's $\frac{23}{24}$). Also, there's supposed to be some sort of relation between the columns of $A = [T]_C$.
Did I do something wrong? Do I even have the right method? Did I make a mistake somewhere? Is there a way to check my calculations?
