Let $F$ : $\Bbb R^3$ → $\Bbb R^3$ be a linear transformation with $F$($1$, $0$, $0$) = ($1, 1, −1$), F($1, 1, 0$) = ($2, −1, 0$) and $F$($1, 1, 1$) = ($3, 0, −1$). Let $G$ : $\Bbb R^3$ → $\Bbb R^3$ be orthogonal projection onto the line through ($1, 2, a$) for $a$ ∈ $\Bbb R$. Find the value of $a$ for which $G ◦ F$ = $0$. Note that $G ◦ F = 0$ means that $G ◦ F(x, y, z) = 0$ for all ($x, y, z$) ∈ $\Bbb R^3$
Working -
I thought this question involved first finding the standard matrix of each linear transformation. So I found the standard matrix for $F$ but I am not sure on how to find the standard matrix for $G$. Does $G$ even have a standard matrix? Am I going about this question the correct way? Any advice is greatly appreciated. Feel free to edit my question for clarity. Thank you
Denote by $P_a$ the plane orthogonal to the line $L_a$ having $u_a=(1,2,a)$ for direction vector.
As $G$ is the orthogonal projector onto the line $L_a$ you have $G \circ F=0$ if and only if $F(\mathbb R^3) \subset P_a$. Also, $P_a$ has for equation $x+2y+az=0$ and $F(\mathbb R^3) = \text{span}\{(1, 1, −1), (2, −1, 0), (3, 0, −1)\}$.
Now you need to have $$\begin{cases} 1+2.1+a.(-1) = 0 \\ 2+2.(-1)+a.0=0 \\ 3+2.0+a.(-1)=0 \end{cases}$$
Which has $\color{red}{a=3}$ as a unique solution. Conversely, you can verify that for $a=3$, the image of $F$ is included in the plane $P_3 \equiv x+2y+3z=0$, hence $G \circ F=0$.
Conclusion: $a=3$ is the unique solution.
And do not compute the matrices if not required!