I'm not that sure on how to reflect the Shape $A$. I know where the line $x = 1$ but I don't know which way to reflect it. To the left or up? Thanks, I know it seems a bit easy.
[Math] Reflect the Shape A in the line x = 1.
linear algebra
Related Solutions
Hint. Suppose that the required line intersects the given line at the point with parameter $a$, that is, $$(1+a,\,1-a,\,2a)\ .$$ Then the direction of the required line is $$(1+a,\,1-a,\,2a)-(0,1,2)=(1+a,\,-a,\,2a-2)\ .$$ This must be perpendicular to the direction of the given line, which is $(1,-1,2)$. Therefore $$(1+a,\,-a,\,2a-2)\cdot(1,-1,2)=0\ .$$ Can you use this equation to find $a$ and then finish the problem?
Good luck!
Alternative method. Let ${\bf u}=(1,-1,2)$ be the direction of the given line. A vector in the plane of both lines is $${\bf v}=(1,1,0)-(0,1,2)=(1,0,-2)\ .$$ A vector perpendicular to the plane containing both lines is $${\bf n}={\bf u}\times{\bf v}$$ and the required line is perpendicular to both $\bf u$ and $\bf n$, so its direction is $${\bf u}\times{\bf n}={\bf u}\times({\bf u}\times{\bf v})\ .$$
When a question asks you to find a matrix representing a linear transformation $T$ that is only described geometrically, your task is to figure out how that $T$ transforms a basis for your domain.
Unfortunately I can't find a good image on Google Images to describe reflection through a line in $\Bbb R^3$ (and my pgfplots-fu is still pretty basic), but I'll try to describe what it means. Consider an arbitrary point in $\Bbb R^3$. Now connect that point to the $y$-axis by a line segment that is orthogonal to the $y$-axis. Extend that line segment past $y$ by the same length as the distance from the point to the $y$-axis. The far end of that line segment is then at the point that is the reflection of your point across the $y$-axis.
Let's see how this affects the standard basis $\{\hat x, \hat y, \hat z\}$. $T(\hat x)$ goes back along the $x$-axis, goes through the $y$-axis at the origin, and then to $-\hat x$. Thus $T(\hat x) = -\hat x$. Likewise $T(\hat z) = -\hat z$. $\hat y$ on the other hand is unaffected by this transformation -- it's its own reflection across the $y$-axis.
Thus, if $T(x) = Ax$ where $A$ is a $3\times 3$ matrix, then $$\begin{matrix}A \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 0\end{bmatrix}, & A \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, & A \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ -1\end{bmatrix}\end{matrix}$$
Therefore $$A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}$$
The difference between reflecting through a line vs a plane in $\Bbb R^3$ is comparable to reflecting through the origin vs a line in $\Bbb R^2$. Go back and look up the geometric properties of even and odd functions if you don't remember how these reflections work in $\Bbb R^2$ (note however that you can still reflect through the origin in $\Bbb R^3$).
Try to visualize each of these reflections in $\Bbb R^2$ and $\Bbb R^3$. Then consider how you think reflections would work in $\Bbb R^n$ for other values of $n$.
Best Answer
If you were to fold the graph on the line you're reflecting through, the original shape and the reflected shape should line up exactly.
This could apply to any line: vertical, horizontal or otherwise.