I'm doing pretty basic linear algebra practice right now and came across this problem:
Find the distance from the vector $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ to the plane $x_1 + x_3 = 0$.
My initial attempt:
I found the normal vector of the plane, which is $\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$. I then tried finding the projection of the original vector $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ onto the normal vector $\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$, so I got $\begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$.
I then subtracted the original vector from this projection onto the plane and got $\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$.
So, I found the length of the vector $\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$ which is $\sqrt{6}$.
But, this wasn't the right answer, and I'm not sure why my method doesn't work. Shouldn't the length of an orthogonal projection of a vector onto a plane be the distance between the vector and the plane? I don't believe I made any simple algebraic mistakes either. Using this formula:
The distance from $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is $$ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$
I got $2\sqrt{2}$.
All of your computations are correct, the issue is mostly conceptual.
Let $V$ denote the subspace of $\mathbb R^3$ given by the plane $x_1 + x_3 = 0.$ Then the projection of $(1,2,3)$ onto this plane is $\operatorname{proj}_V(1,2,3) = (-1,2,1),$ as you calculated. Note that this projection vector is in the plane $x_1 + x_3 = 0$ (the first component plus the third component is zero).
You also found that $(1,2,3) = \operatorname{proj}_V(1,2,3) + (2,0,2).$ So $(2,0,2)$ is the vector going from $\operatorname{proj}_V(1,2,3)$ to the point $(1,2,3)$ and so the length of this vector will be the distance from the point to the plane. Draw a picture to see this a bit more clearly!