Distance between a point and a plane I was just working on some review textbook problems in James Stewart's Multivariable Calculus when I encountered a problem that looked like the following:


  Find the distance between the point $(1,2,3)$ and the $xz$-plane.


I've seen questions that ask for the shortest distance between a point and a plane, but I haven't seen this sort of problem yet.

The way I tackled this problem is by using the distance formula like this:

I called the point $A(1,2,3)$ and called the $xz$-plane as $T(x,0,z)$ and then did $||AT||=\sqrt{(x-1)^2+(-2)^2+(z-3)^2}=\sqrt{(x-1)^2+4+(z-3)^2}$. I'm not sure how to proceed after this. (Maybe this is as far as it goes, because there are infinitely many distances from a point to a plane, depending on what $x$ and $z$ are?)

Any hints would be appreciated. Thanks.
In $\mathbf{R}^3$, the distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is simply $$\frac{|ax_1 + by_1 + cz_1 +d|}{\sqrt{a^2+b^2+c^2}}$$ This is derived in Stewart and is quite easy using the projection of any vector from $(x_1,y_1, z_1)$ to any point in the plane onto a normal vector to the plane. As noted in the comment above, distance usually means shortest distance.
$$
{\cal F}\left(\vec{r}\right)
=
{1 \over 2}\,\left(\vec{r} - \vec{P}\right)^{2} - \mu\,\vec{r}\cdot\hat{y}\,,
\quad
0={\partial{\cal F}\left(\vec{r}\right) \over \partial\vec{r}}
=
\left(\vec{r} - \vec{P}\right) - \mu\,\hat{y}
\quad\Longrightarrow\quad
\vec{r} = \vec{P} + \mu\,\hat{y}
$$

$$
0 = \vec{r}\cdot\hat{y} = \vec{P}.\cdot\hat{y} + \mu
\quad\Longrightarrow\quad
\mu = -\vec{P}\cdot\hat{y}
\quad\Longrightarrow\quad
\vec{r} = \vec{P} - \vec{P}\cdot\hat{y}\ \hat{y}
$$

$$
?
=
\left\vert\,\vec{r} - \vec{P}\,\right\vert
=
\left\vert\,\vec{P}\cdot\hat{y}\ \hat{y}\,\right\vert
=
\left\vert\,P_{y}\,\right\vert
=
\color{#ff0000}{\Large 2}
$$