Theory

Lines and planes

Lines
Given a point identified with a vector ${\mathbf{r}}_0$ and a direction vector $\mathbf{v}$, the line through ${\mathbf{r}}_0$ in direction $\mathbf{v}$ is given parametrically by the formula:

$$\mathbf{r}(t)={\mathbf{r}}_0+t\mathbf{v}$$

Given two points ${\mathbf{r}}_0$ and ${\mathbf{r}}_1$, the line passing through these two points is given parametrically by:

$$\mathbf{r}(t)=(1-t){\mathbf{r}}_0+t{\mathbf{r}}_1$$

The line segment from ${\mathbf{r}}_0$ to ${\mathbf{r}}_1$ corresponds to $t\in [0,1]$. The point $\mathbf{r}(t)$ may be considered to be a weighted average of ${\mathbf{r}}_0$ and ${\mathbf{r}}_1$ with weights $1-t$ and $t$.

Planes
A plane in ${\mathbb{R}}^3$ can be determined as the points $(x,y,z)$ satisfying a single linear equation, such as:

$$ax+by+cz+d=0$$

Other formulations of this equation can display more meaning.

A plane is determined in vector form by the data of a single point ${\mathbf{r}}_0$ through which it passes together with a normal vector $\mathbf{n}$ that is perpendicular to the plane. The equation of the plane can be formed by simply stating that vectors in the plane from ${\mathbf{r}}_0$ are perpendicular to $\mathbf{n}$:

$$(\mathbf{r}-{\mathbf{r}}_0)\cdot \mathbf{n}=0$$

If we have components $\mathbf{r}=(x,y,z)$ and ${\mathbf{r}}_0=(x_0,y_0,z_0)$ and $\mathbf{n}=(a,b,c)$, then this formula can be rewritten as:

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

This equation implies that the $a$, $b$, $c$ in the linear equation of the plane are just the components of its normal vector $\mathbf{n}$.

Distances

Point to point
The distance between $\mathbf{p}$ and $\mathbf{q}$ (more accurately, between the point identified by $\mathbf{p}$ and that identified by $\mathbf{q}$) is $|\mathbf{p}-\mathbf{q}|$.

Point to line
The vector $\mathbf{u}$ from a point $\mathbf{p}$ to a line $\mathbf{r}(t)={\mathbf{r}}_0-t\mathbf{v}$, namely to the point on the line closest to $\mathbf{p}$, is given by taking the vector from $\mathbf{p}$ to ${\mathbf{r}}_0$ and removing the component parallel to $\mathbf{v}$:

$$\mathbf{u}=({\mathbf{r}}_0-\mathbf{p})-(({\mathbf{r}}_0-\mathbf{p})\cdot {\mathbf{e}}_{\mathbf{v}}){\mathbf{e}}_{\mathbf{v}}$$

The distance from $\mathbf{p}$ to $\mathbf{r}$ is then $|\mathbf{u}|$.

Projection gives minimal distance

One can verify the formula for $\mathbf{u}$ by minimizing the squared distance from $\mathbf{p}$ to $\mathbf{r}(t)$ using the standard calculus technique.

Point to plane
The vector $\mathbf{u}$ from a point $\mathbf{p}$ to the plane through ${\mathbf{r}}_0$ with normal vector $\mathbf{n}$ (that is, to the nearest point on the plane) is given by projecting $\mathbf{p}-{\mathbf{r}}_0$ onto $\mathbf{n}$, that is:

$$\mathbf{u}=((\mathbf{p}-{\mathbf{r}}_0)\cdot {\mathbf{e}}_{\mathbf{n}}){\mathbf{e}}_{\mathbf{n}}$$

So the distance from $\mathbf{p}$ to this plane is:

$$|\mathbf{u}|=|(\mathbf{p}-{\mathbf{r}}_0)\cdot {\mathbf{e}}_{\mathbf{n}}|=(\mathbf{p}-{\mathbf{r}}_0)\cdot \frac{\mathbf{n}}{|\mathbf{n}|}$$

Plane to plane