Distances Between Linear Objects in Three-Dimensional Space

We will now look at some ways to calculate distances between various linear objects in $\mathbb{R}^3$. Before we do though, we should note that by "distance" we mean the shortest distance between two objects

The Distance Between a Point and a Plane

Theorem 1: Let $\Pi$ be the plane defined by the equation $Ax + By + Cz + D = 0$. Then the distance between the point $P(x_0, y_0, z_0)$ and $\Pi$ is given by the formula: $\mathrm{distance} = \frac{\mid Ax_0 + By_0 + Cz_0 + D \mid}{\sqrt{A^2 + B^2 + C^2} }$.

• Proof: Let $\Pi$ be the plane described above. Then $\vec{n} = (A, B, C)$ is a normal to $\Pi$. Let $P(x, y, z)$ be any point on the plane. Then the vector $\vec{PP_0} = (x_0 - x, y_0 - y, z_0 - z)$ is a vector from the point $P$ to $P_0$.

• We can see that the distance $D$ from $P_0$ to $\Pi$ is given as the scalar projection of the vector $\vec{PP_0}$ on $\vec{n}$ (or the length of the vector projection of $\vec{PP_0}$ onto $\vec{n}$), that is:

(1)

\begin{align} D = \mathrm{comp}_{\vec{n}} \vec{PP_0} \\ D = \frac{\mid \vec{n} \cdot \vec{PP_0} \mid}{\| \vec{n} \|} \\ D = \frac{\mid (A, B, C) \cdot (x_0 - x, y_0 - y, z_0 - z) \mid}{\sqrt{A^2 + B^2 + C^2}} \\ D = \frac{\mid A(x_0 - x) + B(y_0 - y) + C(z_0 - z) \mid}{\sqrt{A^2 + B^2 + C^2}} \\ D = \frac{\mid Ax_0 + By_0 + Cz_0 + D \mid}{\sqrt{A^2 + B^2 + C^2}} \quad \blacksquare \end{align}

Example 1

Calculate the distance between the plane $x + 2y + 3z + 4 = 0$ and the point $( 5, 1, 4)$.

Using the formula from above, we get that:

(2)

\begin{align} \quad \mathrm{distance} = \frac{\mid 1(5) + 2(1) + 3(4) + 4 \mid}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{23}{\sqrt{14}} \end{align}

The Distance Between Two Parallel Planes

Theorem 2: If $\Pi_1$ and $\Pi_2$ are two parallel planes, then the distance between $\Pi_1$ and $\Pi_2$ can be calculated by taking any point on either plane and calculating the distance from that point to the other plane.

• Proof: Let $P$ and $Q$ be two different points on $\Pi_1$, and let the distance from $P$ to $\Pi_2$ be $D_1$ and let the distance from $Q$ to $\Pi_2$ be $D_2$. Suppose that $D_1 \neq D_2$. Then this implies that $\Pi_1$ and $\Pi_2$ are not parallel which is a contradiction. Therefore $D_1 = D_2$. $\blacksquare$

Example 2

Calculate the distance between the planes $x + 2y + 3z + 4 = 0$ and $2x + 4y + 6z + 5 = 0$.

We note that these two planes are parallel since their normals, $(1, 2, 3)$ and $(2, 4, 6)$ are scalar multiples of each other. Consider the point $(-1, 0, -1)$ on the first plane. Then from theorem 2, we have that the distance between these two planes are:

(3)

\begin{align} \quad \mathrm{distance} = \frac{\mid 2(-1) + 4(0) + 6(-1) + 5 \mid}{\sqrt{2^2 + 4^2 + 6^2}} = \frac{ \mid -2 - 6 + 5 \mid}{\sqrt{56}} = \frac{3}{\sqrt{56}} \end{align}