3.19 DistancesHL ONLY

1. Distance Between a Point and a Line

Calculating the shortest distance from a point to a line requires finding the perpendicular foot on the line.

Methodology:

  1. Define the foot point $P$ using the parametric equation of line $L$.
  2. Construct the vector $\overrightarrow{AP}$.
  3. Since the shortest path is perpendicular, set the dot product of $\overrightarrow{AP}$ and the line's direction vector $\vec{b}$ to zero ($\overrightarrow{AP} \cdot \vec{b} = 0$).
  4. Solve for $\lambda$, find the coordinates of $P$, and calculate the distance $d(A,P)$.

EXAMPLE 1 (Point to Line Calculation)

Determine the distance between $A(1,2,3)$ and line $L: \vec{r} = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$.

Foot $P$ has coordinates $(5+3\lambda, 7+2\lambda, 9+\lambda)$.
Calculate the vector $\overrightarrow{AP}$:
$\overrightarrow{AP} = \begin{pmatrix} 5+3\lambda-1 \\ 7+2\lambda-2 \\ 9+\lambda-3 \end{pmatrix} = \begin{pmatrix} 4+3\lambda \\ 5+2\lambda \\ 6+\lambda \end{pmatrix}$
Set $\overrightarrow{AP} \cdot \vec{b} = 0$:
$$\begin{aligned} 3(4+3\lambda) + 2(5+2\lambda) + 1(6+\lambda) &= 0 \\ 12 + 9\lambda + 10 + 4\lambda + 6 + \lambda &= 0 \\ 14\lambda + 28 &= 0 \\ 14\lambda &= -28 \\ \lambda &= -2 \end{aligned}$$
Substitute $\lambda = -2$ to find point $P(-1, 3, 7)$.
Calculate the distance:
$d(A,P) = \sqrt{(-1-1)^2 + (3-2)^2 + (7-3)^2} = \sqrt{4 + 1 + 16} = \mathbf{\sqrt{21}}$

2. Distance Between Two Lines

If the lines are parallel: Select a point on Line 1 and calculate its distance to Line 2 using the point-to-line method.
If the lines are skew: The shortest distance is a line segment perpendicular to both lines. This establishes two foot points, $P$ and $Q$.

EXAMPLE 2 (Distance Between Skew Lines)

Find the distance between $L_1: \vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}$ and $L_2: \vec{r} = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$.

Define the points: $P(1+4\mu, 2+5\mu, 3)$ and $Q(5+3\lambda, 7+2\lambda, 9+\lambda)$.
Find the vector $\overrightarrow{PQ}$:
$\overrightarrow{PQ} = \begin{pmatrix} 4+3\lambda-4\mu \\ 5+2\lambda-5\mu \\ 6+\lambda \end{pmatrix}$
Set the dot products of $\overrightarrow{PQ}$ with both direction vectors to zero:
$\overrightarrow{PQ} \cdot \vec{b}_1 = 0 \implies 4(4+3\lambda-4\mu) + 5(5+2\lambda-5\mu) + 0 = 0 \implies 22\lambda - 41\mu = -41$

$\overrightarrow{PQ} \cdot \vec{b}_2 = 0 \implies 3(4+3\lambda-4\mu) + 2(5+2\lambda-5\mu) + 1(6+\lambda) = 0 \implies 14\lambda - 22\mu = -28$
Solving the system yields $\lambda = -\dfrac{41}{15}$ and $\mu = -\dfrac{7}{15}$.
Substitute these values to find the points and compute the distance $|\overrightarrow{PQ}|$.

3. Distance Between a Point and a Plane

Calculating the distance from a point to a plane uses a similar method. The perpendicular line from the point to the plane is parallel to the plane's normal vector $\vec{n}$.

EXAMPLE 3 (Point to Plane Calculation)

Find the distance between $A(3,4,6)$ and Plane $\Pi: 2x + 3y + 5z = 10$.

Key point: The line through $A$ is parallel to the normal vector $\vec{n} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$.
The equation of this line is $\vec{r} = \begin{pmatrix} 3 \\ 4 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$.
Point $P$ has coordinates $(3+2\lambda, 4+3\lambda, 6+5\lambda)$. Since $P$ lies on the plane, substitute these coordinates into the plane equation:
$$\begin{aligned} 2(3+2\lambda) + 3(4+3\lambda) + 5(6+5\lambda) &= 10 \\ 6 + 4\lambda + 12 + 9\lambda + 30 + 25\lambda &= 10 \\ 48 + 38\lambda &= 10 \\ 38\lambda &= -38 \iff \lambda = -1 \end{aligned}$$
This yields point $P(1,1,1)$.
Calculate the distance:
$d(A,P) = \sqrt{(1-3)^2 + (1-4)^2 + (1-6)^2} = \sqrt{(-2)^2 + (-3)^2 + (-5)^2} = \sqrt{4+9+25} = \mathbf{\sqrt{38}}$

4. Distance Between a Line and a Plane

This applies only if the line is parallel to the plane. We find the distance from a point $A$ on line $L$ to the plane $\Pi$.

A(3,4,6) P n Line L Plane Π

EXAMPLE 4 (Line to Plane Calculation)

Consider Line $L: \vec{r} = \begin{pmatrix} 3 \\ 4 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 2 \\ 0 \end{pmatrix}$ and Plane $\Pi: 2x + 3y + 5z = 10$.

Since the line is parallel to the plane, we find the distance from point $A(3,4,6)$ on line $L$ to plane $\Pi$.
This uses the same calculation as Example 3.
$d(L, \Pi) = d(A, P) = \mathbf{\sqrt{38}}$