3.19 Distances (HL)

1. Distance Between a Point and a Line

Calculating the absolute shortest distance tracking from a stationary coordinate point to a geometric line necessitates finding the precise perpendicular "foot" junction resting perfectly on the line.

Methodology:

  1. Define the unknown foot coordinate point $P$ operating mathematically upon the parametric form of line $L$.
  2. Construct the directional connection vector mapping $\overrightarrow{AP}$.
  3. Since the shortest path tracks exactly perpendicularly, set the spatial dot product between connecting vector $\overrightarrow{AP}$ and the line's inherent direction vector $\vec{b}$ decisively to zero ($\overrightarrow{AP} \cdot \vec{b} = 0$).
  4. Solve the formulated algebraic sequence identifying specific parameter $\lambda$, extract precise coordinates for foot $P$, and calculate the Pythagorean distance $d(A,P)$.

EXAMPLE 1 (Point to Line Calculation)

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

Foot $P$ maps parametrically to $(5+3\lambda, 7+2\lambda, 9+\lambda)$.
The required connecting trajectory $\overrightarrow{AP}$ evaluates strictly via point subtraction:
$\overrightarrow{AP} = \begin{pmatrix} 5+3\lambda-1 \\ 7+2\lambda-2 \\ 9+\lambda-3 \end{pmatrix} = \begin{pmatrix} 4+3\lambda \\ 5+2\lambda \\ 6+\lambda \end{pmatrix}$
Executing perpendicular constraint rules calculates $\overrightarrow{AP} \cdot \vec{b} = 0$:
$$\begin{aligned} 3(4+3\lambda) + 2(5+2\lambda) + 1(6+\lambda) &= 0 \\ 12 + 9\lambda + 10 + 4\lambda + 6 + \lambda &= 0 \\ 14\lambda + 28 &= 0 \\ 14\lambda &= -28 \\ \lambda &= -2 \end{aligned}$$
Substituting back resolves exact coordinates for foot point $P(-1, 3, 7)$.
Measuring distance via geometric boundaries yields:
$d(A,P) = \sqrt{(-1-1)^2 + (3-2)^2 + (7-3)^2} = \sqrt{4 + 1 + 16} = \mathbf{\sqrt{21}}$

2. Distance Between Two Lines

If Lines act Parallel: Select any arbitrary anchor point cleanly off Line 1 and calculate distance relative cleanly to Line 2 utilizing the precise point-to-line methodology outlined above.
If Lines act Skew: The shortest dimensional gap traces a continuous line acting simultaneously perfectly perpendicular to both structural geometries. This establishes two interacting foot nodes ($P$ and $Q$).

EXAMPLE 2 (Distance Between Skew Lines)

Analyze distances interacting spanning $L_1: \vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} 4 \\ 5 \\ 0 \end{pmatrix}$ against $L_2: \vec{r} = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$.

Designate respective intersection parameter nodes: $P(1+4\mu, 2+5\mu, 3)$ operating against $Q(5+3\lambda, 7+2\lambda, 9+\lambda)$.
Determine structural connecting linkage $\overrightarrow{PQ}$:
$\overrightarrow{PQ} = \begin{pmatrix} 4+3\lambda-4\mu \\ 5+2\lambda-5\mu \\ 6+\lambda \end{pmatrix}$
Establish dual perpendicular restrictions mapped securely against respective directional matrices:
$\overrightarrow{PQ} \cdot \vec{b}_1 = 0 \implies 4(4+3\lambda-4\mu) + 5(5+2\lambda-5\mu) + 0 = 0 \implies 22\lambda - 41\mu = -41$

$\overrightarrow{PQ} \cdot \vec{b}_2 = 0 \implies 3(4+3\lambda-4\mu) + 2(5+2\lambda-5\mu) + 1(6+\lambda) = 0 \implies 14\lambda - 22\mu = -28$
The resulting system solves evaluating $\lambda = -\dfrac{41}{15}$ and $\mu = -\dfrac{7}{15}$.
Substituting numerical coefficients extracts the specific location points, granting exact capability to compute target distance length $|\overrightarrow{PQ}|$.

3. Distance Between a Point and a Plane

Calculating the minimal interval space bounding a fixed spatial point against an extensive continuous plane mirrors prior logical mechanics perfectly. The perpendicular projection axis intersecting the designated target bounds automatically operates perfectly parallel extending adjacent to the plane's identified normal vector $\vec{n}$.

EXAMPLE 3 (Point to Plane Calculation)

Assess distance metrics tracing between coordinate $A(3,4,6)$ evaluating strictly against internal Plane $\Pi: 2x + 3y + 5z = 10$.

Key point: The connecting vector axis $AP$ adheres rigidly parallel tracing the normal factor $\vec{n} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$.
The projection equation builds as $\vec{r} = \begin{pmatrix} 3 \\ 4 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$.
Extracting coordinate variables produces node $P(3+2\lambda, 4+3\lambda, 6+5\lambda)$. Since $P$ resides mathematically inside plane bounds, integrate cleanly into Cartesian boundaries:
$$\begin{aligned} 2(3+2\lambda) + 3(4+3\lambda) + 5(6+5\lambda) &= 10 \\ 6 + 4\lambda + 12 + 9\lambda + 30 + 25\lambda &= 10 \\ 48 + 38\lambda &= 10 \\ 38\lambda &= -38 \iff \lambda = -1 \end{aligned}$$
Coordinate mapping yields footprint node $P(1,1,1)$.
Calculated resulting length establishes absolute 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 geometric case occurs only if the line is strictly parallel to the plane. We systematically find the distance of an arbitrary fixed point $A$ stationed on the line $L$ evaluating against the surface boundaries of 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 $\Pi: 2x + 3y + 5z = 10$.

It is established that the line acts entirely parallel traversing over the plane structure.
We just target and extract the strict geometric distance mapping point $A(3,4,6)$ resting upon line $L$ evaluating towards the plane boundary $\Pi$.
This mirrors precisely the calculations generated cleanly within Example 3.
$d(L, \Pi) = d(A, P) = \mathbf{\sqrt{38}}$