Closest points and distances including skew lines | Year 12

01 / LEARN

Understand the idea.

Find the foot, then the length

The shortest segment from a point to an infinite line is perpendicular to the line. Write a variable point $Q = a + t d$ , require $(q - p) \cdot d = 0$ , and solve for the perpendicular foot before taking its distance.

For the plane $n \cdot r = c$ , perpendicular motion follows its normal. The distance from $P$ is $∣ n \cdot p - c ∣ / ∣ n ∣$ . The absolute value makes the length non-negative regardless of which side contains $P$ .

Skew is different from parallel

Nonparallel lines in three dimensions may be skew, with no common point. Their shortest joining segment is perpendicular to both direction vectors. Use separate parameters for the two lines and impose both zero dot-product conditions.

A common perpendicular direction is $d_{1} \times d_{2}$ . Projecting the anchor displacement onto it gives the shortest distance, but not the closest points. If the directions are parallel, this cross product vanishes and a point-to-line calculation is needed instead.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find the distance of (1,2,3) from the x axis.

Worked solution

1
Choose a variable point on the x axis.
$Q = (t, 0, 0), \vec{Q P} = (1 - t, 2, 3)$
Every point on the axis has zero $y$ and $z$ coordinates.
2
Require a perpendicular joining vector.
$(1 - t, 2, 3) \cdot (1, 0, 0) = 1 - t = 0$
The shortest displacement to an infinite line is perpendicular to its direction.
3
Find the foot and its distance.
$Q = (1, 0, 0), d = \sqrt{0^{2} + 2^{2} + 3^{2}} = \sqrt{13}$
The perpendicular segment varies only in the $y$ and $z$ directions.

Answer

$\sqrt{13}$

EXAMPLE 02

Worked example 2 — oblique skew lines

Find the closest points and shortest distance for $ℓ_{1} = s (1, 1, 0)$ and $ℓ_{2} = (0, 1, 1) + t (1, 0, 1)$ .

Worked solution

1
Write separate points on the two lines.
$\begin{aligned} P & = (s, s, 0) \\ Q & = (t, 1, 1 + t) \\ \vec{P Q} & = (t - s, 1 - s, 1 + t) \end{aligned}$
Independent line parameters allow the closest point on each line to be determined.
2
Impose perpendicularity to both directions.
$\begin{aligned} \vec{P Q} \cdot (1, 1, 0) & = t + 1 - 2 s = 0 \\ \vec{P Q} \cdot (1, 0, 1) & = 2 t - s + 1 = 0 \end{aligned}$
A shortest connecting segment has no component along either line.
3
Solve the simultaneous parameter equations.
$\begin{aligned} t & = 2 s - 1 \\ 2 (2 s - 1) - s + 1 & = 0 \\ 3 s & = 1 \end{aligned} s = \frac{1}{3}, t = - \frac{1}{3}$
Substitution eliminates $t$ and determines both perpendicular feet.
4
Calculate the closest points and displacement.
$\begin{aligned} P & = (\frac{1}{3}, \frac{1}{3}, 0) \\ Q & = (- \frac{1}{3}, 1, \frac{2}{3}) \\ \vec{P Q} & = (- \frac{2}{3}, \frac{2}{3}, \frac{2}{3}) \end{aligned}$
Substitute each solved parameter into its own line equation.
5
Take the perpendicular length.
$d = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} = \frac{2}{\sqrt{3}}$
The joining vector is perpendicular to both nonparallel lines, so its magnitude is their shortest separation.

Answer

$P = (1 / 3, 1 / 3, 0)$ , $Q = (- 1 / 3, 1, 2 / 3)$ ; shortest distance $2 / \sqrt{3}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the distance from (0,0,5) to z=2.

Show answer & reasoning

$3$

The plane $z = 2$ has vertical normal. The perpendicular foot of $(0, 0, 5)$ is $(0, 0, 2)$ , so the distance is $∣ 5 - 2 ∣ = 3$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the closest point on $(t, 0, 0)$ to $(3, 4, 5)$ and its distance.
2 marks
Find the distance from $(1, 2, 3)$ to $x + y + z = 0$ .
2 marks
Find the foot of the perpendicular from $(1, 2, 3)$ to $x + y + z = 0$ .
2 marks
Find the closest points and distance between $ℓ_{1} = s (1, 1, 0)$ and $ℓ_{2} = (0, 2, 1) + t (1, 0, 1)$ .
4 marks
Find the distance between $(s, 0, 0)$ and $(t, 3, 4)$ .
2 marks
For line $(1, 0, 0) + t (1, 1, 0)$ and $P (0, 1, 2)$ , find the closest $t$ and distance.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the closest point on $(t, 0, 0)$ to $(3, 4, 5)$ and its distance.
2 marks
Perpendicularity requires $t = 3$ .
The remaining displacement is $(0, 4, 5)$ .
$Q = (3, 0, 0)$ , distance $\sqrt{41}$ .
Find the distance from $(1, 2, 3)$ to $x + y + z = 0$ .
2 marks
Use $\frac{∣ 6 ∣}{\sqrt{3}}$ .
Simplify to $2 \sqrt{3}$ .
$2 \sqrt{3}$
Find the foot of the perpendicular from $(1, 2, 3)$ to $x + y + z = 0$ .
2 marks
Subtract $[\frac{6}{3}] (1, 1, 1)$ from the point.
The result lies in the plane.
$(- 1, 0, 1)$
Find the closest points and distance between $ℓ_{1} = s (1, 1, 0)$ and $ℓ_{2} = (0, 2, 1) + t (1, 0, 1)$ .
4 marks
The joining vector is $(t - s, 2 - s, 1 + t)$ . Perpendicularity gives $t + 2 - 2 s = 0$ and $2 t - s + 1 = 0$ .
Solve to obtain $s = 1, t = 0$ . The joining vector is $(- 1, 1, 1)$ , which has length $\sqrt{3}$ .
$P = (1, 1, 0)$ and $Q = (0, 2, 1)$ ; distance $\sqrt{3}$ .
Find the distance between $(s, 0, 0)$ and $(t, 3, 4)$ .
2 marks
The lines are parallel to the $x$ axis.
The perpendicular displacement is $(0, 3, 4)$ .
$5$
For line $(1, 0, 0) + t (1, 1, 0)$ and $P (0, 1, 2)$ , find the closest $t$ and distance.
2 marks
The dot product $(1 + t, t - 1, - 2) \cdot (1, 1, 0) = 2 t$ vanishes at 0.
Then the squared distance is $1 + 1 + 4 = 6$ .
$t = 0$ , distance $\sqrt{6}$ .

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

In this lesson

Understand the idea.

Find the foot, then the length

Skew is different from parallel

See the method in action.

Worked example 1

Choose a variable point on the x axis.

Require a perpendicular joining vector.

Find the foot and its distance.

Worked example 2 — oblique skew lines

Write separate points on the two lines.

Impose perpendicularity to both directions.

Solve the simultaneous parameter equations.

Calculate the closest points and displacement.

Take the perpendicular length.

Try it yourself.

Check your understanding

Keep practising