Closest points and distances including skew lines | Year 12

01 / LEARN

The idea, step by step

Find the foot, then the length

For P and line $q = a + t d$ , the closest point satisfies $(q - p) \cdot d = 0$ . Solve for t, find Q, and calculate PQ. This checks the geometry rather than guessing a parameter.

The distance to plane $n \cdot r = c$ is $∣ n \cdot p - c ∣ / ∣ n ∣$ . To locate the foot, move from P along the normal until the plane equation holds.

Skew is different from parallel

For nonparallel line directions, $d_{1} \times d_{2}$ gives their common perpendicular. Their shortest distance is $∣ (b - a) \cdot (d_{1} \times d_{2}) ∣ / ∣ d_{1} \times d_{2} ∣$ .

For actual closest points, use separate parameters and require their joining vector to be perpendicular to both directions. Parallel lines need a point-to-line method because the cross-product denominator is zero.

A common trap

Every pair of non-intersecting lines is parallel.

Instead: In three dimensions they can be skew: nonparallel and non-intersecting.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

The closest point is (1,0,0).
The perpendicular displacement is (0,2,3).

\sqrt{13}

Example 2

Worked example 2 — oblique skew lines

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

Let P=(s,s,0) and Q=(t,1,1+t). Then $\vec{P Q} = (t - s, 1 - s, 1 + t)$ .
Require the joining vector to be perpendicular to both line directions: $t + 1 - 2 s = 0$ and $2 t - s + 1 = 0$ .
Solving gives $s = 1 / 3$ and $t = - 1 / 3$ . Hence $P = (1 / 3, 1 / 3, 0)$ and $Q = (- 1 / 3, 1, 2 / 3)$ .
The joining vector is $(- 2 / 3, 2 / 3, 2 / 3)$ , whose magnitude is $2 / \sqrt{3}$ . Its dot product with each direction is zero.

P = (1 / 3, 1 / 3, 0)

Q = (- 1 / 3, 1, 2 / 3)

; shortest distance

2 / \sqrt{3}

Check your understanding

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

Show answer

$3$

Move along the z normal from height 5 to height 2.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 $∣ 6 ∣ / \sqrt{3}$ .
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 $[6 / 3] (1, 1, 1)$ 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 $\sqrt{6}$ .

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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.

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