Distance from a point to a line, and between two lines – Points, Straight Lines and Shapes in 3D (Senior 5 Mathematics) | Ganzaa

Points, Straight Lines and Shapes in 3D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 3D

5 lessons

Lesson 3 of 5

Distance from a point to a line, and between two lines

40 min

The shortest cable

An electricity line runs along one side of a valley and a water pipe crosses the valley higher up, in a different direction. An engineer wants the shortest possible connecting bar between them, for a support or a sensor. The bar must meet both at right angles, and its length is the distance between two skew lines. This lesson finds that distance, and the simpler distance from a point to a line, with the vector and mixed products of unit 7.

Definition

The distance from a point $P$ to a line $L$ is the length $P H$ , where $H$ , the foot of the perpendicular from $P$ , is the point of $L$ with $\overrightarrow{PH} \perp L$ . It is the smallest distance from $P$ to any point of $L$ .

The distance between two lines is the smallest distance between a point of one and a point of the other. For skew lines it is the length of the common perpendicular $P Q$ , the segment perpendicular to both lines.

Important

Point to line. If $L$ passes through $A$ with direction $d$ , then

d(P, L) = \frac{\left|\overrightarrow{AP} \times d\right|}{|d|} .

Reason. The parallelogram built on $\overrightarrow{AP}$ and $d$ has area $|\overrightarrow{AP} \times d|$ . Its base is $∣ d ∣$ and its height is the distance from $P$ to $L$ . The foot $H = A + t\,d$ is found from $\overrightarrow{PH} \cdot d = 0$ , which gives $t = \frac{\overrightarrow{AP} \cdot d}{|d|^2}$ .

Between two lines. Let $L_{1}$ pass through $A_{1}$ with direction $d_{1}$ , and $L_{2}$ through $A_{2}$ with direction $d_{2}$ .

If the lines are parallel, their distance is the distance from $A_{2}$ to $L_{1}$ .
If not, $n = d_1 \times d_2$ is perpendicular to both, and

d(L_1, L_2) = \frac{\left|\overrightarrow{A_1A_2} \cdot (d_1 \times d_2)\right|}{|d_1 \times d_2|} = \frac{\left|[\overrightarrow{A_1A_2}, d_1, d_2]\right|}{|d_1 \times d_2|} .

Reason. The common perpendicular has direction $n$ . Projecting any segment $A_{1} A_{2}$ joining the lines onto $n$ gives exactly the length of the common perpendicular. A result of 0 means the lines intersect.

Worked example 1: distance and foot of the perpendicular

Find the distance from $P (1, 2, 3)$ to the line $L : r = (0, 1, - 1) + t (1, 2, 2)$ , and the foot of the perpendicular.

Solution. $A = (0, 1, - 1)$ , $\overrightarrow{AP} = (1, 1, 4)$ , $d = (1, 2, 2)$ , $∣ d ∣ = 3$ .

\overrightarrow{AP} \times d = (1 \cdot 2 - 4 \cdot 2,\ 4 \cdot 1 - 1 \cdot 2,\ 1 \cdot 2 - 1 \cdot 1) = (-6, 2, 1), \qquad d(P, L) = \frac{\sqrt{41}}{3} \approx 2.13 .

Foot: $t = \frac{\overrightarrow{AP} \cdot d}{|d|^2} = \frac{1 + 2 + 8}{9} = \frac{11}{9}$ , so $H = \left(\frac{11}{9}, \frac{31}{9}, \frac{13}{9}\right)$ . Check: $\overrightarrow{PH} = \frac{1}{9}(2, 13, -14)$ , $\overrightarrow{PH} \cdot d = \frac{2 + 26 - 28}{9} = 0$ ✓ and $PH = \frac{\sqrt{369}}{9} = \frac{\sqrt{41}}{3}$ . ✓

Worked example 2: distance between two skew lines

Find the distance between $L_{1} : r = (1, 2, 3) + t (1, - 1, 2)$ and $L_{3} : r = (0, 1, 1) + s (1, 1, 0)$ , shown to be skew in lesson 2.

Solution. $n = d_1 \times d_3 = (1, -1, 2) \times (1, 1, 0) = (-2, 2, 2)$ , $|n| = \sqrt{12} = 2\sqrt{3}$ . $\overrightarrow{A_1A_3} = (-1, -1, -2)$ , and $\overrightarrow{A_1A_3} \cdot n = 2 - 2 - 4 = -4$ . So

d(L_1, L_3) = \frac{|-4|}{2\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.15 .

Worked example 3: the ends of the common perpendicular

Find the points $P$ on $L_{1}$ and $Q$ on $L_{3}$ (worked example 2) such that $P Q$ is perpendicular to both lines.

Solution. $P = (1 + t,\ 2 - t,\ 3 + 2t)$ and $Q = (s,\ 1 + s,\ 1)$ , so $\overrightarrow{PQ} = (s - 1 - t,\ s - 1 + t,\ -2 - 2t)$ .

\overrightarrow{PQ} \cdot (1, -1, 2) = -6t - 4 = 0, \qquad \overrightarrow{PQ} \cdot (1, 1, 0) = 2s - 2 = 0 .

So $t = -\frac{2}{3}$ , $s = 1$ : $P\left(\frac{1}{3}, \frac{8}{3}, \frac{5}{3}\right)$ , $Q (1, 2, 1)$ , and $PQ = \left|\left(\frac{2}{3}, -\frac{2}{3}, -\frac{2}{3}\right)\right| = \frac{2\sqrt{3}}{3}$ , as in worked example 2. ✓

Worked example 4: two parallel lines

Find the distance between $L : r = (1, 0, 2) + t (2, 1, - 2)$ and $M : r = (4, 2, 0) + s (2, 1, - 2)$ .

Solution. The directions are equal, so the lines are parallel and the mixed-product formula cannot be used ( $d_1 \times d_2 = 0$ ). Use the point $B (4, 2, 0)$ of $M$ : $\overrightarrow{AB} = (3, 2, -2)$ , $\overrightarrow{AB} \times d = (-2, 2, -1)$ , so $d(L, M) = \frac{3}{3} = 1$ .

Warning

The skew-lines formula divides by $|d_1 \times d_2|$ , which is 0 for parallel lines. Check the directions first.
Divide by $∣ d ∣$ in the point-to-line formula; $|\overrightarrow{AP} \times d|$ alone is an area, not a distance.
A distance is never negative: take the absolute value of the mixed product.

Tip

The skew-line formula is a "shadow": the length of $\overrightarrow{A_1A_2}$ projected onto the common normal $n$ , just like the scalar projection of unit 7.

Khan Academy India: Distance between skew lines (intuition), 3D Geometry (about 8 min). Watch why the shortest segment between skew lines is perpendicular to both, and how projecting onto d₁ × d₂ gives its length.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Find the distance from $(2, - 1, 3)$ to the line $r = (1, 0, 1) + t (2, 2, 1)$ .
Find the foot of the perpendicular from the origin to $r = (3, 1, 2) + t (1, - 1, 0)$ , and the distance from $O$ to the line.
Find the distance from $(3, 5, 12)$ to the $x$ -axis.
Find the distance between the parallel lines $r = (2, 1, 0) + t (1, 2, 2)$ and $r = (0, 3, 1) + s (1, 2, 2)$ .
Find the distance between the skew lines $r = t (1, 1, 0)$ and $r = (1, 0, 1) + s (0, 1, 1)$ .
Use the mixed product to show that $r = (1, 2, 3) + t (1, - 1, 2)$ and $r = (5, 1, 6) + s (2, 1, - 1)$ intersect.
Explain why the shortest segment joining two skew lines must be perpendicular to both of them.

Show answersHide answers

$\overrightarrow{AP} = (1, -1, 2)$ , $\overrightarrow{AP} \times d = (-5, 3, 4)$ , distance $\frac{\sqrt{50}}{3} = \frac{5\sqrt{2}}{3} \approx 2.36$ .
$t = -\frac{(3, 1, 2) \cdot (1, -1, 0)}{2} = -1$ : foot $(2, 2, 2)$ , distance $2\sqrt{3} \approx 3.46$ .
The foot is $(3, 0, 0)$ , so the distance is $\sqrt{5^2 + 12^2} = 13$ .
$\overrightarrow{AB} = (-2, 2, 1)$ , $\overrightarrow{AB} \times d = (2, 5, -6)$ , distance $\frac{\sqrt{65}}{3} \approx 2.69$ .
$n = (1, - 1, 1)$ , $\overrightarrow{A_1A_2} \cdot n = (1, 0, 1) \cdot (1, -1, 1) = 2$ , distance $\frac{2}{\sqrt{3}} \approx 1.15$ .
$\overrightarrow{A_1A_2} = (4, -1, 3)$ , $d_1 \times d_2 = (-1, 5, 3)$ , mixed product $- 4 - 5 + 9 = 0$ : the distance is 0 and the directions are not parallel, so the lines meet (at $(3, 0, 7)$ ).
If the segment $P Q$ were not perpendicular to $L_{1}$ , moving $P$ along $L_{1}$ towards the foot of the perpendicular from $Q$ would shorten it, just as the foot gives the shortest distance from a point to a line; the same holds for $L_{2}$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram of skew lines (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every distance and foot was checked by computer.
Video: Distance between skew lines (intuition) | 3D Geometry | Grade 12 | Math, Khan Academy India, embedded from YouTube.

1The shortest cable

The shortest cable