Distances to Lines and Planes

How far is a point from a plane? How close do two skew lines come? “Distance” here always means the shortest distance, which is measured along a perpendicular. Every formula on this page comes from one of two ideas you already know: vector projections and the area of a parallelogram from the cross product.

Key ideas

The big idea: project onto the normal

Suppose $Q$ is any point on a plane and $P$ is a point off it. The vector $\overrightarrow{QP}$ usually points off at an angle. The part of it that points straight away from the plane, along the normal $\vec{n}$, is exactly the distance you want. That’s the length of the scalar projection of $\overrightarrow{QP}$ onto $\vec{n}$:

$$d = \frac{\left|\overrightarrow{QP} \cdot \vec{n}\right|}{|\vec{n}|}$$

The absolute value is there because a distance can’t be negative.

The distance $d$ is the length of the projection of $\overrightarrow{QP}$ onto the normal $\vec{n}$.

Point to a line in 2-space

For the line $Ax + By + C = 0$, the normal is $\vec{n} = [A, B]$. Take $P(x_1, y_1)$ and any point $Q(x_0, y_0)$ on the line. Then

$$\overrightarrow{QP} \cdot \vec{n} = A(x_1 - x_0) + B(y_1 - y_0) = Ax_1 + By_1 - (Ax_0 + By_0) = Ax_1 + By_1 + C$$

because $Ax_0 + By_0 = -C$ ($Q$ is on the line). Dividing by $|\vec{n}|$ gives the standard formula:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

The nice thing: you never actually need to find $Q$. Substitute the point into the left side of the equation, take the absolute value, and divide by the length of the normal.

Point to a plane

Exactly the same argument in 3-space: for the plane $Ax + By + Cz + D = 0$ and the point $P(x_1, y_1, z_1)$,

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Between parallel planes (or parallel lines in 2-space)

Parallel planes are the same distance apart everywhere. Pick any point on one plane and use the point-to-plane formula with the other. If the two equations have the same normal, $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$, this simplifies to

$$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$$

The same idea works for two parallel lines in 2-space.

Point to a line in 3-space

In 3-space a line has no single normal, so use the cross product instead. Let $Q$ be a point on the line with direction $\vec{m}$, and let $\theta$ be the angle between $\overrightarrow{QP}$ and $\vec{m}$. From the right triangle in the figure, $d = |\overrightarrow{QP}| \sin\theta$. Since $|\overrightarrow{QP} \times \vec{m}| = |\overrightarrow{QP}|\,|\vec{m}| \sin\theta$,

$$d = \frac{\left|\overrightarrow{QP} \times \vec{m}\right|}{|\vec{m}|}$$

Another way to see it: $\left|\overrightarrow{QP} \times \vec{m}\right|$ is the area of the parallelogram formed by the two vectors, and area $\div$ base $=$ height.

The distance $d$ is the height of the parallelogram formed by $\overrightarrow{QP}$ and $\vec{m}$.

Between skew lines

Let the lines be $\vec{r} = \vec{r}_1 + t\vec{m}_1$ and $\vec{r} = \vec{r}_2 + s\vec{m}_2$, through points $P_1$ and $P_2$. The shortest segment between skew lines is perpendicular to both lines, so it runs in the direction

$$\vec{n} = \vec{m}_1 \times \vec{m}_2$$

Project the connecting vector $\overrightarrow{P_1P_2}$ onto $\vec{n}$:

$$d = \frac{\left|\overrightarrow{P_1P_2} \cdot (\vec{m}_1 \times \vec{m}_2)\right|}{|\vec{m}_1 \times \vec{m}_2|}$$

This is really the point-to-plane formula again: the two skew lines lie in parallel planes with normal $\vec{n}$, and $d$ is the gap between those planes. If the answer is $0$, the lines actually intersect. (For parallel lines, $\vec{m}_1 \times \vec{m}_2 = \vec{0}$, so use the point-to-line formula instead.)

Summary

Distance	Formula
point to line in 2-space	$\dfrac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$
point to plane	$\dfrac{\lvert Ax_1 + By_1 + Cz_1 + D \rvert}{\sqrt{A^2 + B^2 + C^2}}$
point to line in 3-space	$\dfrac{\lvert \overrightarrow{QP} \times \vec{m} \rvert}{\lvert \vec{m} \rvert}$
skew lines	$\dfrac{\lvert \overrightarrow{P_1P_2} \cdot (\vec{m}_1 \times \vec{m}_2) \rvert}{\lvert \vec{m}_1 \times \vec{m}_2 \rvert}$

Worked examples

Example 1: Point to a line in 2-space

Find the distance from $P(5, 3)$ to the line $3x - 4y + 8 = 0$.

Solution.

$$d = \frac{|3(5) - 4(3) + 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|11|}{5} = \frac{11}{5} = 2.2$$

Check with a projection. The point $Q(0, 2)$ is on the line ($0 - 8 + 8 = 0$). Then $\overrightarrow{QP} = [5, 1]$ and $\vec{n} = [3, -4]$:

$$\frac{|[5, 1] \cdot [3, -4]|}{|[3, -4]|} = \frac{|15 - 4|}{5} = \frac{11}{5}$$

Same answer ✓.

Example 2: Point to a plane, and parallel planes

• (a) Find the distance from $P(2, -1, 4)$ to the plane $2x - y + 2z - 3 = 0$.
• (b) Find the distance between the planes $2x - y + 2z - 3 = 0$ and $4x - 2y + 4z + 9 = 0$.

Solution.

(a)

$$d = \frac{|2(2) - (-1) + 2(4) - 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|10|}{\sqrt{9}} = \frac{10}{3} \approx 3.33$$

(rounded to two decimal places).

(b) The normals $[2, -1, 2]$ and $[4, -2, 4]$ are parallel, so the planes are parallel. Pick a point on the first plane: with $x = z = 0$, $-y - 3 = 0$, so $Q(0, -3, 0)$. Its distance to the second plane:

$$d = \frac{|4(0) - 2(-3) + 4(0) + 9|}{\sqrt{4^2 + (-2)^2 + 4^2}} = \frac{15}{\sqrt{36}} = \frac{15}{6} = 2.5$$

Check with the shortcut: divide the second equation by $2$ to match normals: $2x - y + 2z + 4.5 = 0$. Then $d = \dfrac{|-3 - 4.5|}{3} = \dfrac{7.5}{3} = 2.5$ ✓.

Example 3: Point to a line in 3-space

Find the distance from $P(3, 1, -2)$ to the line $\vec{r} = [1, 0, 1] + t[2, -1, 2]$.

Solution. Take $Q(1, 0, 1)$ on the line and $\vec{m} = [2, -1, 2]$. Then $\overrightarrow{QP} = [2, 1, -3]$, and

$$\begin{aligned} \overrightarrow{QP} \times \vec{m} &= [2, 1, -3] \times [2, -1, 2] \\ &= [1(2) - (-3)(-1),\ (-3)(2) - 2(2),\ 2(-1) - 1(2)] \\ &= [-1, -10, -4] \end{aligned}$$ $$d = \frac{\sqrt{(-1)^2 + (-10)^2 + (-4)^2}}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{\sqrt{117}}{3} = \frac{3\sqrt{13}}{3} = \sqrt{13} \approx 3.61$$

(rounded to two decimal places).

Example 4: Between skew lines

In intersections of lines and planes, Example 2 showed that these lines are skew. Find the distance between them.

$$L_1: \vec{r} = [1, 0, 2] + t[2, 1, -1] \qquad L_2: \vec{r} = [0, 4, 1] + s[1, -1, 2]$$

Solution. A vector perpendicular to both lines:

$$\vec{n} = [2, 1, -1] \times [1, -1, 2] = [1(2) - (-1)(-1),\ (-1)(1) - 2(2),\ 2(-1) - 1(1)] = [1, -5, -3]$$

Connect a point on each line: $P_1(1, 0, 2)$ and $P_2(0, 4, 1)$ give $\overrightarrow{P_1P_2} = [-1, 4, -1]$.

$$d = \frac{|[-1, 4, -1] \cdot [1, -5, -3]|}{\sqrt{1^2 + (-5)^2 + (-3)^2}} = \frac{|-1 - 20 + 3|}{\sqrt{35}} = \frac{18}{\sqrt{35}} \approx 3.04$$

(rounded to two decimal places). The answer isn’t $0$, which confirms the lines don’t meet.

Common mistakes

Dropping the absolute value. The dot product can be negative, depending on which side of the plane the point is on. Distance is always positive (or zero), so take the absolute value.

Forgetting to divide by the length of the normal. $|Ax_1 + By_1 + Cz_1 + D|$ on its own is not the distance unless $|\vec{n}| = 1$. Divide by $\sqrt{A^2 + B^2 + C^2}$.

Leaving the equation in the wrong form. The formula needs $Ax + By + Cz + D = 0$. For a plane written as $2x + 3y + 6z = 14$, the constant is $D = -14$, not $14$.

Using the shortcut for planes whose normals don’t match. $\dfrac{|D_1 - D_2|}{|\vec{n}|}$ only works when both equations use the same normal. Scale one equation first, or just pick a point on one plane.

Using a dot product for point-to-line distance in 3-space. A line in 3-space has no single normal, so the point-to-plane formula doesn’t apply. Use $\dfrac{|\overrightarrow{QP} \times \vec{m}|}{|\vec{m}|}$.

Using the skew-line formula on parallel lines. If the lines are parallel, $\vec{m}_1 \times \vec{m}_2 = \vec{0}$ and you’d divide by zero. Find the distance from a point on one line to the other line instead.

Practice

1. (Warm-up) Find the distance from $(1, 2)$ to the line $6x + 8y - 2 = 0$.

Solution

$$d = \frac{|6(1) + 8(2) - 2|}{\sqrt{36 + 64}} = \frac{20}{10} = 2$$

2. (Warm-up) Find the distance from the origin to the plane $2x + 3y + 6z = 14$.

Solution

Rewrite as $2x + 3y + 6z - 14 = 0$:

$$d = \frac{|0 + 0 + 0 - 14|}{\sqrt{4 + 9 + 36}} = \frac{14}{7} = 2$$

3. (Core) Find the distance from $(4, -2, 1)$ to the plane $x - 2y + 2z + 5 = 0$.

Solution

$$d = \frac{|4 - 2(-2) + 2(1) + 5|}{\sqrt{1 + 4 + 4}} = \frac{15}{3} = 5$$

4. (Core) Find the distance between the parallel lines $3x + 4y - 10 = 0$ and $3x + 4y + 15 = 0$.

Solution

The point $(2, 1)$ is on the first line: $6 + 4 - 10 = 0$. Its distance to the second line:

$$d = \frac{|3(2) + 4(1) + 15|}{\sqrt{9 + 16}} = \frac{25}{5} = 5$$

(Shortcut: same normal, so $d = \dfrac{|-10 - 15|}{5} = 5$.)

5. (Core) Find the distance between the planes $x + 2y - 2z = 4$ and $-2x - 4y + 4z = 10$.

Solution

Divide the second equation by $-2$: $x + 2y - 2z = -5$. Now both planes have normal $[1, 2, -2]$, and in the form $Ax + By + Cz + D = 0$ the constants are $D_1 = -4$ and $D_2 = 5$:

$$d = \frac{|-4 - 5|}{\sqrt{1 + 4 + 4}} = \frac{9}{3} = 3$$

6. (Core) Find the distance from $P(2, 3, -1)$ to the line $\vec{r} = [0, 1, 1] + t[1, 2, 2]$.

Solution

$Q(0, 1, 1)$ is on the line, so $\overrightarrow{QP} = [2, 2, -2]$, and $\vec{m} = [1, 2, 2]$.

$$[2, 2, -2] \times [1, 2, 2] = [2(2) - (-2)(2),\ (-2)(1) - 2(2),\ 2(2) - 2(1)] = [8, -6, 2]$$$$d = \frac{\sqrt{64 + 36 + 4}}{\sqrt{1 + 4 + 4}} = \frac{\sqrt{104}}{3} = \frac{2\sqrt{26}}{3} \approx 3.40$$

(rounded to two decimal places).

7. (Core) Find the distance from $P(6, -1)$ to the line $\vec{r} = [1, 2] + t[3, 4]$.

Solution

Convert to scalar form. The direction $[3, 4]$ gives normal $[4, -3]$, so $4x - 3y + C = 0$. With $(1, 2)$: $4 - 6 + C = 0$, so $C = 2$.

$$d = \frac{|4(6) - 3(-1) + 2|}{\sqrt{16 + 9}} = \frac{29}{5} = 5.8$$

8. (Challenge) Show that the lines $L_1: \vec{r} = [0, 0, 1] + t[1, 1, 0]$ and $L_2: \vec{r} = [2, 0, 3] + s[0, 1, 1]$ are skew by finding the distance between them.

Solution

The directions aren’t parallel. A common perpendicular:

$$\vec{n} = [1, 1, 0] \times [0, 1, 1] = [1(1) - 0(1),\ 0(0) - 1(1),\ 1(1) - 1(0)] = [1, -1, 1]$$

$\overrightarrow{P_1P_2} = [2, 0, 2]$, so

$$d = \frac{|[2, 0, 2] \cdot [1, -1, 1]|}{\sqrt{3}} = \frac{4}{\sqrt{3}} \approx 2.31$$

(rounded to two decimal places). The lines aren’t parallel and the distance isn’t $0$, so they never meet: they’re skew.

9. (Challenge) An engineer models a sloped solar panel as part of the plane $2x + 2y + z = 12$, with distances in metres. A sensor is mounted at $S(5, 4, 9)$.

• (a) How far is the sensor from the plane of the panel?
• (b) Find the point on the plane closest to the sensor.

Solution

(a)

$$d = \frac{|2(5) + 2(4) + 9 - 12|}{\sqrt{4 + 4 + 1}} = \frac{15}{3} = 5 \text{ m}$$

(b) Follow the normal from $S$: $\vec{r} = [5, 4, 9] + t[2, 2, 1]$. Substitute into the plane:

$$2(5 + 2t) + 2(4 + 2t) + (9 + t) = 12 \quad\Rightarrow\quad 27 + 9t = 12 \quad\Rightarrow\quad t = -\frac{5}{3}$$

The closest point is

$$\left(5 - \frac{10}{3},\ 4 - \frac{10}{3},\ 9 - \frac{5}{3}\right) = \left(\frac{5}{3},\ \frac{2}{3},\ \frac{22}{3}\right)$$

Check: $\dfrac{10}{3} + \dfrac{4}{3} + \dfrac{22}{3} = \dfrac{36}{3} = 12$ ✓. Also, the distance from $S$ is $|t|\,|\vec{n}| = \dfrac{5}{3}(3) = 5$ m ✓, matching part (a).