Lesson 5 of 5Lines & Planes

Unit 6 · Lesson 5 of 5

Distances in 3-Space

Shortest distances are measured along a perpendicular. Use the scalar projection onto a normal for a point and a plane (and parallel planes), the cross product for a point and a line, and both together for two skew lines.

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Key things to remember

1Point to plane: $d = \frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$ (in 2-space, $\frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ ).
2Parallel planes with the same $A$ , $B$ , $C$ : $d = \frac{∣ D _{1} - D _{2} ∣}{A ^{2} + B ^{2} + C ^{2}}$ .
3Point to line in 3-space: $d = \frac{∣ P _{0} Q \times m ∣}{∣ m ∣}$ .
4Skew lines: $d = \frac{∣ P _{1} P _{2} \cdot ( m _{1} \times m _{2} ) ∣}{∣ m _{1} \times m _{2} ∣};$ $d = 0$ means the lines intersect.
5The foot of the perpendicular from $Q$ to a plane is where the line $r = OQ + k n$ meets the plane.

Engineers routing pipes, ducts and cables through a building need to know the clearance between two of them: how close they come where one passes over another. Two straight pipes that are not parallel and do not touch are skew lines, and their clearance is the shortest distance between them. In every case in this lesson the shortest distance is measured along a perpendicular: drop straight down from a point to a plane, or find the one segment that is perpendicular to both skew lines. Vectors find that perpendicular for you: the normal of a plane, or a cross product.

From a point to a plane

Fig 1A point Q off a plane, a point P₀ in the plane and the vector P₀Q. The projection of P₀Q onto the normal n is the shortest distance d = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²).

Let $Q (x_{1}, y_{1}, z_{1})$ be the point and $P_{0} (x_{0}, y_{0}, z_{0})$ any point of the plane $A x + B y + C z + D = 0$ . The distance from $Q$ to the plane is the length of the part of $P_{0} Q$ that points along the normal, which is the absolute value of the scalar projection: $d = \frac{∣ n \cdot P _{0} Q ∣}{∣ n ∣}$ . Expanding the numerator gives $A x_{1} + B y_{1} + C z_{1} - (A x_{0} + B y_{0} + C z_{0}),$ and since $P_{0}$ is in the plane, $- (A x_{0} + B y_{0} + C z_{0}) = D$ . The choice of $P_{0}$ drops out completely.

d = \frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}} d_{2-space} = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

Point to plane, and the same idea for a point and a line Ax + By + C = 0 in 2-space.

Two parallel planes are a constant distance apart, so pick any point on one and use the formula with the other. If the equations are written with the SAME coefficients $A$ , $B$ , $C$ , this simplifies to $d = \frac{∣ D _{1} - D _{2} ∣}{A ^{2} + B ^{2} + C ^{2}}$ . A line parallel to a plane works the same way: use any point of the line.

VideoWorked examples finding the shortest distance from a point to a plane.Watch on YouTube

From a point to a line in 3-space

A line in 3-space has no single normal vector (every direction perpendicular to $m$ is normal to it), so the formula above does not apply. Use area instead. Let $P_{0}$ be a point on the line, $m$ its direction and $Q$ the point. The parallelogram with sides $m$ and $P_{0} Q$ has area $∣ P_{0} Q \times m ∣$ . Its base is $∣ m ∣$ , and its height is exactly the perpendicular distance from $Q$ to the line, so divide.

d = \frac{∣ P _{0} Q \times m ∣}{∣ m ∣}

Distance from a point Q to the line through P₀ with direction m.

Between two skew lines

Let the lines pass through $P_{1}$ and $P_{2}$ with directions $m_{1}$ and $m_{2}$ . The vector $n = m_{1} \times m_{2}$ is perpendicular to both lines, so each line lies in a plane with normal $n$ , and those two planes are parallel. The distance between the lines is the distance between those planes: project $P_{1} P_{2}$ onto $n$ . A result of $0$ means the lines actually intersect. If $m_{1} \times m_{2} = 0$ , the lines are parallel, and the distance is the distance from any point of one line to the other line.

d = \frac{∣ P _{1} P _{2} \cdot ( m _{1} \times m _{2} ) ∣}{∣ m _{1} \times m _{2} ∣}

Distance between skew lines through P₁ and P₂ with directions m₁ and m₂.

Distance between	Formula	Key idea
Point and plane	$\frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$	Project onto the normal
Two parallel planes	$\frac{∣ D _{1} - D _{2} ∣}{A ^{2} + B ^{2} + C ^{2}}$ (same $A$ , $B$ , $C$ )	Point on one, plane formula for the other
Point and line (2-space)	$\frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$	Project onto the normal
Point and line (3-space)	$\frac{∣ P _{0} Q \times m ∣}{∣ m ∣}$	Parallelogram height = area ÷ base
Two skew lines	$\frac{∣ P _{1} P _{2} \cdot ( m _{1} \times m _{2} ) ∣}{∣ m _{1} \times m _{2} ∣}$	Project onto the common perpendicular

Summary of the distance formulas. Every one measures along a perpendicular.

Finding the foot of the perpendicular

To find WHERE the shortest segment from

Q

meets a plane, build the line through

Q

in the direction of the normal,

r = OQ + k n

, and intersect it with the plane as in the last lesson. The point you get is the foot of the perpendicular, and its distance from

Q

is a built-in check on the formula.

VideoFinds the shortest distance between two skew lines in 3-space.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Find the distance from

Q (2, - 1, 3)

to the plane

2 x - y + 2 z + 4 = 0

Example 2

Find the point where the perpendicular from

A (- 5, 3, 7)

meets the plane

v = (0, 0, 2) + t (- 1, 1, 3) + s (2, 0, - 3),

and the distance from

A

to the plane.

Example 3

Find the distance from

Q (3, 1, - 2)

to the line

r = (1, 0, 2) + t (1, - 1, 1)

Example 4

Find the distance between the skew lines

L_{1} : r = (1, 0, 2) + t (1, - 1, 1)

and

L_{2} : r = (0, 1, - 1) + s (2, 1, 0)

Where marks get lost

Leaving out the absolute value and reporting a negative distance. The sign only tells you which side of the plane the point is on.
Substituting into a plane written as $3 x + 5 y - 2 z = 6$ without first moving the $6$ across. In the formula $D = - 6$ , not $6$ .
Using $\frac{∣ D _{1} - D _{2} ∣}{∣ n ∣}$ for $x + y + z = 1$ and $2 x + 2 y + 2 z = 5$ without first scaling one equation so the coefficients match.
Using a dot product instead of a cross product for a point and a line in 3-space. $∣ P_{0} Q \times m ∣$ measures the parallelogram's area; the dot product measures the part of $P_{0} Q$ that runs ALONG the line, not across it.
Applying the skew-line formula to parallel lines, where $m_{1} \times m_{2} = 0$ and the formula divides by zero. Use the point-to-line distance instead.

Flashcards 7

Distance from a point to a plane

Distance from a point to a line in 2-space

Distance between parallel planes

Distance from a point to a line in 3-space

+ 3 more

Practice

Find the distance from

(1, 2, 3)

to the plane

2 x + y - 2 z + 9 = 0

, to 2 decimal places.

Find the distance from the point

(4, - 2)

to the line

3 x - 4 y + 12 = 0

in 2-space.

Find the distance between the parallel planes

2 x - y + 2 z + 4 = 0

and

2 x - y + 2 z - 5 = 0

Find the distance from the origin to the line

r = (1, 2, 2) + t (0, 0, 1)

, to 2 decimal places.

Find the distance between the skew lines

r = t (1, 0, 0)

(the

x

-axis) and

r = (0, 0, 5) + s (0, 1, 0)

Find the positive value of

D

for which the plane

2 x - y + 2 z + D = 0

3

units from the origin.

Along which direction is the shortest distance from a point to a plane measured?

For two lines,

m_{1} \times m_{2} = 0

. What should you do to find the distance between them?

What is the name of the point where the perpendicular from a point meets a plane? (the ___ of the perpendicular)

Match each distance to the formula used to find it.

A. $\frac{∣ P _{0} Q \times m ∣}{∣ m ∣}$
B. $\frac{∣ P _{1} P _{2} \cdot ( m _{1} \times m _{2} ) ∣}{∣ m _{1} \times m _{2} ∣}$
C. $\frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
D. $\frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$

Point to plane

Point to line in 2-space

Point to line in 3-space

Between skew lines

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