Distance from point to plane (HL only)

IB Mathematics: Analysis and Approaches HL· Topic 3: Geometry & Trigonometry, 3.12· 20 min read

1. Derivation of the Distance Formula★★☆☆☆HL only⏱ 8 min

The shortest distance from any point to a plane is always the perpendicular distance, measured along the line parallel to the plane's normal vector. We derive the formula starting from general forms of a plane and point.

📘 Definition

Perpendicular Distance

$D$

The shortest non-negative distance between a point and a plane, measured along the line perpendicular to the plane through the point.

🔬 Derivation

Goal:

Derive the distance formula from a point $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz + d = 0$

Starting from:

Plane $Π : a x + b y + cz + d = 0$ , point $P (x_{0}, y_{0}, z_{0})$ , $Q (x_{1}, y_{1}, z_{1})$ is any point on $Π$ , normal vector $n = a b c$

1
Distance $D$ equals the absolute scalar projection of $P Q$ onto $n$ :
2
$D = \left| \overrightarrow{PQ} \cdot \frac{\mathbf{n}}{|\mathbf{n}|} \right|$
3
Since $Q$ lies on $Π$ , $a x_{1} + b y_{1} + c z_{1} + d = 0 ⟹ d = - a x_{1} - b y_{1} - c z_{1}$ . Expand the dot product:
4
$\overrightarrow{PQ} \cdot \mathbf{n} = a(x_1 - x_0) + b(y_1 - y_0) + c(z_1 - z_0) = -(ax_0 + by_0 + cz_0 + d)$
5
Substitute back, take absolute value, and use $∣ n ∣ = a^{2} + b^{2} + c^{2}$

Result:

We get the final formula: $D = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$ , where the absolute value ensures non-negative distance.

Exam tip:

Examiners often award 1 mark just for writing the correct distance formula, even if you make an arithmetic mistake later.

2. Calculating Distance: Step-by-Step★★☆☆☆HL only⏱ 6 min

To use the formula, always first rearrange your plane equation into the general form $a x + b y + cz + d = 0$ , then substitute. The worked example below demonstrates the full process:

📐 Worked Example

Find the perpendicular distance from $P (1, 2, - 1)$ to the plane $2 x - y + 3 z = 4$

1
1. Rearrange to general form $a x + b y + cz + d = 0$ :
2
$2x - y + 3z - 4 = 0 \implies a=2, b=-1, c=3, d=-4$
3
1. Identify point coordinates: $x_{0} = 1, y_{0} = 2, z_{0} = - 1$
4
1. Calculate the numerator:
5
$|(2)(1) + (-1)(2) + (3)(-1) + (-4)| = |2 - 2 - 3 - 4| = 7$
6
1. Calculate the denominator (magnitude of normal):
7
$\sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$
8
1. Simplify for final distance:
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$D = \frac{7}{\sqrt{14}} = \frac{\sqrt{14}}{2} \approx 1.87$

✓ Quick check

Test your understanding: What is the distance from the origin $(0, 0, 0)$ to the plane $x + 2 y - 2 z + 6 = 0$ ?

What is the correct distance?
- 1
- 2
- 3
- 6
Reveal answer
2 —
Correct: $D = \frac{∣0 + 0 - 0 + 6∣}{1 ^{2} + 2 ^{2} + ( - 2 ) ^{2}} = \frac{6}{3} = 2$

3. Special Cases: Distance Between Parallel Planes★★★☆☆HL only⏱ 6 min

A common exam problem asks for the distance between two parallel planes. This reduces to finding the distance from any point on the first plane to the second plane, as shown below:

📐 Worked Example

Find the distance between parallel planes $Π_{1} : 2 x + 2 y - z + 1 = 0$ and $Π_{2} : 2 x + 2 y - z + 4 = 0$

1
1. Confirm planes are parallel: their normal vectors are identical $(2, 2, - 1)$ , so they are parallel and do not intersect.
2
1. Find any point on $Π_{1}$ : set $x = 0, y = 0$ , solve for $z$ :
3
$0 + 0 - z + 1 = 0 \implies z=1 \implies P(0, 0, 1) \in \Pi_1$
4
1. Calculate distance from $P$ to $Π_{2}$ using the standard formula:
5
$D = \frac{|2(0) + 2(0) - 1(1) + 4|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{3}{3} = 1$

For planes given in vector form $r \cdot n = p$ , the equivalent distance formula is $D = \frac{∣ r _{0} \cdot n - p ∣}{∣ n ∣}$ , where $r_{0}$ is the position vector of the point.

Exam tip:

Always confirm planes are parallel before calculating distance: intersecting non-parallel planes have 0 distance.

4. Common Pitfalls

Wrong move:

Forgetting the absolute value in the numerator

Why:

Distance is a non-negative scalar, IB examiners penalize missing absolute value

Correct move:

Always take the absolute value of the numerator after substitution

Wrong move:

Using the wrong sign for $d$ from an incorrectly rearranged plane equation

Why:

If the constant term is left on the wrong side of the equals sign, the numerator will be wrong

Correct move:

Always rearrange to $a x + b y + cz + d = 0$ before identifying coefficients

Wrong move:

Calculating distance between non-parallel intersecting planes

Why:

Only parallel planes have a constant non-zero distance between them

Correct move:

First confirm normals are scalar multiples before calculating distance between two planes

Wrong move:

Leaving answers in unsimplified form with unrationalized denominators

Why:

IB mark schemes require fully simplified exact answers, so you lose marks for unsimplified radicals

Correct move:

Always rationalize denominators and simplify fractions fully before writing your final answer

5. Quick Reference Cheatsheet

Scenario	Formula
Point $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz + d = 0$	$D = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Point $r_{0}$ to plane $r \cdot n = p$	$D = \frac{∣ r _{0} \cdot n - p ∣}{∣ n ∣}$
Distance between parallel planes	Find any point on one plane, calculate distance to the other plane

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Find distance from origin to plane
2021 · 2
Distance in 3D coordinate problem

Going deeper

Formula sheetIB AA HL 3D Geometry Formula SummaryOfficial formula sheet reference

What's Next

Mastering distance from a point to a plane is a critical foundation for more advanced 3D geometry topics in IB AA HL. This skill is the basis for calculating distance between skew lines, finding closest points of intersection between lines and planes, and solving volume problems for 3D solids bounded by planes. It is frequently combined with vector line equations in extended response exam questions, so building fluency now will help with harder problems later.