# Distance from point to plane (HL only)

> IB Mathematics: Analysis and Approaches HL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-distance-from-point-to-plane/

This module covers derivation and application of the formula for the shortest perpendicular distance between a point and a plane in 3D space, a core HL-only skill for IB AA HL exams.

**Prerequisites:** [Cartesian equation of a plane](https://www.owlsprep.com/study/ib-math-aa-hl-u3-plane-cartesian-equation/); [Vector dot product and scalar projection](https://www.owlsprep.com/study/ib-math-aa-hl-u2-vector-dot-product/)

## Learning objectives

- Derive the formula for perpendicular distance from a point to a plane
- Calculate distance between a point and a plane in Cartesian/vector form
- Solve problems involving distance between parallel planes

## Derivation of the Distance Formula

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.

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

*Notation:* $D$

**Derivation:** Derive the distance formula from a point $(x_0, y_0, z_0)$ to plane $ax + by + cz + d = 0$

*Starting from:* Plane $\Pi: ax + by + cz + d = 0$, point $P(x_0, y_0, z_0)$, $Q(x_1, y_1, z_1)$ is any point on $\Pi$, normal vector $\mathbf{n} = \begin{pmatrix}a \\ b \\ c\end{pmatrix}$

1. Distance $D$ equals the absolute scalar projection of $\overrightarrow{PQ}$ onto $\mathbf{n}$:
2. $$D = \left| \overrightarrow{PQ} \cdot \frac{\mathbf{n}}{|\mathbf{n}|} \right|$$
3. Since $Q$ lies on $\Pi$, $ax_1 + by_1 + cz_1 + d = 0 \implies d = -ax_1 - by_1 - cz_1$. 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 $|\mathbf{n}| = \sqrt{a^2 + b^2 + c^2}$

*Conclusion:* We get the final formula: $D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$, where the absolute value ensures non-negative distance.

> **tip**
>
> Always remember the absolute value in the numerator: losing it is a common 1-mark deduction in IB exams.

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

## Calculating Distance: Step-by-Step

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

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

1. 1. Rearrange to general form $ax + by + cz + d = 0$:
2. $$2x - y + 3z - 4 = 0 \implies a=2, b=-1, c=3, d=-4$$
3. 2. Identify point coordinates: $x_0 = 1, y_0=2, z_0=-1$
4. 3. Calculate the numerator:
5. $$|(2)(1) + (-1)(2) + (3)(-1) + (-4)| = |2 - 2 - 3 - 4| = 7$$
6. 4. Calculate the denominator (magnitude of normal):
7. $$\sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$$
8. 5. Simplify for final distance:
9. $$D = \frac{7}{\sqrt{14}} = \frac{\sqrt{14}}{2} \approx 1.87$$

**Check your understanding**

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

1. What is the correct distance?

   - 1
   - 2
   - 3
   - 6

   *Why:* Correct: $D = \frac{|0 + 0 - 0 + 6|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{6}{3} = 2$

## Special Cases: Distance Between Parallel Planes

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 $\Pi_1: 2x + 2y - z + 1 = 0$ and $\Pi_2: 2x + 2y - z + 4 = 0$

1. 1. Confirm planes are parallel: their normal vectors are identical $(2, 2, -1)$, so they are parallel and do not intersect.
2. 2. Find any point on $\Pi_1$: 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. 3. Calculate distance from $P$ to $\Pi_2$ 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 $\mathbf{r} \cdot \mathbf{n} = p$, the equivalent distance formula is $D = \frac{|\mathbf{r}_0 \cdot \mathbf{n} - p|}{|\mathbf{n}|}$, where $\mathbf{r}_0$ is the position vector of the point.

> **Exam tip:** Always confirm planes are parallel before calculating distance: intersecting non-parallel planes have 0 distance.

## Common pitfalls

- **Wrong:** Forgetting the absolute value in the numerator
  - Why it fails: Distance is a non-negative scalar, IB examiners penalize missing absolute value
  - Correct: Always take the absolute value of the numerator after substitution
- **Wrong:** Using the wrong sign for $d$ from an incorrectly rearranged plane equation
  - Why it fails: If the constant term is left on the wrong side of the equals sign, the numerator will be wrong
  - Correct: Always rearrange to $ax + by + cz + d = 0$ before identifying coefficients
- **Wrong:** Calculating distance between non-parallel intersecting planes
  - Why it fails: Only parallel planes have a constant non-zero distance between them
  - Correct: First confirm normals are scalar multiples before calculating distance between two planes
- **Wrong:** Leaving answers in unsimplified form with unrationalized denominators
  - Why it fails: IB mark schemes require fully simplified exact answers, so you lose marks for unsimplified radicals
  - Correct: Always rationalize denominators and simplify fractions fully before writing your final answer

## Cheatsheet

| Scenario | Formula |
| --- | --- |
| Point $(x_0,y_0,z_0)$ to plane $ax+by+cz+d=0$ | $D = \frac{\|ax_0 + by_0 + cz_0 + d\|}{\sqrt{a^2 + b^2 + c^2}}$ |
| Point $\mathbf{r}_0$ to plane $\mathbf{r} \cdot \mathbf{n} = p$ | $D = \frac{\|\mathbf{r}_0 \cdot \mathbf{n} - p\|}{\|\mathbf{n}\|}$ |
| Distance between parallel planes | Find any point on one plane, calculate distance to the other plane |

## 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.

- [Statistics & Probability](https://www.owlsprep.com/study/ib-math-aa-hl-u4-overview/)
- [Data types and sampling](https://www.owlsprep.com/study/ib-math-aa-hl-u4-data-types-and-sampling/)

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