# Vector equations of lines and planes (HL only)

> IB Mathematics: Analysis and Approaches HL · IB AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-vector-equations-of-lines-and/

This sub-topic covers how to represent lines and planes in 3D space using vector notation, including multiple forms for each. You will learn to find intersections and calculate angles between these core 3D objects.

**Prerequisites:** [Vectors and the scalar dot product](https://www.owlsprep.com/study/ib-math-aa-hl-vectors-dot-product/); [Cross product of 3D vectors](https://www.owlsprep.com/study/ib-math-aa-hl-cross-product-of-vectors/)

## Learning objectives

- Derive and use vector equations of lines in 2D and 3D space
- Write equations of planes in vector, parametric and Cartesian forms
- Find intersections between lines and planes in 3D
- Calculate angles between lines and planes using dot product
- Distinguish between parallel, intersecting and coincident planes

## Vector Equation of a Line in 3D

**Vector equation of a line** — General form for a line passing through a point with position vector $\mathbf{a}$, with direction vector $\mathbf{d}$, where $t \in \mathbb{R}$ is a scalar parameter that generates all points on the line.

*Notation:* $\mathbf{r} = \mathbf{a} + t\mathbf{d}$

*Example:* A line through $(1,2,3)$ in direction $(2,-1,0)$ is $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}2\\-1\\0\end{pmatrix}$

Two other common forms are also used: parametric form (splitting into separate $x, y, z$ components) and Cartesian form (eliminating the parameter $t$ to write a single equality of fractions).

**Worked example:** Find the Cartesian equation of the line with vector equation $\mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + t\begin{pmatrix}1\\3\\-2\end{pmatrix}$

1. Write the line in parametric form by splitting into components:

   $$x = 2 + t,\quad y = -1 + 3t,\quad z = 4 - 2t$$
2. Isolate $t$ from each equation:

   $$t = x - 2,\quad t = \frac{y+1}{3},\quad t = \frac{z-4}{-2}$$
3. Equate all expressions for $t$ to get the final Cartesian form:

   $$\frac{x - 2}{1} = \frac{y + 1}{3} = \frac{z - 4}{-2}$$

> **exam_tip**
>
> Always check that your direction vector components match the denominators in Cartesian form to avoid sign errors.

## Vector Equations of a Plane

**Normal form of a plane** — A plane with normal vector $\mathbf{n}$ (perpendicular to the plane) has this form, where $\frac{d}{|\mathbf{n}|}$ is the perpendicular distance from the origin to the plane. All points with position vector $\mathbf{r}$ on the plane satisfy this equation.

*Notation:* $\mathbf{r} \cdot \mathbf{n} = d$

Three forms of the plane equation are common in IB exams: normal vector form, Cartesian form (expanded from normal form), and parametric form (using two direction vectors in the plane). If you are given three points on the plane, find two direction vectors by subtracting position vectors, then use the cross product to find the normal vector.

**Worked example:** Find the Cartesian equation of the plane passing through points $A(1,0,1)$, $B(2,1,3)$ and $C(1,2,3)$

1. Find two direction vectors lying on the plane by subtracting point A from B and C:

   $$\vec{AB} = \begin{pmatrix}1\\1\\2\end{pmatrix}, \quad \vec{AC} = \begin{pmatrix}0\\2\\2\end{pmatrix}$$
2. Calculate the normal vector as the cross product of the two direction vectors:

   $$\mathbf{n} = \vec{AB} \times \vec{AC} = -2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} = \begin{pmatrix}-2\\-2\\2\end{pmatrix}$$
3. Find $d$ by substituting the position vector of point A into normal form:

   $$d = \mathbf{a} \cdot \mathbf{n} = (1)(-2) + 0(-2) + (1)(2) = 0$$
4. Expand and simplify to get Cartesian form:

   $$-2x -2y + 2z = 0 \implies x + y - z = 0$$

> **tip**
>
> You can always simplify the normal vector by dividing by a non-zero scalar to make the Cartesian equation easier to work with.

## Intersections of Lines and Planes

There are three possible outcomes when finding the intersection of a line and a plane: the line intersects the plane at exactly one point, the line is parallel and never intersects, or the entire line lies on the plane. For two planes, they either intersect along a line or are parallel (distinct or coincident).

**Worked example:** Find the intersection point of the line $\mathbf{r} = \begin{pmatrix}1\\2\\1\end{pmatrix} + t\begin{pmatrix}2\\-1\\3\end{pmatrix}$ and the plane $2x + 3y - z = 7$

1. Write the parametric equations for the line:

   $$x = 1 + 2t,\quad y = 2 - t,\quad z = 1 + 3t$$
2. Substitute the parametric expressions into the plane equation:

   $$2(1+2t) + 3(2-t) - (1+3t) = 7$$
3. Expand and solve for $t$:

   $$2 + 4t + 6 - 3t - 1 - 3t = 7 \implies 7 - 2t = 7 \implies t = 0$$
4. Substitute $t=0$ back to find the intersection point:

   $$(1, 2, 1)$$

> **info**
>
> If solving for $t$ gives a contradiction (e.g. $0=5$), the line is parallel and does not intersect the plane. If you get $0=0$, the entire line lies on the plane.

## Angles Between Lines and Planes

All angles between 3D vector objects are calculated using the dot product, with different formulas depending on what you are finding the angle between. Angles are always between $0^\circ$ and $90^\circ$ in this topic.

**Angle between two planes** — The acute angle between two planes is equal to the acute angle between their normal vectors, calculated with the formula:

*Example:* $\cos\theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}||\mathbf{n_2}|}$

**Worked example:** Find the angle between the planes $x + y + z = 1$ and $2x - y + 3z = 2$, to the nearest degree.

1. Extract the normal vectors from the Cartesian equations:

   $$\mathbf{n_1} = \begin{pmatrix}1\\1\\1\end{pmatrix}, \quad \mathbf{n_2} = \begin{pmatrix}2\\-1\\3\end{pmatrix}$$
2. Calculate the dot product and magnitudes of the normals:

   $$\mathbf{n_1} \cdot \mathbf{n_2} = 4, \quad |\mathbf{n_1}| = \sqrt{3}, \quad |\mathbf{n_2}| = \sqrt{14}$$
3. Substitute into the formula to find $\theta$:

   $$\cos\theta = \frac{|4|}{\sqrt{3}\sqrt{14}} \approx 0.617 \implies \theta \approx 52^\circ$$

> **exam_tip**
>
> For the angle between a line and a plane, use $\sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$, not $\cos\theta$. The angle is complementary to the angle between the direction and normal.

## Common pitfalls

- **Wrong:** Confusing position and direction vectors when writing a line equation
  - Why it fails: Swapping these gives an incorrect line through the origin instead of the given point
  - Correct: Always keep the position vector of your known point as the constant term, and multiply the direction vector by the parameter.
- **Wrong:** Forgetting the absolute value in angle formulas
  - Why it fails: A negative dot product will give you the obtuse supplementary angle, instead of the required acute angle between objects
  - Correct: Always take the absolute value of the dot product when calculating angles between lines and planes.
- **Wrong:** Assuming planes with the same normal are the same plane
  - Why it fails: Planes with the same normal are parallel but can be distinct if the constant term $d$ is different
  - Correct: Check the value of $d$ from $\mathbf{r} \cdot \mathbf{n} = d$ to confirm if parallel planes are coincident or distinct.
- **Wrong:** Calculating direction of intersection line with dot product of normals
  - Why it fails: The direction of the intersection line must be perpendicular to both normals, which requires cross product not dot product
  - Correct: The direction vector of the line of intersection of two planes is always the cross product of the two normals.
- **Wrong:** Swapping normal component order when converting to Cartesian plane equation
  - Why it fails: This mixes up the coefficients of $x, y, z$ and gives an incorrect plane
  - Correct: Always match the first component of the normal to $x$, the second to $y$, and the third to $z$.

## Cheatsheet

| Object | Formula | Notes |
| --- | --- | --- |
| Line in 3D | $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ | Cartesian: $\frac{x-a_x}{d_x} = \frac{y-a_y}{d_y} = \frac{z-a_z}{d_z}$ |
| Plane (normal form) | $\mathbf{r} \cdot \mathbf{n} = d$ | Cartesian: $n_x x + n_y y + n_z z = d$ |
| Plane (parametric) | $\mathbf{r} = \mathbf{a} + s\mathbf{d_1} + t\mathbf{d_2}$ | Convert via $\mathbf{n} = \mathbf{d_1} \times \mathbf{d_2}$ |
| Angle between two lines | $\cos\theta = \frac{\|\mathbf{d_1} \cdot \mathbf{d_2}\|}{\|\mathbf{d_1}\|\|\mathbf{d_2}\|}$ | $\mathbf{d_1}, \mathbf{d_2}$ = direction vectors |
| Angle between two planes | $\cos\theta = \frac{\|\mathbf{n_1} \cdot \mathbf{n_2}\|}{\|\mathbf{n_1}\|\|\mathbf{n_2}\|}$ | $\mathbf{n_1}, \mathbf{n_2}$ = normal vectors |
| Angle between line and plane | $\sin\theta = \frac{\|\mathbf{d} \cdot \mathbf{n}\|}{\|\mathbf{d}\|\|\mathbf{n}\|}$ | $\mathbf{d}$ = line direction, $\mathbf{n}$ = plane normal |

## What's next

Mastering vector equations of lines and planes is the foundation for all advanced 3D vector topics tested in IB AA HL. The next core skill is calculating distances from points to lines and planes, which relies entirely on your ability to write correct vector equations for these objects. This topic also prepares you for more extended problems involving intersections of multiple planes, which regularly appear as high-mark questions in Paper 1 and Paper 2 of the exam.

- [Intersections and angles between lines and planes (HL only)](https://www.owlsprep.com/study/ib-math-aa-hl-u3-intersections-and-angles-between-lines/)
- [Distance from point to plane (HL only)](https://www.owlsprep.com/study/ib-math-aa-hl-u3-distance-from-point-to-plane/)

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