# Intersections and angles between lines and planes (HL only)

> IB Mathematics AA HL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-intersections-and-angles-between-lines/

This HL-only module covers core 3D vector geometry: finding intersections of lines with planes, and calculating angles between lines & planes, and between two intersecting planes, with exam-style worked examples.

**Prerequisites:** [Vector equation of lines in 3D](https://www.owlsprep.com/study/ib-math-aa-hl-u3-vector-equation-lines/); [Vector and Cartesian equations of planes](https://www.owlsprep.com/study/ib-math-aa-hl-u3-planes-equations/)

## Learning objectives

- Classify the relative position of a line and a plane
- Calculate the intersection point of a non-parallel line and plane
- Find the acute angle between a line and a plane
- Find the acute dihedral angle between two intersecting planes

## Intersection of a Line and a Plane

**Three possible cases for line-plane intersection** — Given a line $l$ with parametric form and plane $\text{\Pi}$ with Cartesian equation, substitution gives a linear equation in the line parameter $t$: 1. Unique solution for $t$ = one intersection point, 2. $0t = k, k \neq 0$ = line parallel to plane, no intersection, 3. $0t = 0$ = line lies entirely on the plane, infinite intersections

**Worked example:** Find the intersection point of the line $\vec{r} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + t\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}$ and plane $2x + y - 2z = 5$

1. Write parametric equations for the line from the vector form:
2. $$x = 1 + 2t, \quad y = 2 - t, \quad z = 3 + t$$
3. Substitute into the plane's Cartesian equation:
4. $$2(1+2t) + (2-t) - 2(3 + t) = 5$$
5. Expand and simplify to solve for $t$:
6. $$2 + 4t + 2 - t - 6 - 2t = 5 \implies -2 + t = 5 \implies t = 7$$
7. Substitute $t=7$ back into the parametric equations to get the intersection point:
8. $$x = 1 + 14 = 15, \quad y = 2 - 7 = -5, \quad z = 3 + 7 = 10$$
9. Verify the point satisfies the plane equation: $2(15) + (-5) - 2(10) = 30 - 5 - 20 = 5$, which matches, so the intersection is:
10. $$(15, -5, 10)$$

> **Exam tip:** Always check which case you have before concluding there is no intersection or an infinite number of intersections.

## Angle Between a Line and a Plane

**Line-plane angle** — The acute angle $\theta$ (between $0^\circ$ and $90^\circ$) between a line and its orthogonal projection onto the plane. The formula uses the direction vector of the line $\vec{d}$ and the normal vector of the plane $\vec{n}$.

The angle between the line and the plane's normal is $90^\circ - \theta$, so using the dot product formula for the angle between two vectors gives the identity $\cos(90^\circ - \theta) = \sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$. The absolute value ensures we get an acute angle.

**Worked example:** Find the acute angle between the line $\vec{r} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + t\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}$ and plane $2x + y - 2z = 5$

1. Identify the direction vector of the line and normal vector of the plane:
2. $$\vec{d} = \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}, \quad \vec{n} = \begin{pmatrix}2 \\ 1 \\ -2\end{pmatrix}$$
3. Calculate the dot product and magnitudes:
4. $$\vec{d} \cdot \vec{n} = (2)(2) + (-1)(1) + (1)(-2) = 1, \quad |\vec{d}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}, \quad |\vec{n}| = 3$$
5. Substitute into the line-plane angle formula:
6. $$\sin\theta = \frac{|1|}{\sqrt{6} \times 3} = \frac{1}{3\sqrt{6}} \approx 0.136$$
7. Calculate $\theta$:
8. $$\theta = \arcsin(0.136) \approx 7.8^\circ \quad (\text{or } 0.136 \text{ radians})$$

> **warning**
>
> Never use $\cos\theta$ for the line-plane angle! This is the most common mistake in this topic.

## Angle Between Two Intersecting Planes

**Dihedral angle (angle between two planes)** — The acute angle between two intersecting planes, equal to the acute angle between their normal vectors. For planes with normals $\vec{n_1}$ and $\vec{n_2}$, the formula is $\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}$.

**Worked example:** Find the acute angle between planes $\Pi_1: x + y - z = 2$ and $\Pi_2: 2x - y + 3z = 1$

1. Extract the normal vectors from the Cartesian equations:
2. $$\vec{n_1} = \begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}, \quad \vec{n_2} = \begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix}$$
3. Calculate dot product and magnitudes:
4. $$\vec{n_1} \cdot \vec{n_2} = (1)(2) + (1)(-1) + (-1)(3) = -2, \quad |\vec{n_1}| = \sqrt{3}, \quad |\vec{n_2}| = \sqrt{14}$$
5. Substitute into the formula (remember absolute value):
6. $$\cos\theta = \frac{| -2 |}{\sqrt{3} \sqrt{14}} = \frac{2}{\sqrt{42}} \approx 0.309$$
7. Calculate $\theta$:
8. $$\theta = \arccos(0.309) \approx 72^\circ \quad (\text{or } 1.26 \text{ radians})$$

IB exam questions almost always ask for the acute angle between two planes. Only omit the absolute value if explicitly asked for the obtuse angle.

## Common pitfalls

- **Wrong:** Using $\cos\theta$ instead of $\sin\theta$ for the line-plane angle
  - Why it fails: You confuse the angle between the line and the normal with the line-plane angle itself
  - Correct: Memorize that line-plane angle uses $\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$
- **Wrong:** Forgetting the absolute value around the dot product, resulting in an obtuse angle when acute is required
  - Why it fails: The dot product of normals can be negative, which means the angle between the normals is obtuse; the dihedral angle is always acute unless stated otherwise
  - Correct: Always add an absolute value around the dot product in all angle formulas for this topic
- **Wrong:** Stopping after solving for $t$ and not reporting the intersection point coordinates
  - Why it fails: Rushing in the exam, you misread the question which asks for the intersection, not just the parameter
  - Correct: Always substitute the calculated $t$ back into the parametric equation of the line to get the final intersection coordinates
- **Wrong:** Calling any inconsistent result 'no intersection' without checking if the line is on the plane
  - Why it fails: You forget the three possible cases for line-plane intersection
  - Correct: If you get $0t = k$: if $k \neq 0$, it is parallel with no intersection; if $k=0$, the line lies on the plane with infinite intersections

## Cheatsheet

| Concept | Formula/Method | Key Notes |
| --- | --- | --- |
| Line-plane intersection | Substitute $x(t), y(t), z(t)$ into plane equation | 3 cases: 1 solution = 1 point, $0=k≠0$ = parallel, $0=0$ = line on plane |
| Angle: line-plane | $\sin\theta = \frac{\|\vec{d} \cdot \vec{n}\|}{\|\vec{d}\|\|\vec{n}\|}$ | $\vec{d}$ = line direction, $\vec{n}$ = plane normal, acute only |
| Angle: two planes | $\cos\theta = \frac{\|\vec{n_1} \cdot \vec{n_2}\|}{\|\vec{n_1}\|\|\vec{n_2}\|}$ | Dihedral angle, always acute unless stated otherwise |

## What's next

Mastering intersections and angles between lines and planes is a critical foundation for all advanced 3D vector geometry topics that appear frequently on IB AA HL exams. These skills are used to solve problems involving shortest distances between points, lines and planes, and to find the intersection of multiple planes, which are common long questions in both paper 1 and paper 2. They also underpin more advanced HL topics including vector integration in the calculus option, and 3D shape volume problems. Building fluency with these formulas and methods will save you significant time in the exam.

- [Distance from point to plane (HL only)](https://www.owlsprep.com/study/ib-math-aa-hl-u3-distance-from-point-to-plane/)
- [Statistics & Probability](https://www.owlsprep.com/study/ib-math-aa-hl-u4-overview/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-math-aa-hl-u3-intersections-and-angles-between-lines/