# Vectors

> CIE A-Level Further Mathematics · Further Pure 1
> Source: https://www.owlsprep.com/study/cie-9231-u1-vectors/

This sub-topic covers vector geometry in 3D space for CIE A-Level Further Maths FP1. You will learn to represent lines and planes, calculate distances and angles, and solve common intersection problems that appear regularly in Paper 1.

**Prerequisites:** [Basic vector algebra, scalar and vector products](https://www.owlsprep.com/study/cie-9709-pure-vectors/)

## Learning objectives

- Represent lines and planes in 3D space using vector forms
- Calculate angles between lines and between planes
- Find shortest distances between points, lines and planes
- Solve intersection problems for lines and planes
- Apply vector products and scalar products to geometric problems

## Lines in 3D Space

**Vector Equation of a Line** — A line through a fixed point with position vector $\mathbf{a}$, parallel to direction vector $\mathbf{d}$, where $t \in \mathbb{R}$ is a scalar parameter.

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

*Example:* Line through $(1,2,3)$ parallel to $\begin{pmatrix}2\\-1\\0\end{pmatrix}$ is $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}2\\-1\\0\end{pmatrix}$

**Worked example:** Find the vector equation of the line passing through points $A(1, 0, 2)$ and $B(3, 1, -1)$.

1. Take the position vector of point $A$ as the fixed point $\mathbf{a}$:
2. $$\mathbf{a} = \begin{pmatrix}1\\0\\2\end{pmatrix}$$
3. Calculate the direction vector $\mathbf{d} = \overrightarrow{AB}$:
4. $$\mathbf{d} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}3\\1\\-1\end{pmatrix} - \begin{pmatrix}1\\0\\2\end{pmatrix} = \begin{pmatrix}2\\1\\-3\end{pmatrix}$$
5. Substitute into the general line equation:
6. $$\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-3\end{pmatrix}, \quad t \in \mathbb{R}$$

> **tip**
>
> Any non-zero scalar multiple of $\mathbf{d}$ is a valid direction vector, so your answer may look different but still be correct.

## Vector Forms of Planes

**Scalar Product Form of a Plane** — A plane with normal vector $\mathbf{n}$, where $k$ is a constant equal to $\mathbf{a} \cdot \mathbf{n}$ for any fixed point $\mathbf{a}$ on the plane.

*Notation:* \mathbf{r} \cdot \mathbf{n} = k

*Example:* Plane with normal $\begin{pmatrix}1\\2\\-1\end{pmatrix}$ through $(2,0,1)$ gives $k = 2(1) + 0(2) + 1(-1) = 1$, so $\mathbf{r} \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix} = 1$

To find the scalar product form for a plane through three points, first calculate two vectors lying on the plane, then take their vector product to get the normal vector.

**Worked example:** Find the scalar product form of the plane through $P(1,1,0)$, $Q(2,1,3)$ and $R(1,2,-1)$.

1. Find two vectors on the plane:
2. $$\overrightarrow{PQ} = \begin{pmatrix}1\\0\\3\end{pmatrix}, \quad \overrightarrow{PR} = \begin{pmatrix}0\\1\\-1\end{pmatrix}$$
3. Calculate the normal vector as the vector product:
4. $$\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&0&3\\0&1&-1\end{vmatrix} = \begin{pmatrix}-3\\1\\1\end{pmatrix}$$
5. Calculate $k$ by substituting point $P$:
6. $$k = \mathbf{p} \cdot \mathbf{n} = (1)(-3) + (1)(1) + (0)(1) = -2$$
7. Final equation (simplified to Cartesian form):
8. $$\mathbf{r} \cdot \begin{pmatrix}-3\\1\\1\end{pmatrix} = -2 \implies 3x - y - z = 2$$

## Shortest Distances

All distance formulae are derived from projecting a vector onto a direction or normal vector. Standard formulae exist for all common pairs of objects:

- Distance from a point to a line
- Distance from a point to a plane
- Distance between two skew lines
- Distance between two parallel planes

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

1. Use the standard formula for distance from $(x_0,y_0,z_0)$ to plane $ax + by + cz = d$:
2. $$D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$$
3. Substitute the values $a=1, b=2, c=-2, d=1, x_0=2, y_0=1, z_0=3$:
4. $$D = \frac{|(1)(2) + (2)(1) + (-2)(3) - 1|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|-3|}{3} = 1$$
5. The shortest distance is 1 unit.

> **warning**
>
> Always take the absolute value in the numerator, as distance is a non-negative quantity.

## Intersections and Angles

To find intersections, substitute the parametric form of a line into the plane equation, solve for the parameter, then find the intersection point coordinates. For angles, we use the scalar product to find the acute angle between direction or normal vectors.

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

1. The angle between two planes equals the acute angle between their normal vectors:
2. $$\mathbf{n}_1 = \begin{pmatrix}1\\1\\1\end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix}2\\-1\\3\end{pmatrix}$$
3. Use the scalar product formula for acute angle $\theta$:
4. $$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$$
5. Calculate each term:
6. $$\mathbf{n}_1 \cdot \mathbf{n}_2 = 4, \quad |\mathbf{n}_1| = \sqrt{3}, \quad |\mathbf{n}_2| = \sqrt{14}$$
7. Solve for $\theta$:
8. $$\cos\theta = \frac{4}{\sqrt{42}} \approx 0.617 \implies \theta \approx 52^\circ$$

## Common pitfalls

- **Wrong:** Forgetting to take the absolute value for distance or angles
  - Why it fails: Distance is always non-negative, and exams require the acute angle unless stated otherwise
  - Correct: Always add absolute value to the numerator of distance and angle formulae
- **Wrong:** Using $\cos\theta$ for the angle between a line and a plane
  - Why it fails: The angle between a line and plane is complementary to the angle between the line direction and plane normal
  - Correct: Use $\sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ for line-plane angles
- **Wrong:** Assuming all non-parallel 3D lines intersect
  - Why it fails: Skew lines are non-parallel and never intersect, a common exam question setup
  - Correct: Always check for intersection, calculate the shortest distance if lines are skew
- **Wrong:** Not checking plane equation calculations after derivation
  - Why it fails: Small arithmetic errors when calculating $k$ or rearranging lead to wrong answers
  - Correct: Substitute all three original points back into the final plane equation to verify

## Cheatsheet

| Object Type | Vector Form | Key Formula |
| --- | --- | --- |
| Line through $\mathbf{a}$ parallel to $\mathbf{d}$ | $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ | - |
| Plane through $\mathbf{a}$ with normal $\mathbf{n}$ | $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$ | - |
| Distance from $P$ to plane | - | $\frac{\|\mathbf{p} \cdot \mathbf{n} - k\|}{\|\mathbf{n}\|}$ |
| Distance from $P$ to line | - | $\|\overrightarrow{AP} - (\overrightarrow{AP} \cdot \hat{\mathbf{d}})\hat{\mathbf{d}}\|$ |
| Angle between two lines | - | $\cos\theta = \frac{\|\mathbf{d}_1 \cdot \mathbf{d}_2\|}{\|\mathbf{d}_1\|\|\mathbf{d}_2\|}$ |
| Angle between two planes | - | $\cos\theta = \frac{\|\mathbf{n}_1 \cdot \mathbf{n}_2\|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}$ |
| Angle between line and plane | - | $\sin\theta = \frac{\|\mathbf{d} \cdot \mathbf{n}\|}{\|\mathbf{d}\|\|\mathbf{n}\|}$ |

## What's next

Vectors in 3D is a foundational topic for many areas of CIE A-Level Further Mathematics, including 3D kinematics in mechanics, linear algebra, and further vector geometry in Further Pure 2. Mastery of the formulae and methods here is critical for solving larger geometric problems that appear frequently in Paper 1, and accounts for around 10-15% of the FP1 paper marks. The vector techniques you learned in this sub-topic will be extended when you study matrices, linear transformations and more advanced vector problems in later units.

- [Further Pure 2](https://www.owlsprep.com/study/cie-9231-u2-overview/)
- [Hyperbolic Functions](https://www.owlsprep.com/study/cie-9231-u2-hyperbolic-functions/)
- [Further Differentiation and Applications](https://www.owlsprep.com/study/cie-9231-u2-further-differentiation-and-applications/)

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