# Vectors (Edexcel IAL Further Maths FP3)

> Edexcel International A-Level Further Mathematics · FP3 Vectors
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-vectors/

This guide covers all FP3 vector content for Edexcel IAL Further Maths: cross products, triple scalar products, plane equations, and core 3D geometry applications aligned to the 2018 specification.

**Prerequisites:** Dot product and basic vector operations (from P3); 3D coordinate geometry fundamentals (from P3)

## Learning objectives

- Calculate vector cross products and interpret their magnitude as area of parallelograms/triangles
- Compute triple scalar products and use them to find volumes of parallelepipeds
- Derive, convert between, and use both standard forms of plane equations
- Solve core 3D problems: point-to-plane distance, line of intersection of two planes, shortest distance between skew lines

## Vector Cross Product & Triple Scalar Product

**Vector (Cross) Product** — Vector perpendicular to both $\boldsymbol{a}$ and $\boldsymbol{b}$, with magnitude $|\boldsymbol{a}||\boldsymbol{b}|\sin\theta$ (where $\theta$ is the angle between the vectors) and direction given by the right-hand rule. Has the anti-commutative property $\boldsymbol{a} \times \boldsymbol{b} = -(\boldsymbol{b} \times \boldsymbol{a})$.

*Notation:* $\boldsymbol{a} \times \boldsymbol{b}$

*Example:* For $\boldsymbol{a}=2\boldsymbol{i}+\boldsymbol{j}$, $\boldsymbol{b}=\boldsymbol{i}-3\boldsymbol{j}+\boldsymbol{k}$, $\boldsymbol{a} \times \boldsymbol{b} = \boldsymbol{i} - 2\boldsymbol{j} -7\boldsymbol{k}$.

The magnitude $|\boldsymbol{a} \times \boldsymbol{b}|$ equals the area of the parallelogram formed by $\boldsymbol{a}$ and $\boldsymbol{b}$ as adjacent sides. Divide this by 2 to get the area of the triangle formed by the two vectors.

**Triple Scalar Product** — Scalar result of the dot product of $\boldsymbol{a}$ with the cross product of $\boldsymbol{b}$ and $\boldsymbol{c}$, equal to the determinant of the 3×3 matrix formed by the components of $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$. Its magnitude gives the volume of the parallelepiped formed by the three vectors as adjacent edges. Obeys the cyclic identity: $\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = \boldsymbol{b} \cdot (\boldsymbol{c} \times \boldsymbol{a}) = \boldsymbol{c} \cdot (\boldsymbol{a} \times \boldsymbol{b})$.

*Notation:* $\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c})$

**Worked example:** Find the volume of the parallelepiped with edges $\boldsymbol{a}=\boldsymbol{i}+2\boldsymbol{j}$, $\boldsymbol{b}=2\boldsymbol{j}+\boldsymbol{k}$, $\boldsymbol{c}=\boldsymbol{i}-\boldsymbol{j}+3\boldsymbol{k}$.

1. 1. Compute the cross product $\boldsymbol{b} \times \boldsymbol{c}$ using the determinant formula:
2. $$\boldsymbol{b} \times \boldsymbol{c} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 0 & 2 & 1 \\ 1 & -1 & 3 \end{vmatrix} = 7\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}$$
3. 2. Take the dot product of $\boldsymbol{a}$ with the result:
4. $$\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = (1)(7) + (2)(1) + (0)(-2) = 9$$
5. 3. The magnitude of the result is the volume: 9 cubic units.

> **Exam tip:** Sign from cross product only matters for orientation questions: take the magnitude when calculating area or volume to avoid errors.

*Calculator:* allowed

## Equations of Planes

There are two standard forms of plane equations you will need to convert between and apply for FP3 problems, both with equivalent Cartesian forms.

**Normal (Scalar) Form of a Plane** — Where $\boldsymbol{r}$ is the position vector of any point on the plane, $\boldsymbol{n}$ is the normal vector to the plane, and $p = \boldsymbol{a} \cdot \boldsymbol{n}$ for a known point $\boldsymbol{a}$ on the plane. Equivalent Cartesian form: $n_1x + n_2y + n_3z = p$, or $n_1x + n_2y + n_3z - p = 0$ with $d=-p$.

*Notation:* $\boldsymbol{r} \cdot \boldsymbol{n} = p$

**Parametric Form of a Plane** — Where $\boldsymbol{a}$ is the position vector of a fixed point on the plane, $\boldsymbol{b}$ and $\boldsymbol{c}$ are non-parallel direction vectors lying on the plane, and $s, t$ are scalar parameters.

*Notation:* $\boldsymbol{r} = \boldsymbol{a} + s\boldsymbol{b} + t\boldsymbol{c}$

**Worked example:** Find the normal form and Cartesian form of the plane passing through $(1,2,0)$ with direction vectors $\boldsymbol{b}=\boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}$ and $\boldsymbol{c}=2\boldsymbol{i}+3\boldsymbol{j}$.

1. 1. Compute the normal vector as the cross product of the two direction vectors:
2. $$\boldsymbol{n} = \boldsymbol{b} \times \boldsymbol{c} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 1 & -1 & 1 \\ 2 & 3 & 0 \end{vmatrix} = -3\boldsymbol{i} + 2\boldsymbol{j} +5\boldsymbol{k}$$
3. 2. Calculate $p = \boldsymbol{a} \cdot \boldsymbol{n}$ where $\boldsymbol{a} = \boldsymbol{i} + 2\boldsymbol{j}$:
4. $$p = (1)(-3) + (2)(2) + (0)(5) = 1$$
5. 3. Normal form: $\boldsymbol{r} \cdot (-3\boldsymbol{i} + 2\boldsymbol{j} +5\boldsymbol{k}) = 1$ (multiply by -1 for positive leading coefficient if preferred).
6. 4. Cartesian form: Replace $\boldsymbol{r}$ with $x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}$ to get $3x - 2y -5z + 1 = 0$.

> **Exam tip:** If you are given three points on a plane, find two direction vectors between pairs of points, then compute their cross product to get the normal vector for the plane.

*Calculator:* allowed

## 3D Applications: Point-Plane Distance & Plane Intersection

The first two core 3D applications for FP3 use plane equations and vector products to solve distance and intersection problems. The perpendicular distance formula for a point to a plane is provided in your exam formula booklet.

**Worked example:** Find the perpendicular distance of the point $(2, -1, 3)$ from the plane $3x + 2y - z + 4 = 0$.

1. 1. Identify values for the distance formula: $n_1=3, n_2=2, n_3=-1, d=4, \alpha=2, \beta=-1, \gamma=3$.
2. 2. Compute the absolute value of the numerator:
3. $$|3(2) + 2(-1) + (-1)(3) + 4| = |6 - 2 - 3 + 4| = 5$$
4. 3. Compute the magnitude of the normal vector:
5. $$\sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{14}$$
6. 4. Distance = $\frac{5}{\sqrt{14}}$, or rationalized to $\frac{5\sqrt{14}}{14}$.

The line of intersection of two planes is parallel to the cross product of the normal vectors of the two planes, as it is perpendicular to both normals. Find one point lying on both planes to define the full line.

**Worked example:** Find the line of intersection of planes $2x - y + z = 5$ and $x + y - z = 1$, in the form $(\boldsymbol{r} - \boldsymbol{a}) \times \boldsymbol{b} = 0$.

1. 1. Extract normal vectors: $\boldsymbol{n_1}=2\boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}$, $\boldsymbol{n_2}=\boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k}$.
2. 2. Compute the direction vector of the line as the cross product of the normals:
3. $$\boldsymbol{b} = \boldsymbol{n_1} \times \boldsymbol{n_2} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 3\boldsymbol{j} + 3\boldsymbol{k} = 3(\boldsymbol{j} + \boldsymbol{k})$$
4. 3. Find a point on both planes: set $x=2$, then solve to get $y=0, z=1$, so $\boldsymbol{a} = 2\boldsymbol{i} + 0\boldsymbol{j} + \boldsymbol{k}$.
5. 4. Final line equation: $(\boldsymbol{r} - (2\boldsymbol{i} + \boldsymbol{k})) \times (\boldsymbol{j} + \boldsymbol{k}) = 0$.

> **Exam tip:** If setting $x=0$ to find a point on the intersection of two planes leads to inconsistent equations, try setting $y=0$ or $z=0$ instead.

*Calculator:* allowed

## 3D Applications: Shortest Distance Between Skew Lines

**Skew Lines** — Two lines in 3D space that are neither parallel nor intersecting. The shortest distance between them is the length of the line segment that is perpendicular to both lines.

The formula for the shortest distance between skew lines $L_1: \boldsymbol{r} = \boldsymbol{a} + \lambda \boldsymbol{u}$ and $L_2: \boldsymbol{r} = \boldsymbol{b} + \mu \boldsymbol{v}$ is $\frac{|(\boldsymbol{b} - \boldsymbol{a}) \cdot (\boldsymbol{u} \times \boldsymbol{v})|}{|\boldsymbol{u} \times \boldsymbol{v}|}$. The numerator is the volume of the parallelepiped formed by $(\boldsymbol{b}-\boldsymbol{a}), \boldsymbol{u}, \boldsymbol{v}$, divided by the area of the base parallelogram to give the perpendicular height (distance).

**Worked example:** Find the shortest distance between skew lines $L_1: \boldsymbol{r} = \boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k} + \lambda(2\boldsymbol{i} - \boldsymbol{j} + 3\boldsymbol{k})$ and $L_2: \boldsymbol{r} = 2\boldsymbol{i} - \boldsymbol{j} + 2\boldsymbol{k} + \mu(3\boldsymbol{i} + \boldsymbol{j} - \boldsymbol{k})$.

1. 1. Identify values: $\boldsymbol{a}=\boldsymbol{i}+2\boldsymbol{j}+\boldsymbol{k}$, $\boldsymbol{u}=2\boldsymbol{i}-\boldsymbol{j}+3\boldsymbol{k}$, $\boldsymbol{b}=2\boldsymbol{i}-\boldsymbol{j}+2\boldsymbol{k}$, $\boldsymbol{v}=3\boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k}$.
2. 2. Compute the vector between the fixed points of the two lines:
3. $$\boldsymbol{b} - \boldsymbol{a} = \boldsymbol{i} - 3\boldsymbol{j} + \boldsymbol{k}$$
4. 3. Compute the cross product of the direction vectors:
5. $$\boldsymbol{u} \times \boldsymbol{v} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 2 & -1 & 3 \\ 3 & 1 & -1 \end{vmatrix} = -2\boldsymbol{i} + 11\boldsymbol{j} +5\boldsymbol{k}$$
6. 4. Compute the absolute value of the triple scalar product:
7. $$|(\boldsymbol{b} - \boldsymbol{a}) \cdot (\boldsymbol{u} \times \boldsymbol{v})| = |-2 -33 +5| = 30$$
8. 5. Compute the magnitude of the cross product:
9. $$|\boldsymbol{u} \times \boldsymbol{v}| = \sqrt{(-2)^2 + 11^2 + 5^2} = \sqrt{150} = 5\sqrt{6}$$
10. 6. Shortest distance = $\frac{30}{5\sqrt{6}} = \sqrt{6}$ units.

> **Exam tip:** Double-check cross product calculations for skew line problems: arithmetic errors here are the most common cause of lost marks on this topic.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Swapping the order of vectors in a cross product and forgetting the sign flips.
  - Why it fails: The cross product is anti-commutative: $\boldsymbol{a} \times \boldsymbol{b} = -(\boldsymbol{b} \times \boldsymbol{a})$.
  - Correct: Keep track of vector order, or take the magnitude when calculating area/volume so sign does not affect results.
- **Wrong:** Assuming non-intersecting 3D lines are parallel.
  - Why it fails: 3D space has an extra dimension, so lines can be non-parallel and non-intersecting (skew).
  - Correct: First check if direction vectors are scalar multiples to confirm parallelism, before checking for intersection.
- **Wrong:** Using parallel direction vectors for the parametric plane equation $\boldsymbol{r} = \boldsymbol{a} + s\boldsymbol{b} + t\boldsymbol{c}$.
  - Why it fails: Parallel vectors do not define a unique plane, as they only span a line.
  - Correct: Use two non-parallel direction vectors, verified by a non-zero cross product.
- **Wrong:** Forgetting the absolute value in distance formulas.
  - Why it fails: Distance is a non-negative scalar; sign only indicates orientation relative to the plane normal.
  - Correct: Always include absolute value around the numerator in point-plane and skew line distance calculations.
- **Wrong:** Assuming $p$ in $\boldsymbol{r} \cdot \boldsymbol{n} = p$ is the perpendicular distance from the origin.
  - Why it fails: This is only true if $\boldsymbol{n}$ is a unit normal vector.
  - Correct: If you need the origin distance, divide $p$ by the magnitude of $\boldsymbol{n}$, or use the Cartesian distance formula for point $(0,0,0)$.

## Cheatsheet

| Concept | Formula/Rule | Use Case |
| --- | --- | --- |
| Cross Product | $\|\boldsymbol{a} \times \boldsymbol{b}\| = \|\boldsymbol{a}\|\|\boldsymbol{b}\|\sin\theta$ | Find area of parallelogram/triangle, plane normal vectors |
| Triple Scalar Product | $\|\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c})\|$ | Volume of parallelepiped, skew line distance calculations |
| Plane: Normal Form | $\boldsymbol{r} \cdot \boldsymbol{n} = p$ | Point-plane distance checks, verifying points lie on a plane |
| Plane: Parametric Form | $\boldsymbol{r} = \boldsymbol{a} + s\boldsymbol{b} + t\boldsymbol{c}$ | Defining planes from a point and two direction vectors |
| Point-Plane Distance | $\frac{\|n_1\alpha + n_2\beta + n_3\gamma + d\|}{\sqrt{n_1^2 + n_2^2 + n_3^2}}$ | Perpendicular distance from a point to a plane |
| Skew Line Distance | $\frac{\|(\boldsymbol{b} - \boldsymbol{a}) \cdot (\boldsymbol{u} \times \boldsymbol{v})\|}{\|\boldsymbol{u} \times \boldsymbol{v}\|}$ | Shortest distance between two non-parallel, non-intersecting lines |

## What's next

Now that you have mastered FP3 vectors, you can apply these skills to past paper questions, as this topic typically makes up 15-20% of the FP3 exam marks. Vectors are also used in other further math units including FP2 complex number geometry and M3 kinematics, but focus strictly on the pure geometry applications specified for FP3 to avoid off-syllabus content. Practice writing all steps clearly, as method marks are awarded for correct cross and dot product calculations even if you make minor arithmetic errors.

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