Vectors (Edexcel IAL P4)

Q: Do I need to learn the dot product formula for Edexcel IAL P4?

Yes, the dot product formula $a\cdot b = a1b1+a2b2+a3b3$ is not given in the formula booklet, so you must memorise it. The cosine of the angle formula and distance formula are standard and expected to be applied correctly.

Edexcel International A-Level Mathematics· 2018 Specification (Issue 3) Unit P4 (WMA14) §7· 25 min read

1. Vectors, Magnitude and Basic Operations★★☆☆☆⏱ 5 min

✓ Calculator OK

📘 Definition

Vector

A quantity with both magnitude (size) and direction, written in component form as (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) or ((x,y,z)) for 3D vectors.

The magnitude of a vector (\mathbf{a} = (a_1,a_2,a_3)) is calculated as $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ . A unit vector in the direction of (\mathbf{a}) is found by dividing (\mathbf{a}) by its magnitude: $\hat{a} = \frac{a}{∣ a ∣}$ . Vectors can be added component-wise, or multiplied by a scalar to scale their magnitude (and reverse direction if the scalar is negative).

📐 Worked Example

Find the magnitude of the vector (\mathbf{v} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) and state the unit vector in the direction of (\mathbf{v}).

1
Square each component of (\mathbf{v}):
2
$3^2 = 9, \quad (-2)^2 = 4, \quad 6^2 = 36$
3
Sum the squared components:
4
$9 + 4 + 36 = 49$
5
Take the square root to find magnitude:
6
$|\mathbf{v}| = \sqrt{49} = 7$
7
Divide the original vector by its magnitude to get the unit vector:
8
$\hat{\mathbf{v}} = \frac{3}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{6}{7}\mathbf{k}$

2. Position Vectors and Distance Between Points★★☆☆☆⏱ 5 min

✓ Calculator OK

📘 Definition

Position Vector

The vector from the origin O to a point A, written (\overrightarrow{OA} = \mathbf{a}). The vector between points A and B is given by (\overrightarrow{AB} = \mathbf{b} - \mathbf{a}).

The distance between two points A and B is equal to the magnitude of (\overrightarrow{AB}). For points with coordinates ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)), the distance formula is $d = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}$ .

📐 Worked Example

Points A and B have position vectors (\mathbf{a} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) and (\mathbf{b} = 5\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}). Calculate the exact distance between A and B.

1
Calculate (\overrightarrow{AB} = \mathbf{b} - \mathbf{a}):
2
$\overrightarrow{AB} = (5-2)\mathbf{i} + (-4-1)\mathbf{j} + (2 - (-3))\mathbf{k} = 3\mathbf{i} - 5\mathbf{j} + 5\mathbf{k}$
3
Find the magnitude of (\overrightarrow{AB}):
4
$|\overrightarrow{AB}| = \sqrt{3^2 + (-5)^2 + 5^2} = \sqrt{9 + 25 + 25} = \sqrt{59}$
5
The exact distance between A and B is (\sqrt{59}) units.

3. Vector Equations of Straight Lines★★★☆☆⏱ 6 min

✓ Calculator OK

📘 Definition

Vector Line Equation

A straight line can be written in two standard forms: $r = a + t b$ (where (\mathbf{a}) is a point on the line, (\mathbf{b}) is the direction vector, and (t) is a scalar parameter) or $r = c + t (d - c)$ for a line passing through points with position vectors (\mathbf{c}) and (\mathbf{d}).

Direction vectors are not unique: any scalar multiple of a direction vector is also a valid direction vector for the same line. When writing line equations, you may use any point on the line as the fixed position vector.

📐 Worked Example

Write a vector equation for the line L that passes through points C(1, 3, -2) and D(4, -1, 1).

1
Select the position vector of point C as the fixed point on the line:
2
$\mathbf{c} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$
3
Calculate the direction vector (\overrightarrow{CD} = \mathbf{d} - \mathbf{c}):
4
$\overrightarrow{CD} = (4-1)\mathbf{i} + (-1-3)\mathbf{j} + (1 - (-2))\mathbf{k} = 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}$
5
Write the line equation with parameter (t):
6
$\mathbf{r} = (\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) + t(3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k})$

4. Relationships Between Two Lines★★★★☆⏱ 6 min

✓ Calculator OK

Two lines in 3D space can be:

Parallel: their direction vectors are scalar multiples of each other
Intersecting: there exists a single pair of parameters (t,s) that make the position vectors of the two lines equal
Skew: the lines are not parallel, and there is no solution for (t,s) that makes their position vectors equal (skew lines never meet)

📐 Worked Example

Determine if lines (L_1: \mathbf{r} = (2\mathbf{i} + \mathbf{j} + \mathbf{k}) + t(3\mathbf{i} - \mathbf{j} + 2\mathbf{k})) and (L_2: \mathbf{r} = (\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) + s(\mathbf{i} + 2\mathbf{j} + \mathbf{k})) are parallel, intersecting or skew.

1
Check if direction vectors are scalar multiples: direction vectors are (3,-1,2) and (1,2,1). (3/1 \neq -1/2), so lines are not parallel.
2
Set components equal to solve for (t) and (s):
3
$2 + 3t = 1 + s \quad \text{(x-component)}\newline 1 - t = 4 + 2s \quad \text{(y-component)}\newline 1 + 2t = 5 + s \quad \text{(z-component)}$
4
Rearrange first equation to (s = 1 + 3t), substitute into second equation:
5
$1 - t = 4 + 2(1 + 3t) \implies 1 - t = 6 + 6t \implies -7t = 5 \implies t = -\frac{5}{7}, s = -\frac{8}{7}$
6
Substitute values into third component to verify:
7
$\text{Left: } 1 + 2(-\frac{5}{7}) = -\frac{3}{7} \quad \text{Right: }5 + (-\frac{8}{7}) = \frac{27}{7}$
8
Values do not match, so no solution exists. Lines are skew.

5. Scalar Dot Product and Applications★★★★☆⏱ 7 min

✓ Calculator OK

📘 Definition

Scalar (Dot) Product

For vectors (\mathbf{a} = (a_1,a_2,a_3)) and (\mathbf{b} = (b_1,b_2,b_3)), the dot product is $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ . The cosine of the angle (\theta) between two vectors is given by $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ . Non-zero vectors are perpendicular if and only if their dot product equals 0.

📐 Worked Example

Calculate the acute angle between lines (L_1: \mathbf{r} = (2\mathbf{i} + \mathbf{j}) + t(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})) and (L_2: \mathbf{r} = (3\mathbf{i} + 2\mathbf{k}) + s(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})). Give your answer to 1 decimal place in degrees.

1
Extract direction vectors of both lines (do not use position vectors):
2
$\mathbf{d_1} = (1,2,-2), \quad \mathbf{d_2} = (2,-1,2)$
3
Calculate the dot product:
4
$\mathbf{d_1} \cdot \mathbf{d_2} = (1 \times 2) + (2 \times -1) + (-2 \times 2) = 2 - 2 - 4 = -4$
5
Calculate magnitudes of the direction vectors:
6
$|\mathbf{d_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3, \quad |\mathbf{d_2}| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$
7
Take the absolute value of the cosine ratio to find the acute angle:
8
$\cos\theta = \left| \frac{-4}{3 \times 3} \right| = \frac{4}{9}$
9
Calculate the angle using inverse cosine:
10
$\theta = \arccos\left(\frac{4}{9}\right) \approx 63.6^\circ$

6. Common Pitfalls

Wrong move:

Forgetting the dot product formula is not given in the formula booklet.

Why:

You will lose access to all marks for dot product questions if you cannot recall the formula.

Correct move:

Memorise (\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3) before your exam.

Wrong move:

Assuming non-parallel 3D lines must intersect.

Why:

3D lines can be skew, meaning they never meet and are not parallel.

Correct move:

Always solve the system of equations for parameters to confirm intersection, even if lines are not parallel.

Wrong move:

Using position vectors instead of direction vectors when calculating the angle between two lines.

Why:

The angle between lines depends only on their direction, not their position in space.

Correct move:

Extract the direction vectors of both lines before applying the dot product angle formula.

Wrong move:

Omitting the square root when calculating vector magnitude or distance between points.

Why:

Magnitude is the square root of the sum of squared components, not the sum itself.

Correct move:

Always apply the square root when asked for magnitude or distance, unless explicitly asked for the squared value.

Wrong move:

Using the same parameter (e.g. t) for both lines when solving for intersection.

Why:

The parameters for each line are independent, using the same variable leads to incorrect solutions.

Correct move:

Use separate parameters (e.g. t for line 1, s for line 2) when setting up intersection equations.

7. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Magnitude of vector (\mathbf{a} = (x,y,z))	(\|\mathbf{a}\| = \sqrt{x^2 + y^2 + z^2})	Unit vector: (\hat{\mathbf{a}} = \frac{\mathbf{a}}{\|\mathbf{a}\|})
Distance between points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2))	(d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2})	Derives from (\|\overrightarrow{AB}\| = \|\mathbf{b} - \mathbf{a}\|)
Vector line equations	(r = \mathbf{a} + t\mathbf{b}) or (r = \mathbf{c} + t(\mathbf{d} - \mathbf{c}))	(\mathbf{b}) = direction vector, (\mathbf{a}) = position of point on line
Line relationships	Parallel: direction vectors scalar multiples Intersect: solution exists for r₁=r₂ Skew: not parallel, no intersection	Skew lines only exist in 3D, not 2D
Dot product	(\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3)	Not given in formula booklet; memorise
Angle between two lines	(\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})	Use direction vectors, take absolute value for acute angle
Perpendicular vectors	(\mathbf{a}\cdot\mathbf{b} = 0)	Applies only to non-zero vectors

8. Frequently Asked

Do I need to learn the dot product formula for Edexcel IAL P4?

Yes, the dot product formula (a\cdot b = a_1b_1+a_2b_2+a_3b_3) is not given in the formula booklet, so you must memorise it. The cosine of the angle formula and distance formula are standard and expected to be applied correctly.

Are planes or vector cross products tested in P4 vectors?

No, planes and cross products are exclusively in Further Pure 3 (FP3) for Edexcel IAL. All P4 vector questions are limited to points, vectors and straight lines only.

Going deeper

formula_bookEdexcel IAL Maths Formula BookletOfficial formula booklet for P4 exams

What's Next

Now you have mastered the core vector content for Edexcel IAL P4, you can consolidate your learning by practicing full exam-style vector questions, which are typically worth 8-12 marks per paper and combine multiple concepts including line equations, skew line checks and dot product angle calculations. Vectors are often paired with differentiation or integration contexts in later P4 questions, so ensure you can apply vector operations alongside other pure maths skills. Once you are confident with basic vector problems, move on to harder mixed-topic questions and past paper practice to build exam technique, paying close attention to presentation of your working for method marks. Remember that calculators are allowed for all P4 papers, so use your calculator to verify magnitude and dot product calculations to avoid arithmetic errors.

→P4 Differentiation
→P4 Integration