IAL P4 vectors are entirely pattern-based. Once you know the six patterns Edexcel keeps testing, you can identify each question in ten seconds and go straight for the formula.
The six patterns
1. Equation of a line through two points
Given $A$ and $B$, the line is $\mathbf{r} = \mathbf{a} + t(\mathbf{b} - \mathbf{a})$.
Trap: writing the direction vector as $(\mathbf{a} + \mathbf{b})$ (that's the midpoint direction, meaningless here) or $(\mathbf{b} + \mathbf{a})$ order. It's always $(\mathbf{b} - \mathbf{a})$.
2. Angle between two lines
Direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$:
$$\cos\theta = \dfrac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$$
The modulus in the numerator ensures the answer is the acute angle. Miss the modulus and you get the obtuse — and lose the mark.
3. Point of intersection of two lines
Set the two vector equations equal, giving three scalar equations in two parameters ($s$ and $t$). Solve any TWO for $s$ and $t$; verify with the third. If the third fails, the lines are skew.
Trap: not verifying with the third equation. Examiners award a mark specifically for "checks that lines intersect".
4. Shortest distance from a point to a line
Point $P$, line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$. The shortest distance is:
$$d = \dfrac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{d}|}{|\mathbf{d}|}$$
Cross product on top, magnitude of direction on bottom.
5. Equation of a plane through three points
Three points $A, B, C$. The normal to the plane is $\mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})$. Then the plane is $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.
6. Angle between a line and a plane
Direction vector $\mathbf{d}$, plane normal $\mathbf{n}$. The angle $\alpha$ between the line and the plane is the COMPLEMENT of the angle between $\mathbf{d}$ and $\mathbf{n}$:
$$\sin\alpha = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$$
Note the SINE, not cosine. If you use cosine you get the angle to the normal, which is $90° - \alpha$.
The unified worked example
Given $A(1, 2, -1)$, $B(3, 0, 5)$, and the plane $\Pi$: $2x - y + z = 4$.
(a) Line through $A$ and $B$: $\mathbf{r} = (1, 2, -1) + t(2, -2, 6)$.
(b) Angle line makes with plane: $\mathbf{d} = (2, -2, 6)$, $\mathbf{n} = (2, -1, 1)$.
$\mathbf{d} \cdot \mathbf{n} = 4 + 2 + 6 = 12$; $|\mathbf{d}| = \sqrt{44}$; $|\mathbf{n}| = \sqrt{6}$.
$\sin\alpha = \dfrac{12}{\sqrt{44}\sqrt{6}} = \dfrac{12}{\sqrt{264}} \approx 0.738$; $\alpha \approx 47.6°$.
(c) Point of intersection: substitute the line's parametric form into the plane equation.
$2(1+2t) - (2-2t) + (-1+6t) = 4$
$2 + 4t - 2 + 2t - 1 + 6t = 4$
$12t = 5 \Rightarrow t = \dfrac{5}{12}$.
Point: $\left(\dfrac{22}{12}, \dfrac{14}{12}, \dfrac{18}{12}\right) = \left(\dfrac{11}{6}, \dfrac{7}{6}, \dfrac{3}{2}\right)$.
Your drill
Print the six formulae. For every past-paper vectors question you attempt, write which pattern it is BEFORE you start working. If you can name the pattern in under 10 seconds, you'll finish the question in under 5 minutes.
The whole P4 vectors section — often 12-14 marks — becomes a question of naming, not thinking.
Ready to put this into practice?
Real Edexcel-style questions, dark-themed engine, method marks tracked as you go.
Open the formula guide →