10 Vectors — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in 10, read straight through: 5 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section 10. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

10.1 · Use vectors in two dimensions and in three dimensions.

A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ .
Work component by component, keeping the $x$ , $y$ and $z$ entries aligned; subtract the initial point from the final point to form a displacement vector.
For example, from $A(1,2,3)$ to $B(4,0,5)$ the displacement is $\overrightarrow{AB}=(3,-2,2)$ .
A vector has no fixed location, whereas a point does; a common error is to confuse the coordinates of an endpoint with the components of the displacement leading to it.

Worked example

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $2\mathbf{a}-\mathbf{b}$ .

1.First $2\mathbf{a}=(4,-2,6)$ .
2.Subtract corresponding components of $\mathbf{b}$ to get $(4-(-1),-2-4,6-2)=(5,-6,4)$ .

Answer: $2\mathbf{a}-\mathbf{b}=(5,-6,4)$

Common mistakes

Don't change the coordinate order between vectors, so components from different axes are combined.
Don't combine vector components inconsistently, applying a scalar to only one coordinate.

Exam tip. Perform each vector operation component by component and preserve the coordinate order throughout.

10.2 · Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

For $\mathbf{v}=(a,b)$ , its magnitude is $|\mathbf{v}|=\sqrt{a^2+b^2}$ and a direction angle must be stated relative to a specified axis or bearing convention.
A vector of magnitude $r$ at angle $\theta$ anticlockwise from the positive $x$ -axis has components $(r\cos\theta,r\sin\theta)$ ; use signs or a quadrant-aware angle calculation when reversing the process.
For example, magnitude $10$ at $30^\circ$ above the positive $x$ -axis gives $(10\cos30^\circ,10\sin30^\circ)=(5\sqrt3,5)$ .
The value from $\tan^{-1}(b/a)$ alone can select the wrong quadrant; a common error is to report an acute reference angle without checking the component signs.
A unit vector in the direction of non-zero $\mathbf a$ is $\mathbf a/|\mathbf a|$ ; divide every component by the vector's magnitude.

Worked example

A vector has magnitude $14$ and direction $120^\circ$ anticlockwise from the positive $x$ -axis. Write it in exact component form.

1.Use $(14\cos120^\circ,14\sin120^\circ)$ .
2.Since $\cos120^\circ=-\frac12$ and $\sin120^\circ=\frac{\sqrt3}{2}$ , the vector is $(-7,7\sqrt3)$ .

Answer: $(-7,7\sqrt3)$

Common mistakes

Don't normalise a vector by dividing by the sum of its components rather than by its magnitude.
Don't use sine for the horizontal component and cosine for the vertical component despite the stated reference axis.

Exam tip. Draw the direction from the positive x-axis, then use cosine horizontally and sine vertically with correct signs.

10.3 · Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Vector addition combines successive displacements: placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$ makes the resultant from the first tail to the final head equal to $\mathbf{a}+\mathbf{b}$ .
Add corresponding components and multiply every component by a scalar; subtraction is addition of the opposite vector.
The vector $k\mathbf{a}$ is parallel to $\mathbf{a}$ , has magnitude $|k||\mathbf{a}|$ , and points in the reverse direction when $k<0$ .
A common error is to multiply only one component by a scalar or to draw vectors head-to-head instead of using a head-to-tail or parallelogram construction.

The head-to-tail construction places

\mathbf v

at the head of

2\mathbf u

; the direct arrow is their resultant.

Worked example

Let $\mathbf{u}=(4,1)$ and $\mathbf{v}=(-2,5)$ . Find $2\mathbf{u}+\mathbf{v}$ and describe a head-to-tail construction for this resultant.

1.Calculate $2\mathbf{u}=(8,2)$ , then add $\mathbf{v}$ to obtain $(8-2,2+5)=(6,7)$ .
2.Diagrammatically, translate the vectors without rotating them and place the second $\mathbf{u}$ after the first, followed by $\mathbf{v}$ ; the direct closing vector is the sum.

Answer: $2\mathbf{u}+\mathbf{v}=(6,7)$ .; Place two copies of $\mathbf{u}$ and then one copy of $\mathbf{v}$ head-to-tail; the resultant joins the initial tail to the final head.

Common mistakes

Don't reverse a displacement vector in the diagram but leave its algebraic sign unchanged.
Don't add vector diagrams tail-to-tail without completing the parallelogram or head-to-tail construction.

Exam tip. For a resultant, scale each vector first and join them head-to-tail in the order represented algebraically.

10.4 · Understand and use position vectors; calculate the distance between two points represented by position vectors.

The position vector of a point $A$ is $\overrightarrow{OA}$ from a fixed origin $O$ ; if these vectors are $\mathbf{a}$ and $\mathbf{b}$ , then $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .
Find a displacement by subtracting position vectors in final-minus-initial order, then find distance by taking the magnitude of that displacement.
A point dividing $AB$ internally in the fraction $t$ from $A$ to $B$ has position vector $\mathbf{a}+t(\mathbf{b}-\mathbf{a})$ .
Distance is a non-negative scalar, not a vector; a common error is to quote $\mathbf{b}-\mathbf{a}$ as the distance without calculating its magnitude.

Worked example

The position vectors of $A$ and $B$ are $(-1,2,4)$ and $(3,-4,7)$ . Calculate the exact distance $AB$ .

1. $\overrightarrow{AB}=(3-(-1),-4-2,7-4)=(4,-6,3)$ .
2.Hence $AB=|\overrightarrow{AB}|=\sqrt{4^2+(-6)^2+3^2}=\sqrt{61}$ .

Answer: $AB=\sqrt{61}$

Common mistakes

Don't square the coordinates of the two points separately instead of squaring their component differences.
Don't find the difference of position vectors but forget to take its Euclidean magnitude.

Exam tip. Subtract the endpoint position vectors in a consistent order, then square and sum every component for the distance.

10.5 · Use vectors to solve problems in pure mathematics and in context (including forces).

Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.
Choose and state positive coordinate directions, resolve every vector consistently, then equate components or use position-vector relationships.
In a parallelogram with adjacent position vectors $\mathbf{a}$ and $\mathbf{b}$ , the opposite vertex has position vector $\mathbf{a}+\mathbf{b}$ and both diagonals share midpoint $\frac12(\mathbf{a}+\mathbf{b})$ .
A common error is to balance force magnitudes without balancing directions; equal numerical magnitudes do not guarantee equilibrium unless the vector sum is zero.

For the parallelogram with position vectors

\mathbf 0,\mathbf a,\mathbf b,\mathbf a+\mathbf b

, both diagonals share midpoint

(\mathbf a+\mathbf b)/2

Worked example

The points $O,A,B,C$ have position vectors $\mathbf{0},\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b}$ respectively. Use vectors to prove that the diagonals $OC$ and $AB$ bisect each other.

1.The midpoint of $OC$ has position vector $\frac12[\mathbf{0}+(\mathbf{a}+\mathbf{b})]=\frac12(\mathbf{a}+\mathbf{b})$ .
2.The midpoint of $AB$ has position vector $\frac12(\mathbf{a}+\mathbf{b})$ as well.
3.Since the two diagonals have the same midpoint, each bisects the other.

Answer: Both diagonals have midpoint position vector $\frac12(\mathbf{a}+\mathbf{b})$ , so they bisect each other.

Common mistakes

Don't assume collinearity from proportional-looking coordinates without finding one common scalar multiplier.
Don't use a diagram alone as proof and never shows that the two candidate midpoints have the same position vector.

Exam tip. In a vector proof, calculate both relevant position vectors and use their equality to justify the geometric conclusion.