4 exam-style questions plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Vectors can be added with the triangle law and multiplied by scalars, and both operations have geometric meanings. You use them to write routes around shapes in terms of given vectors, show that lines are parallel, prove that points are collinear and find the ratios in which lines divide each other. These are usually 4 to 7 mark proof-style questions.
Midpoint of \(AB\): \(\dfrac{1}{2}(\mathbf{a} + \mathbf{b})\)
\(AP : PB = m : n \Rightarrow \overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}\)
Collinear: parallel vectors with a common point
\(\mathbf{a}\), \(\mathbf{b}\) not parallel: \(\lambda\mathbf{a} + \mu\mathbf{b} = p\mathbf{a} + q\mathbf{b} \Rightarrow \lambda = p\) and \(\mu = q\)
Notes
Adding and subtracting
diagram
Triangle law: \(\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\). In a parallelogram, the diagonal from a corner is the sum of the two sides from that corner.
Reversing a vector changes its sign: \(\overrightarrow{BA} = -\overrightarrow{AB}\)
Any route from \(A\) to \(B\) gives the same vector. With \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
Going from \(A\) to \(B\) via \(O\): \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\)
Scalar multiples and parallel vectors
\(\lambda\mathbf{a}\) is parallel to \(\mathbf{a}\), with magnitude \(|\lambda||\mathbf{a}|\); for \(\lambda \lt 0\) it points the opposite way.
Two vectors are parallel exactly when one is a scalar multiple of the other. The scalar also gives the ratio of their lengths.
Ratios and midpoints
If \(M\) is the midpoint of \(AB\),\(\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB}\), so \(\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\)
If \(AP : PB = 2 : 1\), \(P\) is \(\dfrac{2}{3}\) of the way from \(A\) to \(B\): \(\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB}\)
Collinear points and intersections
\(A\), \(B\) and \(C\) are collinear if, for example, \(\overrightarrow{AC} = k\overrightarrow{AB}\): the vectors are parallel and share the point \(A\)
If \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero and not parallel, and \(\lambda\mathbf{a} + \mu\mathbf{b} = p\mathbf{a} + q\mathbf{b}\), then \(\lambda = p\) and \(\mu = q\)
To find where two lines meet, write the position vector of the meeting point in two ways (one unknown scalar for each line) and compare coefficients.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Express vectors in terms of a and b
4 marksGrade D
Mark the given vectors and ratios on the diagram.
Choose a route made of vectors you know, and add them.
Simplify by collecting the \(\mathbf{a}\) and \(\mathbf{b}\) terms.
Example. In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\), and \(N\) is the point on \(OB\) with \(ON : NB = 1 : 2\). Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\), (a) \(\overrightarrow{OM}\), (b) \(\overrightarrow{MN}\).
Find two vectors that join the points, e.g. \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)
Show that one is a scalar multiple of the other.
State that they are parallel and share a point, so the points are collinear.
Example. Relative to an origin \(O\), the points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(3\mathbf{a} - 2\mathbf{b}\) respectively, where \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel. Prove that \(A\), \(B\) and \(C\) are collinear, and find the ratio \(AB : AC\)
\(\overrightarrow{AC} = -2\overrightarrow{AB}\), so \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are parallel
A1
They share the point \(A\), so \(A\), \(B\) and \(C\) are collinear, and \(AB : AC = 1 : 2\) (with \(A\) between \(B\) and \(C\))
A1
Find where two lines intersect
6 marksGrade A
Write the position vector of the meeting point using a scalar \(\lambda\) along one line.
Write it again using a scalar \(\mu\) along the other line.
Since \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, equate the coefficients of each and solve for \(\lambda\) and \(\mu\)
Example. In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(OA\), and \(N\) is the point on \(AB\) with \(AN : NB = 2 : 1\). The lines \(ON\) and \(BM\) meet at \(X\). Find \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), and the ratio \(OX : XN\).
Reading \(AN : NB = 2 : 1\) as \(\overrightarrow{AN} = \dfrac{1}{2}\overrightarrow{AB}\) instead of \(\dfrac{2}{3}\overrightarrow{AB}\)
Concluding 'collinear' from parallel vectors alone, without mentioning the common point.
Comparing coefficients without stating that \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel.
More tips
Memory tricks
'End minus start': \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
Go round the shape: add each step, and change the sign when you travel against an arrow.
Mark ratios on the diagram as fractions of the whole line before writing any vectors.
Collinear = parallel + a shared point. Say both.
Exam technique
Finish a proof with a clear statement, e.g. 'so \(A\), \(B\) and \(C\) are collinear'.
Simplify each vector fully (collect the \(\mathbf{a}\) and \(\mathbf{b}\) terms) so scalar multiples are easy to spot.
Use a different letter for each unknown scalar (\(\lambda\),\(\mu\)) when writing two routes.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Add vectors using the triangle law\(\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\): follow the route from start to finish.
Grade D
Express a route in terms of given vectorsFor example \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\)
Grade C
Show that two vectors are parallelShow that one is a scalar multiple of the other, e.g. \(4\mathbf{a} - 6\mathbf{b} = 2(2\mathbf{a} - 3\mathbf{b})\)
Grade C
Use a ratio to locate a pointIf \(AP : PB = m : n\), then \(\overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}\)
Grade B
Prove that three points are collinearShow that two of the vectors joining them are parallel and share a point.
Grade A
Find where lines meet by comparing routesWrite the same vector in two ways with unknown scalars, and compare the coefficients of two non-parallel vectors.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The diagram shows a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(M\) is the midpoint of \(AB\) and the point \(P\) lies on \(OB\) such that \(OP : PB = 2 : 3\).
[object Object]
(a) Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).[1]
(b) Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).[2]
(c) Find \(\overrightarrow{AP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Simplify your answer.[2]
The diagram shows a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The points \(M\) and \(N\) are the midpoints of \(OA\) and \(OB\) respectively.
[object Object]
(a) Prove, using vectors, that \(MN\) is parallel to \(AB\) and that \(MN = \tfrac{1}{2}AB\).[4]
(b) Hence show that the area of triangle \(OMN\) is one quarter of the area of triangle \(OAB\).[2]
Show the answer and mark scheme
(a)Answer: \(\overrightarrow{MN} = \tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\overrightarrow{AB}\), so \(MN \parallel AB\) and \(MN = \tfrac{1}{2}AB\).
M1 for \(\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a}\) and \(\overrightarrow{ON} = \tfrac{1}{2}\mathbf{b}\)
A1 for \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{AB}\)
A1* for a complete conclusion: \(\overrightarrow{MN}\) is a scalar multiple of \(\overrightarrow{AB}\), so the lines are parallel, and the length of \(MN\) is half the length of \(AB\)
Worked solution: \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a})\) \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}\), so \(\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{AB}\). Hence \(MN\) is parallel to \(AB\) and \(MN = \tfrac{1}{2}AB\).
(b)Answer: Triangles \(OMN\) and \(OAB\) are similar with scale factor \(\tfrac{1}{2}\), so the ratio of areas is \(\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\).
M1 for identifying that the triangles are similar (corresponding sides in the ratio \(1 : 2\), or angle \(O\) common and \(MN \parallel AB\))
A1* for area scale factor \(\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\), or \(\tfrac{1}{2} \times \tfrac{1}{2}OA \times \tfrac{1}{2}OB \sin O = \tfrac{1}{4} \times \tfrac{1}{2}OA \times OB\sin O\)
Worked solution: \(OM = \tfrac{1}{2}OA\), \(ON = \tfrac{1}{2}OB\) and angle \(MON\) = angle \(AOB\). \(\text{Area } OMN = \tfrac{1}{2} \times \tfrac{1}{2}OA \times \tfrac{1}{2}OB \times \sin AOB = \tfrac{1}{4}\left(\tfrac{1}{2} \times OA \times OB \times \sin AOB\right) = \tfrac{1}{4}\,\text{Area } OAB\)
Question 3Hard7 marks
The diagram shows a parallelogram \(OABC\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OC} = \mathbf{c}\). The diagonals \(OB\) and \(AC\) intersect at the point \(X\).
[object Object]
(a) Given that \(\overrightarrow{OX} = \lambda\overrightarrow{OB}\) and \(\overrightarrow{AX} = \mu\overrightarrow{AC}\), find two expressions for \(\overrightarrow{OX}\) in terms of \(\lambda\), \(\mu\), \(\mathbf{a}\) and \(\mathbf{c}\).[3]
(b) Hence prove that the diagonals of the parallelogram bisect each other.[3]
(c) Explain why it is valid to compare the coefficients of \(\mathbf{a}\) and \(\mathbf{c}\) in part (b).[1]
(b)Answer: Comparing coefficients: \(\lambda = 1 - \mu\) and \(\lambda = \mu\), so \(\lambda = \mu = \tfrac{1}{2}\): \(X\) is the midpoint of both diagonals.
M1 for equating the coefficients of \(\mathbf{a}\) and of \(\mathbf{c}\): \(\lambda = 1 - \mu, \ \lambda = \mu\)
A1 for \(\lambda = \mu = \tfrac{1}{2}\)
A1* for the conclusion that \(X\) is the midpoint of \(OB\) and of \(AC\), so the diagonals bisect each other
Worked solution: \(\lambda\mathbf{a} + \lambda\mathbf{c} = (1 - \mu)\mathbf{a} + \mu\mathbf{c}\). Since \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel, compare coefficients: \(\lambda = 1 - \mu\) and \(\lambda = \mu\), giving \(\lambda = \mu = \tfrac{1}{2}\). So \(\overrightarrow{OX} = \tfrac{1}{2}\overrightarrow{OB}\) and \(\overrightarrow{AX} = \tfrac{1}{2}\overrightarrow{AC}\): \(X\) is the midpoint of each diagonal.
(c)Answer: \(\mathbf{a}\) and \(\mathbf{c}\) are non-zero and not parallel, so \(p\mathbf{a} + q\mathbf{c} = r\mathbf{a} + s\mathbf{c}\) implies \(p = r\) and \(q = s\).
B1 for \(\mathbf{a}\) and \(\mathbf{c}\) are non-zero and not parallel (so neither can be written as a multiple of the other)
Worked solution: If \(p\mathbf{a} + q\mathbf{c} = r\mathbf{a} + s\mathbf{c}\) with \(p \ne r\), then \(\mathbf{a} = \frac{s - q}{p - r}\mathbf{c}\), making \(\mathbf{a}\) parallel to \(\mathbf{c}\), which is false. So the coefficients must be equal.