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Or take a quick quiz: 11 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Vectors let you solve problems about shapes and about real situations such as forces and motion. In Pure you prove properties of quadrilaterals and other shapes; in context you find resultant forces, use equilibrium and describe the positions of moving objects with \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\). These are multi-step questions worth 5 to 10 marks, closely linked to Paper 2 Mechanics.
Vector from \(A\) to \(B\): \(\mathbf{r}_B - \mathbf{r}_A\)
Parallelogram: \(\overrightarrow{AB} = \overrightarrow{DC}\); rhombus: also \(|\overrightarrow{AB}| = |\overrightarrow{AD}|\)
\(\mathbf{F} = m\mathbf{a}\)
Notes
Forces
diagram
The resultant of several forces is their vector sum. Find its magnitude by Pythagoras and its direction by trigonometry.
A particle is in equilibrium when the resultant force is zero: \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\)
In equilibrium the forces add to \(\mathbf{0}\): drawn nose to tail they close up. Here \(\mathbf{F}_3 = 2\mathbf{i} + 2\mathbf{j}\)
Newton's second law works with vectors too: \(\mathbf{F} = m\mathbf{a}\) (Paper 2).
Motion with constant velocity
diagram
Velocity \(\mathbf{v}\) is a vector; speed is its magnitude, \(|\mathbf{v}|\)
Position at time \(t\):\(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\). Write it as one vector with \(\mathbf{i}\) and \(\mathbf{j}\) components in terms of \(t\)
The vector from ship \(A\) to ship \(B\) is \(\mathbf{r}_B - \mathbf{r}_A\); the distance between them is its magnitude.
With \(\mathbf{i}\) east and \(\mathbf{j}\) north: '\(B\) is due north of \(A\)' means equal \(\mathbf{i}\)-components, with \(B\) having the larger \(\mathbf{j}\)-component; '\(B\) is north-east of \(A\)' means \(\overrightarrow{AB}\) has equal, positive components.
At \(t = 2\) both ships have \(\mathbf{i}\)-component 1, so \(Q\) is due north of \(P\), 4 km away.
Proving properties of shapes
Parallelogram: one pair of opposite sides are equal vectors (e.g. \(\overrightarrow{AB} = \overrightarrow{DC}\)).
Rhombus: a parallelogram with two adjacent sides of equal length.
Trapezium: one pair of opposite sides parallel (one a scalar multiple of the other) but of different lengths.
Kite: two pairs of adjacent sides of equal length.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, A1* = the solution must be fully correct, with every step shown.
Use equilibrium to find a force
4 marksGrade C
Write the sum of all the forces equal to \(\mathbf{0}\)
Add the known forces and make the unknown force the subject.
Find its magnitude with Pythagoras, giving units.
Example. A particle is in equilibrium under three forces, \(\mathbf{F}_1 = (3\mathbf{i} - 4\mathbf{j})\) N, \(\mathbf{F}_2 = (-5\mathbf{i} + 2\mathbf{j})\) N and \(\mathbf{F}_3\). Find \(\mathbf{F}_3\) and its magnitude.
Write each position vector in terms of \(t\) using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\)
Translate the condition (e.g. 'due north') into an equation between components.
Solve for \(t\), then answer in context (a clock time, a distance).
Example. [\(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north, and distances are in km.] At noon, ship \(P\) is at \((-3\mathbf{i} + 5\mathbf{j})\) km and moves with constant velocity \((2\mathbf{i} + \mathbf{j})\) km h-1. At the same time, ship \(Q\) is at \((7\mathbf{i} + 3\mathbf{j})\) km and moves with constant velocity \((-3\mathbf{i} + 4\mathbf{j})\) km h-1. (a) Find the speed of \(Q\). (b) Find the time at which \(Q\) is due north of \(P\), and the distance between the ships at that time.
Due north: \(-3 + 2t = 7 - 3t\) M1, so \(t = 2\), i.e. at 2 pm
A1
At \(t = 2\): \(\mathbf{r}_P = \mathbf{i} + 7\mathbf{j}\) and \(\mathbf{r}_Q = \mathbf{i} + 11\mathbf{j}\), so the ships are 4 km apart
A1
Prove that a quadrilateral is a rhombus
4 marksGrade B
Show that one pair of opposite sides are equal vectors, so the shape is a parallelogram.
Show that two adjacent sides have equal lengths.
Conclude with the property that makes it a rhombus.
Example. The points \(A(1, 1, 2)\),\(B(4, 5, 2)\), \(C(4, 8, 6)\) and \(D(1, 4, 6)\) are the vertices of the quadrilateral \(ABCD\). Show that \(ABCD\) is a rhombus.
A parallelogram with two adjacent sides equal is a rhombus
A1*
Don’t lose marks
Adding the starting position and the velocity without multiplying the velocity by \(t\)
Giving a velocity when the question asks for a speed (a magnitude), or leaving out units.
Proving a parallelogram but stopping before checking adjacent sides for a rhombus.
Treating \(t\) as a clock time: \(t = 2\) hours after noon is 2 pm.
More tips
Memory tricks
Equilibrium: add the known forces; the missing force is the negative of that sum.
Write positions as a single vector in \(t\): \((x_0 + v_xt)\mathbf{i} + (y_0 + v_yt)\mathbf{j}\)
Compass words to components: due north or south means equal \(\mathbf{i}\)-components; due east or west means equal \(\mathbf{j}\)-components.
3-4-5 and 5-12-13 triangles appear often in speeds and forces.
Exam technique
Read the bracketed information at the start of the question: it gives the directions of \(\mathbf{i}\) and \(\mathbf{j}\) and the units.
Finish with a conclusion in context, e.g. 'at 2 pm the ships are 4 km apart'.
In proofs, state which property you have shown (equal vectors, equal lengths) and what it means for the shape.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade D
Find a resultant forceAdd the force vectors: \(\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \cdots\)
Grade C
Use equilibrium: resultant force is zeroSet the sum of all the forces equal to \(\mathbf{0}\) and solve for the unknown force.
Grade C
Find position with constant velocity\(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\), where \(\mathbf{r}_0\) is the position when \(t = 0\)
Grade B
Find speed and direction from velocitySpeed is \(|\mathbf{v}|\); the direction is an angle or bearing found with trigonometry.
Grade B
Prove the type of a quadrilateralCompare the sides as vectors (parallel, equal) and as lengths.
Grade A
Solve for unknown times or constantsFor example, 'due north' means the \(\mathbf{i}\)-components of the two position vectors are equal.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Two forces \(\mathbf{F}_1 = (-3\mathbf{i} + 4\mathbf{j})\text{ N}\) and \(\mathbf{F}_2 = (-8\mathbf{i} - 7\mathbf{j})\text{ N}\) act on a particle \(P\).
(a) Find the resultant force \(\mathbf{R}\) acting on \(P\).[1]
(b) Find the magnitude of \(\mathbf{R}\), giving your answer to 3 significant figures.[2]
(c) Find the angle between the direction of \(\mathbf{R}\) and the vector \(\mathbf{i}\). Give your answer in degrees to one decimal place.[2]
Worked solution: \(|\mathbf{R}| = \sqrt{(-11)^2 + (-3)^2} = \sqrt{130} = 11.4\) N (3 s.f.)
(c)Answer: \(164.7^\circ\) °
M1 for \(\tan\theta = \frac{3}{11}\) or \(\cos\theta = \frac{-11}{|\mathbf{R}|}\) oe
A1 for \(164.7^\circ\)
Worked solution: \(\tan^{-1}\frac{3}{11} = 15.3^\circ\); \(\mathbf{R}\) has a negative \(\mathbf{i}\) component, so the angle with \(\mathbf{i}\) is \(180^\circ - 15.3^\circ = 164.7^\circ\)
Question 2Medium8 marks
[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) act on a particle \(P\), where \(\mathbf{F}_1 = (5\mathbf{i} - 3\mathbf{j})\text{ N}\) and \(\mathbf{F}_2 = (p\mathbf{i} + q\mathbf{j})\text{ N}\), and \(p\) and \(q\) are constants. The resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is \((9\mathbf{i} - \mathbf{j})\text{ N}\). The particle is in equilibrium.
(a) Find the values of \(p\) and \(q\).[2]
(b) Find \(\mathbf{F}_3\).[2]
(c) Find the magnitude of \(\mathbf{F}_3\), giving your answer to 3 significant figures.[2]
(d) Find the angle between the direction of \(\mathbf{F}_3\) and the vector \(\mathbf{j}\). Give your answer in degrees to one decimal place.[2]
M1 for \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\)
A1 for \((-9\mathbf{i} + \mathbf{j})\text{ N}\)
Worked solution: In equilibrium the resultant force is zero: \(\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(9\mathbf{i} - \mathbf{j}) = (-9\mathbf{i} + \mathbf{j})\text{ N}\)
(c)Answer: \(\sqrt{82} = 9.06\) N
M1 for \(\sqrt{(-9)^2 + 1^2}\)
A1 for awrt 9.06 (exact value \(\sqrt{82}\)) (N)
Worked solution: \(|\mathbf{F}_3| = \sqrt{(-9)^2 + 1^2} = \sqrt{82} = 9.06\) N (3 s.f.)
(d)Answer: \(83.7^\circ\) °
M1 for \(\cos\theta = \frac{1}{|\mathbf{F}_3|}\) or \(\tan^{-1}9\) with correct use of the quadrant
A1 for \(83.7^\circ\)
Worked solution: \(\cos\theta = \frac{1}{\sqrt{82}} \Rightarrow \theta = 83.7^\circ\)
Question 3Hard9 marks
[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] A particle \(P\) of mass 5 kg is at rest on a smooth horizontal surface. Two horizontal forces \(\mathbf{F}_1 = (-\mathbf{i} - 6\mathbf{j})\text{ N}\) and \(\mathbf{F}_2 = (2p\mathbf{i} + (3 - 3p)\mathbf{j})\text{ N}\), where \(p\) is a constant, are then applied to \(P\). The resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) acts in the direction of \(\mathbf{i} - 2\mathbf{j}\).
(a) Find the value of \(p\).[3]
(b) Find the speed of \(P\) 5 seconds after the forces are applied, giving your answer to 3 significant figures.[4]
(c) Find the distance travelled by \(P\) in these 5 seconds, giving your answer to 3 significant figures.[2]
M1 for using the direction: \(-2(2p - 1) = -3p - 3\) (or \(\mathbf{R} = \lambda(\mathbf{i} - 2\mathbf{j})\))
A1 for \(p = 5\)
Worked solution: \(\mathbf{R} = (2p - 1)\mathbf{i} - (3p + 3)\mathbf{j}\) Since \(\mathbf{R}\) is parallel to \(\mathbf{i} - 2\mathbf{j}\), its components are in the ratio \(1 : -2\): \(-2(2p - 1) = -3p - 3\), so \(p = 5\). Then \(\mathbf{R} = (9\mathbf{i} - 18\mathbf{j})\text{ N}\), which is \(9(\mathbf{i} - 2\mathbf{j})\), in the required direction.
(b)Answer: \(9\sqrt{5} = 20.1\) m s−1
M1 for using \(\mathbf{F} = m\mathbf{a}\): \(\mathbf{a} = \frac{1}{5}(9\mathbf{i} - 18\mathbf{j})\)
A1 for \(\mathbf{a} = (1.8\mathbf{i} - 3.6\mathbf{j})\text{ m s}^{-2}\)
M1 for \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with \(\mathbf{u} = \mathbf{0}\) and \(t = 5\), then finding the magnitude
A1 for awrt 20.1 (exact value \(9\sqrt{5}\)) (m s−1)
M1 for \(\mathbf{s} = \tfrac{1}{2}\mathbf{a}t^2\) (or \(s = \tfrac{1}{2}(u + v)t\) with speeds) and its magnitude
A1 for awrt 50.3 (m)
Worked solution: The motion is in a straight line from rest, so \(s = \tfrac{1}{2}(0 + v)t = \tfrac{1}{2} \times 9\sqrt{5} \times 5 = 50.3\) m (3 s.f.)