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The magnitude of a vector is its length, found by Pythagoras, and its direction can be given by an angle with an axis or by a unit vector. You convert between component form and magnitude–direction form, which is used heavily in Mechanics for forces and velocities. Questions are short, typically 2 to 4 marks.
The magnitude is the hypotenuse, \(\sqrt{x^2 + y^2}\), and \(\tan\theta = \dfrac{y}{x}\)
A magnitude is never negative. In Mechanics, the magnitude of a velocity is the speed.
Unit vectors
A unit vector has magnitude 1. The unit vector in the direction of \(\mathbf{a}\) is \(\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}\)
A vector of magnitude \(m\) in the direction of \(\mathbf{a}\) is \(m\hat{\mathbf{a}}\); in the opposite direction it is \(-m\hat{\mathbf{a}}\)
Direction in 2D
diagram
The angle \(\theta\) between \(x\mathbf{i} + y\mathbf{j}\) and the positive \(x\)-axis satisfies \(\tan\theta = \dfrac{y}{x}\). Sketch the vector first: the calculator's \(\tan^{-1}\) only gives angles between \(-90^\circ\) and \(90^\circ\)
Sketch first: \(3\mathbf{i} - 4\mathbf{j}\) points into the fourth quadrant, \(53.1^\circ\) below the positive x-axis (\(\tan\alpha = \dfrac{4}{3}\)).
From magnitude \(r\) and angle \(\theta\) measured anticlockwise from \(\mathbf{i}\): \(r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}\)
Bearings (Mechanics) are measured clockwise from north, usually the \(\mathbf{j}\) direction.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, ft = follow through: you still earn the mark if you use your own earlier wrong answer correctly.
Find a magnitude and a unit vector
3 marksGrade C
Square each component, add, and take the square root.
Divide every component by the magnitude.
Example.\(\mathbf{a} = 4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\). Find \(|\mathbf{a}|\) and the unit vector in the direction of \(\mathbf{a}\)
Convert between components and magnitude–direction form
4 marksGrade B
Components: multiply the magnitude by \(\cos\) and \(\sin\) of the angle from \(\mathbf{i}\)
Direction: sketch the vector, find an acute angle with \(\tan^{-1}\), then describe it relative to the axis.
Example. (a) A force of magnitude 10 N acts at an angle of \(150^\circ\) anticlockwise from \(\mathbf{i}\). Write the force in the form \((a\mathbf{i} + b\mathbf{j})\) N, giving exact values. (b) Find the angle between the vector \(3\mathbf{i} - 4\mathbf{j}\) and the positive \(x\)-axis.
(b) The vector points into the fourth quadrant; \(\tan\alpha = \dfrac{4}{3}\)
M1
\(\alpha = 53.1^\circ\) below the positive \(x\)-axis (clockwise), equivalent to \(306.9^\circ\) anticlockwise
A1
Find an unknown component from a magnitude
3 marksGrade A
Write the sum of the squared components equal to the magnitude squared.
Solve, keeping both square roots unless the question rules one out.
Example. The vector \(\mathbf{v} = p\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\), where \(p\) is a constant, has magnitude 6. Find the possible values of \(p\)
Show the model answerHide the model answer
\(p^2 + 2^2 + (-4)^2 = 6^2\)
M1
\(p^2 = 16\)
A1
\(p = 4\) or \(p = -4\)
A1
Don’t lose marks
Writing \((-4)^2\) as \(-16\)
Giving the calculator's angle without checking which quadrant the vector is in.
Dividing only some of the components by the magnitude when finding a unit vector.
Giving only the positive value when an unknown is squared.
More tips
Memory tricks
Check a unit vector: the squares of its components add up to 1.
Always sketch a 2D vector before finding its angle.
Exam technique
State clearly which angle you have found, e.g. 'below the positive \(x\)-axis' or 'with the positive \(y\)-axis'.
Leave magnitudes as exact surds (e.g. \(\sqrt{29}\)) unless a decimal is asked for.
In Mechanics, the magnitude of a velocity is a speed and needs units, e.g. m s-1.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Find the magnitude of a 2D vector\[\begin{aligned}|3\mathbf{i} - 4\mathbf{j}| &= \sqrt{3^2 + (-4)^2} \\ &= 5\end{aligned}\]
Grade C
Find a unit vector in a given directionDivide the vector by its magnitude: \(\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}\)
Grade C
Convert magnitude and angle to componentsA vector of magnitude \(r\) at angle \(\theta\) anticlockwise from \(\mathbf{i}\) is \(r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}\)
Grade B
Find the direction of a 2D vectorUse \(\tan\) with a sketch to find the angle with the positive \(x\)-axis in the correct quadrant.
Grade A
Find unknowns from a given magnitudeSet the sum of the squared components equal to the magnitude squared and solve, keeping both roots unless one is ruled out.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The vector \(\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in the directions of the positive \(x\)- and \(y\)-axes.
(a) Find \(|\mathbf{v}|\).[2]
(b) Find the angle between \(\mathbf{v}\) and the positive \(x\)-axis. Give your answer in degrees to one decimal place.[2]
Show the answer and mark scheme
(a)Answer: \(|\mathbf{v}| = 5\)
M1 for \(\sqrt{3^2 + (-4)^2}\)
A1 for 5
Worked solution: \(|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = 5\)
(b)Answer: \(53.1^\circ\) °
M1 for \(\tan\theta = \frac{4}{3}\) or \(\cos\theta = \frac{3}{5}\) oe
A1 for \(53.1^\circ\)
Worked solution: \(\tan\theta = \frac{4}{3} \Rightarrow \theta = 53.1^\circ\) (the vector points below the positive \(x\)-axis)
Question 2Medium5 marks
In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in the directions of the positive \(x\)- and \(y\)-axes respectively.
(a) The vector \(\mathbf{b}\) has magnitude 12 and makes an angle of \(300^\circ\) with the positive \(x\)-axis, measured anticlockwise. Find \(\mathbf{b}\) in the form \(p\mathbf{i} + q\mathbf{j}\), giving the exact values of \(p\) and \(q\).[2]
(b) The vector \(\mathbf{a} = -4\mathbf{i} + 3\mathbf{j}\). Find the angle that \(\mathbf{a}\) makes with the positive \(x\)-axis, measured anticlockwise from the positive \(x\)-axis. Give your answer in degrees to one decimal place.[3]
M1 for \(\tan^{-1}\frac{3}{4}\) (\(36.9^\circ\)) or \(\tan^{-1}\frac{4}{3}\)
M1 for using the quadrant correctly, e.g. \(180^\circ - 36.9^\circ\)
A1 for \(143.1^\circ\)
Worked solution: \(\mathbf{a}\) lies in the second quadrant. \(\tan^{-1}\frac{3}{4} = 36.9^\circ\) Angle \(= 180^\circ - 36.9^\circ = 143.1^\circ\)
Question 3Hard8 marks
Relative to a fixed origin \(O\), the points \(A\) and \(B\) are such that\[\overrightarrow{OA} = 4\mathbf{i} + \mathbf{j}, \qquad \overrightarrow{OB} = -5\mathbf{i} + \mathbf{j}\]
(a) Find the size of angle \(AOB\). Give your answer in degrees to one decimal place.[4]
(b) Find the area of triangle \(OAB\).[2]
(c) The point \(C\) is such that \(OACB\) is a parallelogram. Write down the position vector of \(C\) and the area of the parallelogram.[2]
Show the answer and mark scheme
(a)Answer: \(154.7^\circ\) °
M1 for finding the lengths \(OA = \sqrt{17}\), \(OB = \sqrt{26}\) and \(AB = 9\)
M1 for the cosine rule: \(\cos AOB = \frac{OA^2 + OB^2 - AB^2}{2 \times OA \times OB}\)
A1 for \(\cos AOB = \frac{-38}{2\sqrt{17}\sqrt{26}}\) oe
A1 for \(154.7^\circ\)
Worked solution: \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = -9\mathbf{i}\) \(OA = \sqrt{17}, \quad OB = \sqrt{26}, \quad AB = \sqrt{81}\) \(\cos AOB = \frac{17 + 26 - 81}{2\sqrt{17}\sqrt{26}} = -0.9037\ldots\) so angle \(AOB = 154.7^\circ\)
(b)Answer: \(4.5\) (square units)
M1 for \(\tfrac{1}{2} \times \sqrt{17} \times \sqrt{26} \times \sin AOB\) with their angle
Worked solution: \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OA} + \overrightarrow{OB} = -\mathbf{i} + 2\mathbf{j}\) The diagonal \(AB\) splits the parallelogram into two triangles congruent to \(OAB\), so the area is \(2 \times 4.5 = 9\)