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Additional Mathematics

Vectors

PDF
Matthew Williams
|May 16, 2026|5 min read
Dot ProductKinematicsPaper 01Paper 02Scalar ProductSection 2Vectors

Vector notation, addition, subtraction and scalar multiplication, magnitude, unit vectors, displacement vectors, the scalar product, angle between vectors, and parallel and perpendicular vectors.

A vector has both magnitude and direction. A scalar has magnitude only. Velocity, displacement, and force are vectors; speed, mass, and temperature are scalars.

Notation

Vectors are written in bold (a) in typeset text, or underlined (a‾\underline{a}a​) in handwriting. In two dimensions, a vector is expressed as a column vector

v=(xy)\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}v=(xy​)

or in terms of the standard unit vectors i=(10)\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}i=(10​) and j=(01)\mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}j=(01​) as v=xi+yj\mathbf{v} = x\mathbf{i} + y\mathbf{j}v=xi+yj.

A position vector OP→\overrightarrow{OP}OP describes the location of point PPP relative to the origin OOO.

Operations on Vectors

Addition and subtraction: add or subtract corresponding components.

(ab)+(cd)=(a+cb+d)\begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a+c\\b+d\end{pmatrix}(ab​)+(cd​)=(a+cb+d​)

Scalar multiplication: multiply each component by the scalar kkk. This scales the magnitude by ∣k∣|k|∣k∣ and reverses direction if k<0k < 0k<0.

k(ab)=(kakb)k\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ka\\kb\end{pmatrix}k(ab​)=(kakb​)

Equal vectors have the same magnitude and direction; position does not matter.

Magnitude

The magnitude (length) of v=(xy)\mathbf{v} = \begin{pmatrix}x\\y\end{pmatrix}v=(xy​) is:

∣v∣=x2+y2|\mathbf{v}| = \sqrt{x^2 + y^2}∣v∣=x2+y2​

This is Pythagoras applied to the horizontal and vertical components.

Example

∣(5−12)∣=25+144=169=13\left|\begin{pmatrix}5\\-12\end{pmatrix}\right| = \sqrt{25 + 144} = \sqrt{169} = 13​(5−12​)​=25+144​=169​=13

Direction of a Vector

The direction of a vector is the angle it makes with the positive xxx-axis, measured anticlockwise. For a vector with components (x,y)(x, y)(x,y):

θ=arctan⁡(yx)\theta = \arctan\left(\frac{y}{x}\right)θ=arctan(xy​)

arctan⁡\arctanarctan returns values between −90∘-90^\circ−90∘ and 90∘90^\circ90∘, so the quadrant must be checked. For a vector in Quadrant II or III, add 180∘180^\circ180∘ to the result.

Example

Find the direction of v=−3i+3j\mathbf{v} = -3\mathbf{i} + 3\mathbf{j}v=−3i+3j.

arctan⁡(3−3)=arctan⁡(−1)=−45∘\arctan\left(\dfrac{3}{-3}\right) = \arctan(-1) = -45^\circarctan(−33​)=arctan(−1)=−45∘.

Since x<0x < 0x<0 and y>0y > 0y>0 the vector is in Quadrant II, so direction =−45∘+180∘=135∘= -45^\circ + 180^\circ = 135^\circ=−45∘+180∘=135∘.

Unit Vectors

A unit vector has magnitude 1. To find the unit vector in the direction of a\mathbf{a}a:

a^=a∣a∣\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}a^=∣a∣a​

Example

Find the unit vector in the direction of a=(34)\mathbf{a} = \begin{pmatrix}3\\4\end{pmatrix}a=(34​).

∣a∣=9+16=5|\mathbf{a}| = \sqrt{9 + 16} = 5∣a∣=9+16​=5

a^=15(34)=(3/54/5)\hat{\mathbf{a}} = \frac{1}{5}\begin{pmatrix}3\\4\end{pmatrix} = \begin{pmatrix}3/5\\4/5\end{pmatrix}a^=51​(34​)=(3/54/5​)

Displacement Vectors

The vector from point AAA to point BBB is:

AB→=OB→−OA→=b−a\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}AB=OB−OA=b−a

Equivalently, AB→=B−A\overrightarrow{AB} = B - AAB=B−A (coordinates of BBB minus coordinates of AAA). The order matters: AB→=−BA→\overrightarrow{AB} = -\overrightarrow{BA}AB=−BA.

Example

If A=(2,−1)A = (2, -1)A=(2,−1) and B=(7,4)B = (7, 4)B=(7,4):

AB→=(7−24−(−1))=(55)\overrightarrow{AB} = \begin{pmatrix}7-2\\4-(-1)\end{pmatrix} = \begin{pmatrix}5\\5\end{pmatrix}AB=(7−24−(−1)​)=(55​)

Parallel and Collinear Vectors

Two vectors are parallel if one is a scalar multiple of the other: b=ka\mathbf{b} = k\mathbf{a}b=ka for some scalar kkk.

Three points AAA, BBB, CCC are collinear if AB→\overrightarrow{AB}AB is parallel to AC→\overrightarrow{AC}AC (and both share point AAA).

Example

Show that A=(1,2)A = (1, 2)A=(1,2), B=(3,6)B = (3, 6)B=(3,6), C=(5,10)C = (5, 10)C=(5,10) are collinear.

AB→=(24)\overrightarrow{AB} = \begin{pmatrix}2\\4\end{pmatrix}AB=(24​) and AC→=(48)=2(24)\overrightarrow{AC} = \begin{pmatrix}4\\8\end{pmatrix} = 2\begin{pmatrix}2\\4\end{pmatrix}AC=(48​)=2(24​)

Since AC→=2 AB→\overrightarrow{AC} = 2\,\overrightarrow{AB}AC=2AB and they share point AAA, the three points are collinear.

The Scalar (Dot) Product

The scalar product of a=(a1a2)\mathbf{a} = \begin{pmatrix}a_1\\a_2\end{pmatrix}a=(a1​a2​​) and b=(b1b2)\mathbf{b} = \begin{pmatrix}b_1\\b_2\end{pmatrix}b=(b1​b2​​) is defined two ways:

Component form:

a⋅b=a1b1+a2b2\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2a⋅b=a1​b1​+a2​b2​

Geometric form:

a⋅b=∣a∣∣b∣cos⁡θ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\thetaa⋅b=∣a∣∣b∣cosθ

where θ\thetaθ is the angle between the vectors (0≤θ≤π0 \leq \theta \leq \pi0≤θ≤π).

The result is a scalar (a number), not a vector.

Finding the Angle Between Two Vectors

Combining both definitions:

cos⁡θ=a⋅b∣a∣∣b∣\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}cosθ=∣a∣∣b∣a⋅b​

Example

Find the angle between a=(12)\mathbf{a} = \begin{pmatrix}1\\2\end{pmatrix}a=(12​) and b=(3−1)\mathbf{b} = \begin{pmatrix}3\\-1\end{pmatrix}b=(3−1​).

a⋅b=(1)(3)+(2)(−1)=1\mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-1) = 1a⋅b=(1)(3)+(2)(−1)=1

∣a∣=5|\mathbf{a}| = \sqrt{5}∣a∣=5​, ∣b∣=10|\mathbf{b}| = \sqrt{10}∣b∣=10​

cos⁡θ=15⋅10=150=152⇒θ=arccos⁡ ⁣(152)≈81.9°\cos\theta = \frac{1}{\sqrt{5}\cdot\sqrt{10}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}} \Rightarrow \theta = \arccos\!\left(\tfrac{1}{5\sqrt{2}}\right) \approx 81.9°cosθ=5​⋅10​1​=50​1​=52​1​⇒θ=arccos(52​1​)≈81.9°

Perpendicular Vectors

Two non-zero vectors are perpendicular if and only if their dot product is zero:

a⋅b=0  ⟺  a⊥b\mathbf{a} \cdot \mathbf{b} = 0 \iff \mathbf{a} \perp \mathbf{b}a⋅b=0⟺a⊥b

This follows directly from the geometric form: cos⁡90°=0\cos 90° = 0cos90°=0.

Example

Are p=(4−3)\mathbf{p} = \begin{pmatrix}4\\-3\end{pmatrix}p=(4−3​) and q=(34)\mathbf{q} = \begin{pmatrix}3\\4\end{pmatrix}q=(34​) perpendicular?

p⋅q=(4)(3)+(−3)(4)=12−12=0\mathbf{p} \cdot \mathbf{q} = (4)(3) + (-3)(4) = 12 - 12 = 0p⋅q=(4)(3)+(−3)(4)=12−12=0. Yes, they are perpendicular.

Exam Tip

Paper 02 often asks you to find an unknown component given that two vectors are perpendicular. Set the dot product equal to zero and solve for the unknown.

Properties of the Scalar Product

  • Commutative: a⋅b=b⋅a\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}a⋅b=b⋅a
  • a⋅a=∣a∣2\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2a⋅a=∣a∣2 (useful for finding ∣a∣|\mathbf{a}|∣a∣ without the square root formula when the vector is expressed algebraically)
  • Distributive: a⋅(b+c)=a⋅b+a⋅c\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}a⋅(b+c)=a⋅b+a⋅c (used when expanding expressions involving sums of vectors)

Scalars vs Vectors: Common Examples

ScalarVector
DistanceDisplacement
SpeedVelocity
MassForce
TemperatureAcceleration
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