Vectors

Video lesson4 worked examples

Theory

Component Form & Operations

Vectors can be added or subtracted using directed line segments (nose-to-tail) or by algebraically adding/subtracting their x, y, and z components.

Magnitude

The magnitude (length) of a vector, denoted as |a|, is calculated using Pythagoras' theorem. For a 3D vector $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ , the magnitude is $\sqrt{x^2 + y^2 + z^2}$ .

Journeys (Pathways)

A vector between two points can be built from a “journey” along other vectors, added nose-to-tail. Travelling a vector backwards reverses its sign, so $\overrightarrow{BA} = -\overrightarrow{AB}$ .

The Golden Rule: to add or subtract in component form, work one row (component) at a time. For a journey between two points, take any route along known vectors — adding them nose-to-tail — and reverse the sign of any vector travelled backwards.

⚠️ Common Examiner Traps

Reverse direction: $\overrightarrow{BA}$ is the negative of $\overrightarrow{AB}$ — swapping the letters flips every sign.
Scalar multiplies every component: $2\mathbf{u}$ doubles all of the components, not just the first.
Magnitude of a surd: the magnitude is often not a whole number — simplify the surd rather than rounding, if an exact answer is asked for.
Negative components squared: in a magnitude, $(-3)^2 = 9$ is positive.

Worked examples

Example 1

Adding Vector Components

Given $u = \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix}$ and $v = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}$ , calculate the resultant vector 2u + v.

Step 1: Scalar multiply u by 2: $\begin{pmatrix} 8 \\ -4 \\ 14 \end{pmatrix}$ .

Step 2: Add the components to v: $\begin{pmatrix} 8 + (-1) \\ -4 + 5 \\ 14 + 3 \end{pmatrix}$ .

Answer: $\begin{pmatrix} 7 \\ 1 \\ 17 \end{pmatrix}$ .

Example 2

Calculating Magnitude

Find the magnitude of vector $p = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$ .

Step 1: Apply the magnitude formula: $|p| = \sqrt{2^2 + (-3)^2 + 6^2}$ .

Step 2: Square the numbers: $\sqrt{4 + 9 + 36} = \sqrt{49}$ .

Answer: $|p| = 7$ .

Example 3

Magnitude as a Surd

Find the magnitude of the vector $v = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$ , giving your answer as a surd in its simplest form.

Step 1: Apply the magnitude formula: $|v| = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}$ .

Step 2: Simplify the surd by taking out the largest square factor (9): $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ .

Answer: $|v| = 3\sqrt{5}$ .

Example 4

🔗 Bringing it together (journeys)

ABCD is a parallelogram with $\overrightarrow{AB} = \mathbf{u}$ and $\overrightarrow{AD} = \mathbf{v}$ . Express $\overrightarrow{BD}$ in terms of $\mathbf{u}$ and $\mathbf{v}$ .

Step 1: Plan a journey from B to D along known vectors. Go B → A → D, adding the two steps nose-to-tail: $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD}$ .

Step 2: The first step goes backwards along $\overrightarrow{AB}$ , so $\overrightarrow{BA} = -\mathbf{u}$ , and $\overrightarrow{AD} = \mathbf{v}$ :

\overrightarrow{BD} = -\mathbf{u} + \mathbf{v}

Answer: $\overrightarrow{BD} = \mathbf{v} - \mathbf{u}$ .

Ready to practise?

15 Vectors questions with answers and video solutions