Learn | Grasp Maths

Lesson overview

Geometry - vector methods

A vector is a quantity with both magnitude (size) and direction. Vectors can be represented in component form, for example $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ means 3 units right and 2 units up. Vectors can be added and subtracted by combining their components. Scalar multiplication (multiplying by a number) scales a vector. The magnitude (or length) of a vector $\begin{pmatrix} a \\ b \end{pmatrix}$ is $\sqrt{a^2 + b^2}$ . Position vectors describe the location of points relative to an origin. Using vector methods, we can solve geometric problems: proving points are collinear, finding midpoints, and determining whether lines are parallel.

Vector notation and components

A vector from point A to point B is $\overrightarrow{AB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ . What does this mean?

Component form $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$

5 units horizontally, 3 units vertically

This vector representsA displacement of 5 right, 3 up

From A to B

The top number is the horizontal component. The bottom number is the vertical component.

Adding vectors

Find $\mathbf{u} + \mathbf{v}$ where $\mathbf{u} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ .

Add corresponding components. Horizontal components together, vertical components together.

Subtracting vectors

Find $\mathbf{u} - \mathbf{v}$ where $\mathbf{u} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ .

Subtract corresponding components. Watch the signs carefully.

Scalar multiplication

Find $2\mathbf{u}$ where $\mathbf{u} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ .

Multiply each component by the scalar. This scales the vector without changing its direction.

Magnitude of a vector

Find the magnitude of $\mathbf{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .

Use the formula $|\mathbf{v}| = \sqrt{a^2 + b^2}$ for a vector $\begin{pmatrix} a \\ b \end{pmatrix}$ . This comes from Pythagoras' theorem.

Worked example: Using position vectors to prove collinearity

Points A, B, and C have position vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ , $\mathbf{b} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ , and $\mathbf{c} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$ . Prove that A, B, and C are collinear (lie on the same straight line).

To prove collinearity, we need to show that $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ .
Find $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 - 1 \\ 5 - 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ .
Find $\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = \begin{pmatrix} 5 - 1 \\ 8 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ .
Check if $\overrightarrow{AC} = k \times \overrightarrow{AB}$ for some scalar $k$ .
$\begin{pmatrix} 4 \\ 6 \end{pmatrix} = k \times \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ implies $k = 2$ (since $4 = 2 \times 2$ and $6 = 2 \times 3$ ).
Since $\overrightarrow{AC} = 2 \times \overrightarrow{AB}$ , the vectors are parallel and point in the same direction.
Because they share point A and are parallel, points A, B, and C are collinear.

Try it

Perform vector operations, find magnitudes, and solve geometric problems. Use component form and properties of scalar multiples. Choose the correct answer.

Question 1

Find $\mathbf{u} + \mathbf{v}$ where $\mathbf{u} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ .

💡 Add components: horizontal $4 + (-1) = 3$ , vertical $2 + 3 = 5$ .

Question 2

Find $3\mathbf{w}$ where $\mathbf{w} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ .

💡 Multiply each component by 3: $3 \times 2 = 6$ and $3 \times (-1) = -3$ .

Question 3

Find the magnitude of $\mathbf{v} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ .

💡 $|\mathbf{v}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ . (This is a 5-12-13 right triangle.)

Question 4

If $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$ , find $\mathbf{b} - \mathbf{a}$ .

💡 Subtract components: horizontal $4 - 1 = 3$ , vertical $5 - 2 = 3$ .

Question 5

Vectors $\mathbf{u} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ are parallel. Which statement is true?

💡 $\begin{pmatrix} 2 \\ 4 \end{pmatrix} = 2 \times \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . So $\mathbf{u}$ is twice $\mathbf{v}$ in the same direction.

Common mistakes

Watch for these when working through the lesson.

Treating vectors like scalars and not keeping track of components separately. Always work horizontally and vertically as separate parts.
Forgetting the direction when working with scalar multiples. $2\mathbf{v}$ and $-2\mathbf{v}$ point in opposite directions, even though they have the same magnitude.
Confusing vector subtraction with addition. Remember: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ , not $\mathbf{a} - \mathbf{b}$ . The order matters.

Vectors

Lesson overview

Worked example: Using position vectors to prove collinearity

Try it

Common mistakes

Related topics

Independent practice will plug in here