Mastering Vectors for Your Mathematics Exam

What the examiner is testing

The examiner is testing your ability to navigate geometric problems using vector notation, specifically focusing on expressing position vectors, calculating magnitudes, and proving collinearity or parallelism. Marks are primarily awarded for correct vector expressions and accurate algebraic manipulation.

The method

1. Identify known vectors and points: Clearly label any given vectors (e.g., \(\mathbf{a}\), \(\mathbf{b}\)) and points (e.g., O, A, B). Remember that a position vector \(\vec{OA}\) is usually written as \(\mathbf{a}\).
2. Express required vectors as paths: To find a vector like \(\vec{AB}\), think of a path from A to B using known vectors. For example, \(\vec{AB} = \vec{AO} + \vec{OB}\). Since \(\vec{AO} = -\vec{OA}\), this becomes \(\vec{AB} = -\mathbf{a} + \mathbf{b}\).
3. Substitute and simplify: Replace any position vectors with their column vector or \(\mathbf{i}, \mathbf{j}\) form and combine like terms. For example, if \(\mathbf{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\), then \(\vec{AB} = \begin{pmatrix} -2 \\ -3 \end{pmatrix} + \begin{pmatrix} 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\).
4. Calculate magnitude (if required): For a vector \(\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}\), its magnitude \(|\mathbf{v}|\) is found using Pythagoras' theorem: \(|\mathbf{v}| = \sqrt{x^2 + y^2}\).
5. Check for collinearity or parallelism (if required): Two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel if \(\mathbf{u} = k\mathbf{v}\) for some scalar \(k\). If they share a common point, they are collinear.

Worked example

Given that \(\vec{OA} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}\) and \(\vec{OB} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}\), find the vector \(\vec{AB}\) and its magnitude \(|\vec{AB}|\).

$$ \begin{aligned} \vec{AB} &= \vec{AO} + \vec{OB} \\ &= -\vec{OA} + \vec{OB} \\ &= -\begin{pmatrix} 4 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} -4 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ 5 \end{pmatrix} \\ &= \begin{pmatrix} -4 + 1 \\ -1 + 5 \end{pmatrix} \\ &= \begin{pmatrix} -3 \\ 4 \end{pmatrix} \end{aligned} $$

Now, find the magnitude of \(\vec{AB}\):

$$ \begin{aligned} |\vec{AB}| &= \left| \begin{pmatrix} -3 \\ 4 \end{pmatrix} \right| \\ &= \sqrt{(-3)^2 + 4^2} \\ &= \sqrt{9 + 16} \\ &= \sqrt{25} \\ &= 5 \text{ units} \end{aligned} $$
Sanity check: A vector from (4,1) to (1,5) moves 3 units left and 4 units up. A 3-4-5 right-angled triangle has a hypotenuse of 5. The magnitude is correct.

Worked example: a harder one

OABC is a parallelogram. P is the point on AC such that AP:PC = 1:2.
\(\vec{OA} = 2\mathbf{a}\) and \(\vec{OC} = 3\mathbf{c}\).
Find \(\vec{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\).

The obvious first move might be to find \(\vec{OP}\) directly using \(\vec{OA} + \vec{AP}\) or \(\vec{OC} + \vec{CP}\). This fails initially because we don't know \(\vec{AP}\) or \(\vec{CP}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\) without first finding \(\vec{AC}\).

First, find \(\vec{AC}\):
$$ \begin{aligned} \vec{AC} &= \vec{AO} + \vec{OC} \\ &= -\vec{OA} + \vec{OC} \\ &= -2\mathbf{a} + 3\mathbf{c} \end{aligned} $$
Now, use the ratio AP:PC = 1:2. This means \(\vec{AP} = \frac{1}{1+2}\vec{AC} = \frac{1}{3}\vec{AC}\).
$$ \begin{aligned} \vec{AP} &= \frac{1}{3}(-2\mathbf{a} + 3\mathbf{c}) \\ &= -\frac{2}{3}\mathbf{a} + \frac{3}{3}\mathbf{c} \\ &= -\frac{2}{3}\mathbf{a} + \mathbf{c} \end{aligned} $$
Finally, find \(\vec{OP}\) using \(\vec{OP} = \vec{OA} + \vec{AP}\):
$$ \begin{aligned} \vec{OP} &= 2\mathbf{a} + \left(-\frac{2}{3}\mathbf{a} + \mathbf{c}\right) \\ &= 2\mathbf{a} - \frac{2}{3}\mathbf{a} + \mathbf{c} \\ &= \left(\frac{6}{3} - \frac{2}{3}\right)\mathbf{a} + \mathbf{c} \\ &= \frac{4}{3}\mathbf{a} + \mathbf{c} \end{aligned} $$

Practice

1. Given \(\mathbf{p} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\), find \(2\mathbf{p} + \mathbf{q}\).
2. The vector \(\vec{AB} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}\). Find the magnitude of \(\vec{AB}\).
3. O is the origin. A is the point (3, 7) and B is the point (8, 2). Find the vector \(\vec{AB}\).
4. In the diagram, \(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\). P is a point on AB such that AP:PB = 3:1.
   Find \(\vec{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

Answers:
1. \(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\)
2. 13
3. \(\begin{pmatrix} 5 \\ -5 \end{pmatrix}\)
4. Working for Q4:
First, find \(\vec{AB}\):
$$ \begin{aligned} \vec{AB} &= \vec{AO} + \vec{OB} \\ &= -\mathbf{a} + \mathbf{b} \end{aligned} $$
Since AP:PB = 3:1, \(\vec{AP} = \frac{3}{3+1}\vec{AB} = \frac{3}{4}\vec{AB}\).
$$ \begin{aligned} \vec{AP} &= \frac{3}{4}(-\mathbf{a} + \mathbf{b}) \\ &= -\frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \end{aligned} $$
Finally, find \(\vec{OP}\) using \(\vec{OP} = \vec{OA} + \vec{AP}\):
$$ \begin{aligned} \vec{OP} &= \mathbf{a} + \left(-\frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right) \\ &= \mathbf{a} - \frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \\ &= \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \end{aligned} $$

The three mistakes that lose marks

1. Incorrect direction: Reversing the order of points in a vector without changing the sign. For example, writing \(\vec{BA} = \vec{OA} - \vec{OB}\) instead of \(\vec{BA} = \vec{BO} + \vec{OA} = -\vec{OB} + \vec{OA}\). This often leads to the wrong answer \(\begin{pmatrix} x \\ y \end{pmatrix}\) instead of \(\begin{pmatrix} -x \\ -y \end{pmatrix}\).
2. Confusing position vectors with coordinates: Using the coordinates of a point directly for a vector between two points without subtracting. For example, if A is (3,2) and B is (5,1), incorrectly thinking \(\vec{AB} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\) instead of \(\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 5-3 \\ 1-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\).
3. Errors in ratio division: When a point divides a line segment in a ratio, incorrectly applying the fraction. For example, if P divides AB in ratio 2:3, writing \(\vec{AP} = \frac{2}{3}\vec{AB}\) instead of \(\vec{AP} = \frac{2}{2+3}\vec{AB} = \frac{2}{5}\vec{AB}\).

30-second recap

Vectors describe movement and position using direction and magnitude. Always express unknown vectors as a path using known vectors. Magnitude is found using Pythagoras. Parallel vectors are scalar multiples of each other; if they share a point, they are collinear.