Problem Solving with Vectors (Cambridge (CIE) IGCSE Maths)

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Written by: Jamie Wood

Reviewed by: Dan Finlay

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Vector Problem Solving

What are vector proofs?

Vectors can be used to prove things that are true in geometrical diagrams
- Vector proofs can be used to find additional information that can help us to solve problems

How do I know if two vectors are parallel?

Two vectors are parallel if one is a scalar multiple of the other
- This means if b is parallel to a, then b = ka
  - where k is a constant number (scalar)
For example, $a = (\begin{matrix} 1 \\ 3 \end{matrix})$ and $b = (\begin{matrix} 2 \\ 6 \end{matrix})$
- $(\begin{matrix} 2 \\ 6 \end{matrix}) = 2 \times (\begin{matrix} 1 \\ 3 \end{matrix})$ so $b = 2 a$
- b is a scalar multiple of a, so b is parallel to a
If the scalar multiple is negative, then the vectors are parallel and in opposite directions
- $c = (\begin{matrix} - 3 \\ - 9 \end{matrix}) = - 3 a$
  - c is parallel to a and in the opposite direction

If two vectors factorise with a common bracket, then they are parallel
- They can be written as scalar multiples
For example
- 9a + 6b factorises to 3(3a + 2b)
- 12a + 8b factorises to 4(3a + 2b)
- This means $12 a + 8 b = \frac{4}{3} (9 a + 6 b)$
  - so they are scalar multiples of each other
  - and therefore parallel

How do I know if three points lie on a straight line?

You may be asked to prove that three points lie on a straight line
- Points that lie on a straight line are collinear
To show that the points A , B and C are collinear
- prove that two line segments are parallel
- and show that there is at least one point that lies on both segments
  - This makes them parallel and connected (not parallel and side-by-side)
For example, if you show that $\vec{B C} = 2 \vec{A B}$ then
- the line segments AB and BC are parallel
- and they have a common point, B
  - So A , B and C must be collinear
Similarly, $\vec{A C} = 3 \vec{A B}$ means AC and AB are parallel
- and they have a common point, A
  - so A , B and C must be collinear

If A, B, C are collinear, AB is parallel to AC and BC

How do I use ratios in vector paths?

Convert ratios into fractions
In the example shown, if $A X : X B = 3 : 5$ then
- $\vec{A X} = \frac{3}{8} \vec{A B}$
- $\vec{X B} = \frac{5}{8} \vec{A B}$
  - The ratio 3:5 has 3 + 5 = 8 parts
Always check which ratio you are being asked for
- $\vec{A X} = \frac{3}{5} \vec{X B}$
- $\vec{X B} = \frac{5}{3} \vec{A X}$

Worked Example

The diagram shows trapezium OABC.

$\vec{O A} = 2 a$

$\vec{O B} = c$

AB is parallel to OC, with $\vec{A B} = 3 \vec{O C}$ .

Question Vector Trapezium, IGCSE & GCSE Maths revision notes

(a) Find expressions for vectors $\vec{O B}$ and $\vec{A C}$ in terms of a and c.

$\vec{A B} = 3 \vec{O C}$ and $\vec{O C} = c$ so $\vec{A B} = 3 c$ .

$\begin{array}{rcl} \vec{O B} & = & \vec{O A} + \vec{A B} \\ = & 2 a + 3 c \end{array}$

$\vec{O B} = 2 a + 3 c$

$\begin{array}{rcl} \vec{A C} & = & \vec{A O} + \vec{O C} \\ = & - \vec{O A} + \vec{O C} \\ = & - 2 a + c \end{array}$

$\vec{A C} = c - 2 a$

(b) Point P lies on AC such that AP : PC = 3 : 1.

Find expressions for vectors $\vec{A P}$ and $\vec{O P}$ in terms of a and c.

AP : PC = 3 : 1 means that $\vec{A P} = \frac{3}{3 + 1} \vec{A C} = \frac{3}{4} \vec{A C}$

$\begin{array}{rcl} \vec{A P} & = & \frac{3}{4} \vec{A C} = \frac{3}{4} (- 2 a + c) \end{array}$

$\vec{A P} = - \frac{3}{2} a + \frac{3}{4} c$

$\begin{array}{rcl} \vec{O P} & = & \vec{O A} + \vec{A P} \\ = & 2 a + (- \frac{3}{2} a + \frac{3}{4} c) \end{array}$

$\vec{O P} = \frac{1}{2} a + \frac{3}{4} c$

To show that O, P, and B are colinear (lie on the same line), note that $\vec{O P} = \frac{1}{2} a + \frac{3}{4} c = \frac{1}{4} (2 a + 3 c)$

$\begin{array}{rcl} \vec{O P} & = & \frac{1}{4} (2 a + 3 c) = \frac{1}{4} \vec{O B} \end{array}$

$\vec{O P} = \frac{1}{4} \vec{O B}$ therefore OP is parallel to OB
and so P must lie on the line OB

If $\begin{array}{rcl} \vec{O P} & = & \frac{1}{4} \vec{O B} \end{array}$ then $\begin{array}{rcl} \vec{P B} & = & \frac{3}{4} \vec{O B} \end{array}$

$\vec{O P} : \vec{P B} = 1 : 3$

Last updated: 18 October 2024

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