Finding Vector Paths (Edexcel IGCSE Maths A (Modular))

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A vector path is a path of vectors taking you from a start point to an end point
The following grid is made up entirely of parallelograms
- The vectors a and b defined as marked in the diagram:
  - Any vector that goes horizontally to the right along a side of a parallelogram will be equal to a
  - Any vector that goes up diagonally to the right along a side of a parallelogram will be equal to b

To find the vector between two points
- Count how many times you need to go horizontally to the right
  - This will tell you how many a's are in your answer
- Count how many times you need to go up diagonally to the right
  - This will tell you how many b's are in your answer
- Add the a's and b's together
  - E.g. $\vec{A R} = 2 a + 3 b$
You will have to put a negative in front of the vector if it goes in the opposite direction
- -a is one length horizontally to the left
- -b is one length down diagonally to the left
  - E.g. $\vec{F B} = - b + a$ or $\vec{F B} = a - b$
  - Likewise, $\vec{B F} = - \vec{F B} = - (- b + a) = b - a$

It is possible to describe any vector that goes from one point to another in the above diagram in terms of a and b

Mark schemes will accept different correct paths, as long as the final answer is fully simplified
Check for symmetries in the diagram to see if the vectors given can be used anywhere else

The following diagram consists of a grid of identical parallelograms.

Vectors a and b are defined by $a = \vec{A B}$ and $b = \vec{A F}$ .

Write the following vectors in terms of a and b.

a) $\vec{A E}$

To get from A to E we need to follow vector a four times to the right

$\begin{array}{rcl} \vec{A E} & = & \vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} \\ = & a + a + a + a \end{array}$

$\vec{A E} = 4 a$

b) $\vec{G T}$

There are many ways to get from G to T
One option is to go from G to Q (b twice), and then from Q to T (a three times)

$\begin{array}{rcl} \vec{G T} & = & \vec{G L} + \vec{L Q} + \vec{Q R} + \vec{R S} + \vec{S T} \\ = & b + b + a + a + a \end{array}$

$\vec{G T} = 3 a + 2 b$

c) $\vec{E K}$

There are many ways to get from E to K
One option is to go from E to O (b twice), and then from O to K ( -a four times)

$\begin{array}{rcl} \vec{E K} & = & \vec{E J} + \vec{J O} + \vec{O N} + \vec{N M} + \vec{M L} + \vec{L K} \\ = & b + b - a - a - a - a \end{array}$

$\vec{E K} = 2 b - 4 a$

$- 4 a + 2 b$ is also acceptable

the (exam) results speak for themselves:

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