Finding Vector Paths (Cambridge International IGCSE Mathematics)

Finding vector paths

How do I find the vector between two points?

• 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. $\overset{\rightarrow}{AR}=2a+3b$

• 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. $\overset{\rightarrow}{FB}=-b+a$ or $\overset{\rightarrow}{FB}=a-b$
    • Likewise, $\vec{BF} = - \vec{FB} = - ( - b + a ) = b - a$

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

Exam Hint

• 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

Worked Example The following diagram consists of a grid of identical parallelograms.

Vectors a *and b are defined by $a=\overset{\rightarrow}{AB}$ and $b=\overset{\rightarrow}{AF}$.

Write the following vectors in terms of a and b.

(a) $\overset{\rightarrow}{AE}$

Answer:

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

$\begin{gathered} \vec{AE} & = \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} \\ = a + a + a + a \end{gathered}$

Final answer: $\overset{\rightarrow}{AE}=4a$

(b) $\overset{\rightarrow}{GT}$

Answer:

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)

$\begin{gathered} \vec{GT} & = \vec{GL} + \vec{LQ} + \vec{QR} + \vec{RS} + \vec{ST} \\ = b + b + a + a + a \end{gathered}$

$\overset{\rightarrow}{GT}=3a+2b$

(c) $\overset{\rightarrow}{EK}$

Answer:

*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)

$\begin{gathered} \vec{EK} & = \vec{EJ} + \vec{JO} + \vec{ON} + \vec{NM} + \vec{ML} + \vec{LK} \\ = b + b - a - a - a - a \end{gathered}$

$\overset{\rightarrow}{EK}=2b-4a$