Problem Solving using Vectors (Cambridge International IGCSE Further Maths)

Problem-solving using vectors

What problems may I be asked to solve involving vectors?

• Showing that two lines or vectors are parallel

  • Two vectors are parallel if they are scalar multiples of each other

  • i.e. $a=kb$ where $k$ is a constant

    • See Vector Addition

• Finding the midpoint of two (position) vectors

• Showing that three points are collinear

  • Collinear describes points that lie on the same straight line

    • e.g. The points $( - 2 , - 2 ) , ( 3 , 3 )$ and $( 8 , 8 )$ all lie on the line with equation $y=x$
    • Vectors can be used to show this, and similar, results

• Results concerned with geometric shapes

  • Shapes with parallel lines are often involved

    • e.g. parallelogram, rhombus

  • These often include lines or vectors being split into ratios

    • e.g. The point $Q$ lies on the line $PR$ such that $\overset{\rightarrow}{PQ}:\overset{\rightarrow}{QR}=3:1$

How do I find the midpoint of two vectors?

• If the point $A$ has position vector $a$ and the point $B$ has position vector $b$

  • the position vector of the midpoint of $AB$ is $1 \frac{1}{2} ( a + b )$

• This can be derived by considering

  • $\overset{\rightarrow}{AB}=b-a$

    • using the result from Vector Addition

  • If $M$ is the midpoint of $AB$ then

    • $\overset{\rightarrow}{AM}=\frac{1}{2}\overset{\rightarrow}{AB}$

  • Therefore, the position vector of the midpoint, $\overset{\rightarrow}{OM}$ is

    • $\vec{OM} = \vec{OA} + \vec{AM} = a + 1 \frac{1}{2} ( b - a )$ $\vec{OM} = 1 \frac{1}{2} ( a + b )$

How do I show three points are collinear?

• Three points are collinear if they all lie on the same straight line

• There are two ways to show this for three points, $A,B$ and $C$ say

  • Method 1 Show that $\overset{\rightarrow}{AB}=k\overset{\rightarrow}{AC}$ where $k$ is a constant i.e. show that $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AC}$ are scalar multiples of each other

    • As the vectors are scalar multiples they will have the same direction (and so be parallel)
    • So as both vectors start at point $A$, they must be collinear

  • Method 2 Show that $\overset{\rightarrow}{AB}=k\overset{\rightarrow}{BC}$ AND that point $B$ lies on both the vectors $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{BC}$

• Which method you should use will depend on the information given and how you happen to see the question

How do I solve problems involving geometric shapes?

• Problems involving geometric shapes involve finding paths around the shape using known vectors

  • there will be many other vectors in the shape that are equal and/or parallel to the known vectors

• The following grid is made up entirely of parallelograms, with the vectors $a$ and $b$ defined as marked in the diagram:

• Note the difference between "specific" and "general" vectors

  • The vector $\overset{\rightarrow}{AB}$ in the diagram is specific and refers only to the vector starting at $A$ and ending at $B$

    • However, the vector $a$ is a general vector

      • any vector the same length as $\overset{\rightarrow}{AB}$ and parallel to it is equal to $a$
      • e.g. $\overset{\rightarrow}{RS}=a$

    • Vector $b$ is also a general vector

      • e.g. $\overset{\rightarrow}{GL}=b$

  • There will also be vectors in the diagram that are the same magnitude but have the opposite direction to $a$ or $b$

    • e.g. $\overset{\rightarrow}{ON}=-a,\overset{\rightarrow}{JE}=-b$

• There are also many instances of the vector addition result $\overset{\rightarrow}{FB}=b-a$

  • e.g. $\overset{\rightarrow}{PL}=b-a$

• There are many scalar multiples of the vectors $a$ or $b$

  • e.g. $\overset{\rightarrow}{FI}=3a,\overset{\rightarrow}{IS}=2b,\overset{\rightarrow}{QE}=3(b-a)$

• Using a combination of these it is possible to describe a vector between any two points in terms of $a$ and $b$

Exam Hint

• Diagrams are helpful in vector questions

  • If a diagram has been given, label it and add to it as you progress through a question
  • If a diagram has not been given, draw one, it does not need to be accurate!

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

*To get from A to E 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) $\overset{\rightarrow}{GT}$

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) Point $Z$ is such that it is midpoint of $HM$. Find the vector $\overset{\rightarrow}{PZ}$.

There are many ways to get from P to Z. * One option is to go from P to R($a$ twice), and then from R to Z (*$-b$ one-and-a-half times).

$\begin{gathered} \vec{PZ} & = \vec{PR} + \vec{RZ} \\ = a + a - b - 1 \frac{1}{2} b \end{gathered}$

$\overset{\rightarrow}{PZ}=2a-\frac{3}{2}b$