Problem Solving using Vectors (Cambridge (CIE) IGCSE Additional Maths): Revision Note

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.  $\mathbf{a}=k\mathbf{b}$ 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 $\left(-2,-2\right),\left(3,3\right)$ and $\left(8,8\right)$ 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 $\overrightarrow{PQ}:\overrightarrow{QR}=3:1$

How do I find the midpoint of two vectors?

• If the point $A$ has position vector $\mathbf{a}$ and the point $B$ has position vector $\mathbf{b}$

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

• This can be derived by considering

  • $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$

    • using the result from Vector Addition 

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

    • $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$

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

    • $\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\mathbf{a}+\frac{1}{2}\left(\mathbf{b}-\mathbf{a}\right)$ $\overrightarrow{OM}=\frac{1}{2}\left(\mathbf{a}+\mathbf{b}\right)$ 

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 $\overrightarrow{AB}=k\overrightarrow{AC}$ where $k$ is a constant i.e.  show that $\overrightarrow{AB}$ and $\overrightarrow{AC}$ 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 $\overrightarrow{AB}=k\overrightarrow{BC}$  AND  that point $B$ lies on both the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$

• 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 $\mathbf{a}$ and $\mathbf{b}$ defined as marked in the diagram:

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

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

    • However, the vector $\mathbf{a}$ is a general vector

      • any vector the same length as $\overrightarrow{AB}$ and parallel to it is equal to $\mathbf{a}$

      • e.g.  $\overrightarrow{RS}=\mathbf{a}$

    • Vector $\mathbf{b}$ is also a general vector

      • e.g.  $\overrightarrow{GL}=\mathbf{b}$ 

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

    • e.g.  $\overrightarrow{ON}=-\mathbf{a}\mathit{,}\overrightarrow{\mathit{J}\mathit{E}}\mathit{=}\mathit{-}\mathbf{b}$

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

  • e.g.  $\overrightarrow{PL}=\mathbf{b}-\mathbf{a}$

• There are many scalar multiples of the vectors $\mathbf{a}$ or $\mathbf{b}$

  • e.g.  $\overrightarrow{FI}=3\mathbf{a}\mathit{,}\overrightarrow{\mathit{I}\mathit{S}}=2\mathbf{b}\mathit{,}\overrightarrow{\mathit{Q}\mathit{E}}=3\left(\mathbf{b}-\mathbf{a}\right)$

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