Vector Geometry — Working with Vectors

Vector Geometry

Working with Vectors Through Shapes

GCSE Geometry

What is a vector?

A vector is a journey. It tells you how far to move and in which direction.

For example, \(\overrightarrow{AB}\) means start at A and travel to B. The order of the letters matters.

One idea to remember

Reversing a journey reverses the vector:

\[ \overrightarrow{BA}=-\overrightarrow{AB} \]

Many vector geometry questions are solved by choosing a sensible route through a diagram. Start by asking:

Where am I starting?
Where do I need to finish?
Which labelled sides or lines can take me there?

1. Finding routes through shapes

Suppose you want to travel from A to C, but the useful labels are on the route from A to B and then B to C.

If \(\overrightarrow{AB}=\mathbf a\) and \(\overrightarrow{BC}=\mathbf b\), then the whole journey from A to C is:

\(\overrightarrow{AC}=\mathbf a+\mathbf b\)

You are simply joining two journeys together.

Route tip

If an arrow points the wrong way for your journey, change its sign. For example, if \(\overrightarrow{AB}=\mathbf a\), then \(\overrightarrow{BA}=-\mathbf a\).

Try this

In a triangle, \(\overrightarrow{PQ}=\mathbf p\) and \(\overrightarrow{PR}=\mathbf r\). Find \(\overrightarrow{QR}\).

Show answer and explanation

Solution

Start at Q and finish at R. A useful route is Q to P, then P to R.

\[ \overrightarrow{QR}=\overrightarrow{QP}+\overrightarrow{PR} \]

Since \(\overrightarrow{PQ}=\mathbf p\), the reverse journey is \(\overrightarrow{QP}=-\mathbf p\).

\[ \boxed{\overrightarrow{QR}=\mathbf r-\mathbf p} \]

2. Column vectors

A column vector is another way to describe a movement on a grid.

\[ \begin{pmatrix}4\\-3\end{pmatrix} \]

means 4 squares to the right and 3 squares down.

Reading a column vector

The top number tells you the horizontal movement: right is positive and left is negative.
The bottom number tells you the vertical movement: up is positive and down is negative.

Adding and multiplying column vectors

Work with the top numbers together and the bottom numbers together.

\[ \begin{pmatrix}2\\3\end{pmatrix}+ \begin{pmatrix}1\\5\end{pmatrix} = \begin{pmatrix}3\\8\end{pmatrix} \]

\[ 2\begin{pmatrix}2\\3\end{pmatrix}+3\begin{pmatrix}1\\5\end{pmatrix} = \begin{pmatrix}7\\21\end{pmatrix} \]

Try this

Let \(\mathbf a=\begin{pmatrix}3\\-1\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}-2\\4\end{pmatrix}\). Work out \(2\mathbf a+\mathbf b\).

Show answer and explanation

Solution

\[ 2\mathbf a=\begin{pmatrix}6\\-2\end{pmatrix} \]

\[ 2\mathbf a+\mathbf b= \begin{pmatrix}6\\-2\end{pmatrix}+ \begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}4\\2\end{pmatrix} \]

\[\boxed{\begin{pmatrix}4\\2\end{pmatrix}}\]

3. Finding a vector from one point to another

When coordinates are given, think about the movement from the first point to the second point.

For example, let \(A=(2,1)\) and \(B=(7,4)\).

From 2 to 7 is 5 squares right.
From 1 to 4 is 3 squares up.

So:

\(\overrightarrow{AB}=\begin{pmatrix}5\\3\end{pmatrix}\)

A useful shortcut once the idea makes sense

For each coordinate, do finish minus start. For A to B, subtract A’s coordinates from B’s coordinates.

Try this

C is the point \((5,-2)\) and D is the point \((2,3)\). Find \(\overrightarrow{CD}\) as a column vector.

Show answer and explanation

Solution

From C to D:

5 to 2 is 3 left, so the horizontal movement is \(-3\).
\(-2\) to 3 is 5 up, so the vertical movement is \(5\).

\[\boxed{\overrightarrow{CD}=\begin{pmatrix}-3\\5\end{pmatrix}}\]

4. Vectors in parallelograms

Parallelograms are especially useful in vector questions because opposite sides are parallel and equal.

What that means for vectors

If two opposite sides point in the same direction, they have the same vector. If you travel along one of them in the opposite direction, the sign changes.

Suppose \(\overrightarrow{OA}=2\mathbf a\) and \(\overrightarrow{OB}=2\mathbf b\).

The diagonals of a parallelogram bisect each other, so O is the midpoint of both diagonals. That immediately tells us:

\[ \overrightarrow{OC}=-2\mathbf a,\qquad \overrightarrow{OD}=-2\mathbf b \]

Example

Find \(\overrightarrow{DA}\), \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

Show answer and explanation

Solution

For \(\overrightarrow{DA}\): go D to O, then O to A.

\[ \overrightarrow{DA}=\overrightarrow{DO}+\overrightarrow{OA}=2\mathbf b+2\mathbf a=2\mathbf a+2\mathbf b \]

For \(\overrightarrow{AB}\): go A to O, then O to B.

\[ \overrightarrow{AB}=-2\mathbf a+2\mathbf b=2\mathbf b-2\mathbf a \]

For \(\overrightarrow{AC}\): go A to O, then O to C.

\[ \overrightarrow{AC}=-2\mathbf a-2\mathbf a=-4\mathbf a \]

\[ \boxed{\overrightarrow{DA}=2\mathbf a+2\mathbf b,\quad \overrightarrow{AB}=2\mathbf b-2\mathbf a,\quad \overrightarrow{AC}=-4\mathbf a} \]

Do not try to memorise every parallelogram result.

Mark the start and finish of the vector you need, then build a route using the labelled vectors.

5. Vectors in regular hexagons

A regular hexagon has repeated lengths and repeated directions. This creates lots of equal vectors.

The centre is also very useful: every vertex is the same distance from the centre.

Let \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OB}=\mathbf b\).

Finding \(\overrightarrow{AB}\)

Take the route A to O to B:

\[ \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b=\mathbf b-\mathbf a \]

Using the shape

In a regular hexagon, opposite vertices lie on a straight line through O. So D is directly opposite A and E is directly opposite B:

\[ \overrightarrow{OD}=-\mathbf a,\qquad \overrightarrow{OE}=-\mathbf b \]

Also, each side has the same length as a centre-to-vertex line. Direction is what decides which vector expression is needed.

Try this

Using the diagram above, find \(\overrightarrow{AD}\) in terms of \(\mathbf a\).

Show answer and explanation

Solution

A and D are opposite vertices. Go A to O, then O to D.

\[ \overrightarrow{AO}=-\mathbf a,\qquad \overrightarrow{OD}=-\mathbf a \]

\[ \overrightarrow{AD}=-\mathbf a-\mathbf a=-2\mathbf a \]

\[\boxed{\overrightarrow{AD}=-2\mathbf a}\]

6. Midpoints

If M is the midpoint of AB, then M is exactly halfway from A to B.

\(\overrightarrow{AM}=\dfrac12\overrightarrow{AB}\)

Suppose \(\overrightarrow{OA}=5\mathbf a\), \(\overrightarrow{OB}=3\mathbf b\), and M is the midpoint of AB.

Example

Find \(\overrightarrow{AB}\), \(\overrightarrow{AM}\), and \(\overrightarrow{OM}\).

Show answer and explanation

Solution

First go from A to O and then O to B:

\[ \overrightarrow{AB}=-5\mathbf a+3\mathbf b \]

M is halfway along AB:

\[ \overrightarrow{AM}=\frac12(-5\mathbf a+3\mathbf b)=-\frac52\mathbf a+\frac32\mathbf b \]

Now go O to A, then A to M:

\[ \overrightarrow{OM}=5\mathbf a+\left(-\frac52\mathbf a+\frac32\mathbf b\right)=\frac52\mathbf a+\frac32\mathbf b \]

\[ \boxed{\overrightarrow{AB}=-5\mathbf a+3\mathbf b,\quad \overrightarrow{AM}=-\frac52\mathbf a+\frac32\mathbf b,\quad \overrightarrow{OM}=\frac52\mathbf a+\frac32\mathbf b} \]

7. Proving points are on a straight line

Vector questions sometimes ask you to prove that three points lie on the same straight line.

The key idea is simple: compare two vectors that follow the line. If one vector is a number times the other, they point along the same line.

What you are looking for

\[ \overrightarrow{AB}=k\overrightarrow{AC} \]

for some non-zero number \(k\). This shows the two vectors are parallel. If they also share the point A, then A, B and C lie on one straight line.

Example

Suppose you find:

\[ \overrightarrow{AB}=2\mathbf a+4\mathbf b \]

and

\[ \overrightarrow{AC}=5\mathbf a+10\mathbf b \]

Can you prove that A, B and C are on a straight line?

Show answer and explanation

Solution

Factorise both vectors:

\[ \overrightarrow{AB}=2(\mathbf a+2\mathbf b),\qquad \overrightarrow{AC}=5(\mathbf a+2\mathbf b) \]

So:

\[ \overrightarrow{AC}=\frac52\overrightarrow{AB} \]

The two vectors are scalar multiples, so they are parallel. They both start at A, so they lie on the same line through A.

Therefore A, B and C are collinear.

Good proof wording

State that one vector is a scalar multiple of the other, so the vectors are parallel. Then mention the common point. That completes the straight-line argument.

8. Questions to try

Use the route ideas from the lesson. Try each question before opening the answer.

Question 1

\(\mathbf a=\begin{pmatrix}-2\\3\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}5\\-1\end{pmatrix}\).

Work out:

(a) \(\mathbf a+\mathbf b\)

(b) \(2\mathbf a-\mathbf b\)

Show answer and explanation

Solution

\[ \mathbf a+\mathbf b=\begin{pmatrix}3\\2\end{pmatrix} \]

\[ 2\mathbf a-\mathbf b= \begin{pmatrix}-4\\6\end{pmatrix}-\begin{pmatrix}5\\-1\end{pmatrix} =\begin{pmatrix}-9\\7\end{pmatrix} \]

Question 2

A is \((4,-1)\) and B is \((-2,5)\). Find \(\overrightarrow{AB}\) as a column vector.

Show answer and explanation

Solution

From 4 to \(-2\) is 6 left. From \(-1\) to 5 is 6 up.

\[\boxed{\overrightarrow{AB}=\begin{pmatrix}-6\\6\end{pmatrix}}\]

Question 3

ABCD is a parallelogram. \(\overrightarrow{AB}=\mathbf p\) and \(\overrightarrow{AD}=\mathbf q\).

Find \(\overrightarrow{AC}\) and \(\overrightarrow{DB}\).

Show answer and explanation

Solution

For A to C, travel A to B and then B to C. Since opposite sides of a parallelogram are equal as vectors, \(\overrightarrow{BC}=\mathbf q\).

\[ \overrightarrow{AC}=\mathbf p+\mathbf q \]

For D to B, travel D to A and then A to B:

\[ \overrightarrow{DB}=-\mathbf q+\mathbf p=\mathbf p-\mathbf q \]

Question 4

In regular hexagon ABCDEF with centre O, \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OB}=\mathbf b\). Find \(\overrightarrow{BA}\).

Show answer and explanation

Solution

Go B to O, then O to A:

\[ \overrightarrow{BA}=-\mathbf b+\mathbf a=\mathbf a-\mathbf b \]

Question 5

\(\overrightarrow{OA}=4\mathbf a\), \(\overrightarrow{OB}=2\mathbf b\), and M is the midpoint of AB. Find \(\overrightarrow{OM}\).

Show answer and explanation

Solution

The midpoint is halfway between the two position vectors:

\[ \overrightarrow{OM}=\frac12(4\mathbf a+2\mathbf b)=2\mathbf a+\mathbf b \]

Question 6

You calculate \(\overrightarrow{PQ}=3\mathbf a-6\mathbf b\) and \(\overrightarrow{PR}=5\mathbf a-10\mathbf b\). Explain why P, Q and R lie on a straight line.

Show answer and explanation

Solution

\[ \overrightarrow{PQ}=3(\mathbf a-2\mathbf b),\qquad \overrightarrow{PR}=5(\mathbf a-2\mathbf b) \]

Therefore \(\overrightarrow{PR}=\frac53\overrightarrow{PQ}\). The vectors are scalar multiples, so they are parallel. They both start at P, so P, Q and R are collinear.

9. Main lesson summary

1Read a vector as a journey from the first point to the second.

2Build longer journeys by joining useful routes through the diagram.

3Reverse a vector by changing its sign.

4Use equal and parallel sides in parallelograms and regular hexagons.

5For a midpoint, take half of the vector joining the two endpoints.

6For a straight-line proof, show that one vector is a scalar multiple of another.

Before you finish a vector question, check:

Have you started and finished at the correct points?
Did you reverse the sign when travelling against an arrow?
Can the shape give you an equal or parallel vector?
For coordinates, did you use the movement from the start point to the finish point?
For a straight-line proof, did you state why the vectors are scalar multiples?

Extension: Finding the length of a vector

If a vector is \(\begin{pmatrix}x\\y\end{pmatrix}\), its horizontal and vertical movements form a right-angled triangle. Use Pythagoras to find its length:

\[ \text{length}=\sqrt{x^2+y^2} \]

Try this

Find the length of \(\begin{pmatrix}6\\8\end{pmatrix}\).

Show answer and explanation

Solution

\[ \sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 \]

\[\boxed{10}\]

Vector Geometry: What is a vector and vectors through shapes

What is a vector?

1. Finding routes through shapes

Try this

Solution

2. Column vectors

Adding and multiplying column vectors

Try this

Solution

3. Finding a vector from one point to another

Try this

Solution

4. Vectors in parallelograms

Example

Solution

5. Vectors in regular hexagons

Finding \(\overrightarrow{AB}\)

Using the shape

Try this

Solution

6. Midpoints

Example

Solution

7. Proving points are on a straight line

Example

Solution

8. Questions to try

Solution

Solution

Solution

Solution

Solution

Solution

9. Main lesson summary

Extension: Finding the length of a vector

Try this

Solution

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