Vectors describe movement with both direction and magnitude. They can be added or subtracted to produce resultant vectors. The scalar product can be used to find the angle between vectors.

Part of

A vector describes a movement from one point to another.

A **vector** quantity has both **direction** and magnitude.

(In contrast a scalar quantity has magnitude only - eg, the numbers 1, 2, 3, 4...)

The diagram above represents a vector. The arrow displays its direction, hence this vector can be written as \(\overrightarrow {AB}\), **a**, or \(\begin{pmatrix}
3 \\
4
\end{pmatrix}\).

In print, **a** is written in bold type. In handwriting, the vector is indicated by underlining the letter.

If we reverse the arrow it now points from B to A.

Remember that the arrow describes the direction. So, in this case, the vector is from B to A.

If we move 'backwards' along a vector, it becomes negative, so **a** becomes **-a**. Moving from B to A entails moving 3 units to the left, and 4 down.

So the three ways to write this vector are: \(\overrightarrow {BA}\), **-a** and \(\begin{pmatrix}
-3 \\
-4
\end{pmatrix}\).

If two vectors have the same **magnitude** and **direction**, then they are **equal** regardless of their position.

When adding vectors we follow the rule:

\[\left( \begin{array}{l} a\\ b \end{array} \right) + \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a + c\\ b + d \end{array} \right)\]

Look at the graph below to see the movements between **PQ**, **QR** and **PR**.

Vector \(\overrightarrow {PQ}\) followed by vector \(\overrightarrow {QR}\) represents a movement from **P** to **R**.

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

Written out the vector addition looks like this:

\[\left( \begin{array}{l}2\\5\end{array} \right) + \left( \begin{array}{l}\,\,\,\,\,4\\- 3\end{array} \right) = \left( \begin{array}{l}6\\2\end{array} \right)\]

Subtracting a vector is the same as adding a negative version of the vector (remember that making a vector negative means reversing its direction).

\[\left( \begin{array}{l} a\\ b \end{array} \right) - \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a - c\\ b - d \end{array} \right)\]

Look at the diagram and imagine going from X to Z. How would you write the path in vectors using only the vectors \(\overrightarrow {XY}\) and \(\overrightarrow {ZY}\)?

You could say it is vector \(\overrightarrow {XY}\) followed by a backwards movement along \(\overrightarrow {ZY}\).

So we can write the path from X to Z as:

\[\overrightarrow {XY} - \overrightarrow {ZY} = \overrightarrow {XZ}\]

Written out in numbers it looks like this:

\[\left( \begin{array}{l} 4\\ 2 \end{array} \right) - \left( \begin{array}{l} 1\\ 2 \end{array} \right) = \left( \begin{array}{l} 3\\ 0 \end{array} \right)\]

- Question
If \(x = \left( \begin{array}{l} 1\\ 3 \end{array} \right)\), \(y = \left( \begin{array}{l} - 2\\ 4 \end{array} \right)\) and \(z = \left( \begin{array}{l} - 1\\ - 2 \end{array} \right)\) find:

- \[- y\]
- \[x - y\]
- \[2x + 3z\]

- \(\left( \begin{array}{l}\,\,\,\,\,2\\- 4\end{array} \right)\) Did you remember to change the signs?
- \[\left( \begin{array}{l}\,1\\3\end{array} \right) - \left( \begin{array}{l}- 2\\\,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,1 - - 2\\3 - \,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,\,\,\,\,3\\- 1\end{array} \right)\]
- \[\left( \begin{array}{l}\,1\\3\end{array} \right) + 3\left( \begin{array}{l}- 1\\- 2\end{array} \right) = \left( \begin{array}{l}2\\6\end{array} \right) + \left( \begin{array}{l}- 3\\- 6\end{array} \right) = \left( \begin{array}{l}- 1\\\,\,\,\,0\end{array}\right)\]

A resultant vector is a vector that 'results' from adding two or more vectors together.

To travel from **X** to **Z**, it is possible to move along vector \(\overrightarrow {XY}\) followed by \(\overrightarrow {YZ}\). It is also possible to go directly along \(\overrightarrow {XZ}\).

\(\overrightarrow {XZ}\) is therefore known as the **resultant** of \(\overrightarrow {XY}\) and \(\overrightarrow {YZ}\) .

- Question
Write as single vectors:

**1.**\(f + g\)**2.**\(a + b\)**3.**\(e - b - a\)**1.**\(e\)**2.**\(- c\) (Did you remember the minus sign?)**3.**\(- d\)

Remember: Two vectors are equal if they have the same magnitude and direction, regardless of where they are on the page.

- Question
Triangles

**ABC**and**XYZ**are equilateral.**X**is the midpoint of**AB**,**Y**is the midpoint of**BC**,**Z**is the midpoint of**AC**.\(\overrightarrow {AX} = a\), \(\overrightarrow {XZ} = b\), \(\overrightarrow {AZ} = c\)

Express each of the following in terms of

**a**,**b**and**c**.- \[\overrightarrow {XY}\]
- \[\overrightarrow {YZ}\]
- \[\overrightarrow {XC}\]
- \[\overrightarrow {BZ}\]
- \[\overrightarrow {AC}\]

**c****- a**Remember that \(\overrightarrow {YZ}\) is parallel to \(\overrightarrow {AX}\) and of the same length, but the direction is different.**b + c**(It is also possible to move from**X**to**A**and then on to**C**. This would give the answer**- a + 2c**. How many other answers can you think of?)**b - a**or**2b - c**or**- 2a + c****2c**