Geometric vectors

A vector describes a movement from one point to another.

Vector notation

A vector quantity has both direction and magnitude.

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

The vector from point A to B is labelled a

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.

The vector from point B to A is in the opposite direction

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}.

Vectors a and -a are the same distance and to and from the same points, but in opposite directions

Equal vectors

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

Two vectors with the same magnitude and direction are equal

Adding vectors

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.

Vectors PQ and QR share point Q

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 vectors

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)

Vectors XY and ZY both end at point Y

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:

  1. - y
  2. x - y
  3. 2x + 3z

  1. \left( \begin{array}{l}\,\,\,\,\,2\\- 4\end{array} \right) Did you remember to change the signs?
  2. \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)
  3. \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)

Resultant vectors

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

Triangle made up of vectors XY, YZ and XZ

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
Shape made up of vectors a, b, c, d, e, f and g

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

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

Question
Triangle ABC with internal triangle XYZ and vectors AX=a, XZ=b, AZ=c

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.

  1. \overrightarrow {XY}
  2. \overrightarrow {YZ}
  3. \overrightarrow {XC}
  4. \overrightarrow {BZ}
  5. \overrightarrow {AC}

  1. c
  2. - a Remember that \overrightarrow {YZ} is parallel to \overrightarrow {AX} and of the same length, but the direction is different.
  3. 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?)
  4. b - a or 2b - c or - 2a + c
  5. 2c