Position vectors
For a point P, we call the vector from the origin to the point P the position vector of P.
When P has coordinates (1,4,8) the position vector of P has components \(\left( \begin{array}{l}1\\4\\8\end{array} \right)\).
We say that \(\overrightarrow {OP}=\left( \begin{array}{l}1\\4\\8\end{array} \right)\)
We can use position vectors to calculate the components of a required vector.
Example
P has coordinates (1,4,8) and Q has coordinates (-3,1,-4).
Find the components of vector \(\overrightarrow {PQ}\).
Answer
We have \(\overrightarrow {OP}=\left( \begin{array}{l}1\\4\\8\end{array} \right)\) and \(\overrightarrow {OQ}=\left( \begin{array}{l}-3\\1\\-4\end{array} \right)\).
When we are thinking through a vector calculation we can make a sketch. This sketch may not look anything like an accurate diagram and can even be 2D instead of 3D.
We deduce that \(\overrightarrow {PQ} = \overrightarrow {PO}+ \overrightarrow {OQ}\)
Remember that \(\overrightarrow {PO}\) is the negative of \(\overrightarrow {OP}\)
So \(\overrightarrow {PQ}=\left( \begin{array}{l}-1\\-4\\-8\end{array} \right)+\left( \begin{array}{l}-3\\1\\-4\end{array} \right)\)
\(\overrightarrow {PQ}= \left( \begin{array}{l}-4\\-3\\-12\end{array} \right)\)
More guides on this topic
- Determine the gradient of a straight line
- Circle geometry
- Calculating the volume of a standard solid
- Applying Pythagoras Theorem
- Applying the properties of shapes to determine an angle
- Using similarity
- Working with two-dimensional vectors
- Working with three-dimensional coordinates
- Calculating the magnitude of a vector
- An Approximate History of Co-ordinates