Determining the resultant in three dimensions
Example
Given \(P(1,4,8)\) and \(Q( - 3,1, - 4)\), find \(\overrightarrow {PQ}\).
To do this, think of yourself in the position of point \(P\). How many units in each direction would you have to travel to reach point \(Q\)? A quick way to do this is to subtract the values of the coordinates of \(P\) from the coordinates of \(Q\).
\(\overrightarrow {PQ} = \left( \begin{array}{l}- 3 - 1\\ \,\,\,\,\,1 - 4\\- 4 - 8\end{array} \right) = \left( \begin{array}{l}\,\,\,- 4\\\,\,\,- 3\\- 12\end{array} \right)\)
Take care to subtract the right set of coordinates. You'd get a very different answer if you subtracted \(Q\) from \(P\) instead.
It follows that you can also work out the coordinates of a point if you have the coordinates of another point and the vector that connects them.
Example
Given \(P(1,4,10)\) and \(\overrightarrow {PQ}\) is a representative of vector \(u =\left(\begin{array}{l}\,\,\,\,\,2\\\,\,\,\,\,1\\- 1\end{array} \right)\) find \(Q\).
\(\left(\begin{array}{l}\,\,\,\,\,2\\\,\,\,\,\,1\\- 1\end{array} \right)\) means \(\left( \begin{array}{l}increase\,x\,by\, 2\\increase\,y\,by\, 1\\decrease\,z\,by\, 1\end{array} \right)\)
So if \(P = (1,4,10)\) then \(Q = (3,5,9)\)
The position vector is the vector from the origin to \(P\).
If \(P = (3,4, - 2)\), say, then \(\overrightarrow {OP} = \left( \begin{array}{l}\,\,\,\,\,3\\\,\,\,\,\,4\\- 2 \end{array} \right)\)
\(\overrightarrow {OP}\) is called the position vector of \(P\). We write \(\textbf{p} = \left( \begin{array}{l}\,\,\,\,\,3\\\,\,\,\,\,4\\- 2\end{array} \right)\)
If \(\overrightarrow {PQ} = \left( \begin{array}{l}\,\,\,\,\,5\\\,\,\,\,\,4\\- 2\end{array} \right)\) then the length or magnitude of \(\overrightarrow {PQ}\), written as \(\left| {\overrightarrow {PQ} } \right|\), is given by:
\(\left| {\overrightarrow {PQ} } \right| = \sqrt {{5^2} + {4^2} + {{( - 2)}^2}} = \sqrt {45}=3\sqrt {5}\)
