Using and finding unit vectors
The basic unit vectors are \(i = \left( \begin{array}{l}1\\ 0\\0\end{array} \right)\), \(j=\left(\begin{array}{l}0\\1\\0\end{array}\right)\) and \(k = \left(\begin{array}{l} 0\\ 0\\1\end{array} \right)\)
Any vector can be written in terms of \(i\), \(j\) and \(k\). For example:
\(\left(\begin{array}{l}\,\,\,\,\,\,3\\\,\,\,\,\,4\\- 2\end{array} \right) = \left(\begin{array}{l}\,3\\0\\0\end{array} \right) + \left(\begin{array}{l}\,0\\4\\0\end{array} \right) + \left(\begin{array}{l}\,\,\,\,\,0\\\,\,\,\,\,0\\- 2\end{array} \right)\)
\(= 3\left(\begin{array}{l}1\\0\\0\end{array} \right) + 4\left( \begin{array}{l}0\\1\\0\end{array} \right) - 2\left( \begin{array}{l}0\\0\\1\end{array} \right)\)
\(= 3i + 4j - 2k\)
Question
Express \(\left( \begin{array}{l} \,\,\,\,\,5\\\,\,\,\,\,0\\ - 1\end{array} \right)\) in terms of unit vectors.
\(\left( \begin{array}{l} \,\,\,\,\,5\\\,\,\,\,\,0\\ - 1\end{array} \right) = 5i + 0j - 1k\)
\(= 5i - k\)