A vector in terms of two vectors
If \(\mathbf{u}\) and \(\mathbf{v}\) are non-zero and not parallel, every vector in their plane can be written in exactly one way as \(m\mathbf{u}+n\mathbf{v}\), where \(m\) and \(n\) are numbers.
Method on a grid
In the figure \(\mathbf{u}=\begin{pmatrix}-1\\ 2\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}3\\ 1\end{pmatrix}\). Two copies of \(\mathbf{u}\) take \(S\) to \(K\), and three copies of \(\mathbf{v}\) take \(K\) to \(T\), so \(\overrightarrow{ST}=\overrightarrow{SK}+\overrightarrow{KT}=2\mathbf{u}+3\mathbf{v}\). Check: \(2\begin{pmatrix}-1\\ 2\end{pmatrix}+3\begin{pmatrix}3\\ 1\end{pmatrix}=\begin{pmatrix}7\\ 7\end{pmatrix}\), which is 7 right and 7 up from \(S\) to \(T\).
Negative coefficients and the column check
If a part runs against \(\mathbf{u}\) or \(\mathbf{v}\), its coefficient is negative. For example \(\begin{pmatrix}-5\\ 3\end{pmatrix}=2\mathbf{u}-\mathbf{v}\), because \(2\begin{pmatrix}-1\\ 2\end{pmatrix}-\begin{pmatrix}3\\ 1\end{pmatrix}=\begin{pmatrix}-5\\ 3\end{pmatrix}\).
Without a grid, write \(m\mathbf{u}+n\mathbf{v}\) in columns and solve the two component equations. For \(\begin{pmatrix}7\\ 7\end{pmatrix}\): \(-m+3n=7\) and \(2m+n=7\) give \(m=2\) and \(n=3\).
Check
Count copies of the given vectors, not grid squares: here one copy of \(\mathbf{v}\) is three squares to the right and one square up. The method needs two non-parallel vectors; with parallel \(\mathbf{u}\) and \(\mathbf{v}\) the two lines would never meet.
Create a free account to practise similar questions.
Need help? Join our E-Maths tuition for O-Level success.
Learn more