\[x-u_1 -u_2 , \: y==u_1 -2 u_3 , \: z=u_1 +u_3\]
be a transformation from ordinary Euclidean space to \[(y_1 , u_2 , u_3 )\]
soace.Suppose
\[\mathbf{F} = a \mathbf{i} + b \mathbf{j} +c \mathbf{k}\]
.We can express
\[\mathbf{F}\]
in \[(u_1 , u_2 , u_3 )\]
space.\[\mathbf{F} = \alpha \frac{\partial \mathbf{r}}{\partial u_1}+ \beta \frac{\partial \mathbf{r}}{\partial u_2}+ \gamma \frac{\partial \mathbf{r}}{\partial u_3}\]
In this expression the derivatives are called the unitary base vectors and
\[\alpha , \beta , \gamma\]
are called the contravariant components of \[\mathbf{F}\]
We have
\[\frac{\partial \mathbf{r}}{\partial u_1 }= \mathbf{i} + \mathbf{j} + \mathbf{k} \]
\[\frac{\partial \mathbf{r}}{\partial u_2 }= - \mathbf{i}\]
\[\frac{\partial \mathbf{r}}{\partial u_3 }=-2 \mathbf{i} + \mathbf{k} \]
We solve
\[\mathbf{F} = a \mathbf{i} + b \mathbf{j} +c \mathbf{k}=\alpha (\mathbf{i} + \mathbf{j} + \mathbf{k} )+ \beta (- \mathbf{i}) + \gamma (-2 \mathbf{i} + \mathbf{k}) \]
Equating components
\[a=\alpha - \beta\]
\[b=\alpha - 2 \gamma\]
\[c=\alpha + \gamma\]
The third equation minus the second gives
\[c-b = 3 \gamma \rightarrow \gamma = \frac{c-b}{3}\]
.The second equation plus twice the third gives
\[b+2c= 3 \alpha \rightarrow \alpha = \frac{b+2c}{3}\]
.The first equation then gives
\[\beta =\alpha -a = \frac{c-b}{3} -a = \frac{-3a-b+c}{3}\]
.Then
\[\mathbf{F} = (\frac{b+2c}{3}) \frac{\partial \mathbf{r}}{\partial u_1}+ (\frac{-3a-b+c}{3}) \frac{\partial \mathbf{r}}{\partial u_2}+ (\frac{c-b}{3}) \frac{\partial \mathbf{r}}{\partial u_3}\]