## Expressing a vector in Terms of its Contravariant Components

Let
$x-u_1 -u_2 , \: y==u_1 -2 u_3 , \: z=u_1 +u_3$
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
$\mathbf{F} = (\frac{b+2c}{3}) \frac{\partial \mathbf{r}}{\partial u_1}+ (\frac{-3a-b+c}{3}) \frac{\partial \mathbf{r}}{\partial u_2}+ (\frac{c-b}{3}) \frac{\partial \mathbf{r}}{\partial u_3}$