Let, $\vec{\mathbf{r}}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ be the position vector of a point $P$ in space. Then if we have another coordinate system i.e. $\vec{\mathbf{r}}=\vec{\mathbf{r}}\left ( u_{1},u_{2},u_{3}\right)$. Then, there exist a unit vectors in this coordinate system as well. These unit vectors are the vectors in the curvilinear system.

A reminder here, that we are assuming there is a definite known transformation that takes us from $\left ( x,y,z \right )$ to $\left ( u_{1},u_{2},u_{3} \right )$ coordinate system and vice versa. i.e. $ \left ( x,y,z \right )\leftrightarrow \left ( u_{1},u_{2},u_{3} \right )$ is deifned. The above information is pictorially represented in the following figure:

Now, you can immediately guess that in the curvilinear coordinate system $\left ( u_{1},u_{2},u_{3} \right )$, there are basically two different basis vector by which vectors can be constructed. These two different basis vectors are usually the distinction that is used in defining the

**contravariant**and the**covariant**components of the vectors. One being the usual**tangent vector direction**and the other not so intuitive the**direction of the gradient vector.**
We can then write a vector (let's say $A$) as,

In the tangential direction:

In the gradient direction as:

Notice that the direction vectors are not the unit vectors. These are just the basis vectors. For the unit vector we just need to divide the components of the vectors by its magnitude.

i.e. if we represent unit vector of the tangential components by $e_{i}$, then we have;

And, also if we represent unit vector for the gradient components by $E_{i}$, then we have;

In the both above unit vector components the index $i$ is, $i=1,2,3$. Each index gives the unit vector in the respective components. Thus, we are sucessful in constructing the unit vectors in the curvilinear co-ordinate system.And, also if we represent unit vector for the gradient components by $E_{i}$, then we have;

Source: Schaum's Outline (Vector Analysis)

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