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Proof that each gradient vector grad u_i is Parallel to the of Tangent Vector the Corresponding Coordinate Curve

Take a right handed coordinate system  
\[(u_1 , u_2 , u_3 )\]
.
The normal vector to the surface  
\[u_i = c_i\]
  is the gradient  
\[\mathbf{\nabla} u_i = \frac{\partial u_i}{\partial x} \mathbf{i} + \frac{\partial u_i}{\partial y} \mathbf{j} + \frac{\partial u_i}{\partial z} \mathbf{k}\]

The tangent vector to the coordinate curve  
\[u_i\]
  is  
\[\mathbf{\nabla} r = \frac{\partial r}{\partial u_i} \mathbf{i} + \frac{\partial r}{\partial u_i} \mathbf{j} + \frac{\partial r}{\partial u_i} \mathbf{k}\]

Consider the coordinate curve  
\[u_1\]
  formed by the intersection of the surfaces  
\[u_2 = c_2\]
  and  
\[u_3 = c_3\]
.
The component of the tangent vector parallel to  
\[u_1\]
  is  
\[\frac{\partial r}{\partial u_i}\]
  and this is also perpendicular to the surface normals  
\[\mathbf{\nabla} u_2 , \: \mathbf{\nabla} u_3\]
  to the surfaces  
\[u_2 = c_2 , \:u_3 = c_3\]
  respectively.
Hence the vectors  
\[\frac{\partial r}{\partial u_1} , \: \mathbf{\nabla} u_1\]
  which both point in the direction of increasing  
\[u_1\]
  are parallel. Similarly for  
\[u_2\]
  and  
\[u_3\]
.