Why does the Laplacian operator still maintain its unit vectors i, j, k?

[laramman2]

When two vectors are dotted, the result is a scalar. But why here http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/mathbas/vectors.htm , the del-squared still maintains its unit vectors i, j, k? Isn't it this way ∇² = (∂²/∂x² + ∂²/∂y² + ∂²/∂z²) and not (i∂²/∂x² + j∂²/∂y² + k∂²/∂z²)? Thank you! :)

 

[SteamKing]

When two vectors are dotted, the result is a scalar. But why here http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/mathbas/vectors.htm , the del-squared still maintains its unit vectors i, j, k? Isn't it this way ∇² = (∂²/∂x² + ∂²/∂y² + ∂²/∂z²) and not (i∂²/∂x² + j∂²/∂y² + k∂²/∂z²)? Thank you! :)

There are two types of Laplacian operator: one is scalar and the other is a vector operator.

http://en.wikipedia.org/wiki/Laplacian

http://en.wikipedia.org/wiki/Vector_Laplacian

 

[laramman2]

Thank you sir. :)