Standard unit vector in the positive z direction?

[space-time]

http://tutorial.math.lamar.edu/Classes/CalcIII/CurlDivergence.aspx

In the above link, in the 2nd to last blue box on the page (where it tells how to solve a line integral with respect to a vector field using the curl), it says that the line integral of vector field F with respect to r (with an arrow over the r) equals the iterated integral of the dot product of the curl of F and the standard unit vector in the positive z direction k (with an arrow over the k).

Now when I hear "the standard unit vector in the positive z direction" I generally think of the standard basis vector:

(0,0,1)

Is this the vector that k refers to? I ask this because that seems too simple as well as redundant. If this was the case, they could have just said that the integrand of the iterated integral is simply the z-component of the curl of the vector field.

If this is not what k refers to, then do they potentially mean that k is the unit vector you get when you normalize the vector field F?

If neither of those, then what exactly do they mean by "the standard unit vector in the positive z direction"?

 

[SteamKing]

It's not clear what you are confused about.

The unit vectors which are aligned with the x, y, and z coordinate axes are usually designated i , j , and k , respectively.

In other words, i = (1, 0, 0) ; j = (0, 1, 0) ; k = (0, 0, 1)

 

[space-time]

It's not clear what you are confused about.

The unit vectors which are aligned with the x, y, and z coordinate axes are usually designated i , j , and k , respectively.

In other words, i = (1, 0, 0) ; j = (0, 1, 0) ; k = (0, 0, 1)

So you are saying that the vector k that they refer to in the link I posted is in fact the unit vector (0,0,1)?

I just wanted to make sure of this.

 

[SteamKing]

Yes.