# The Outward Unit Normal Vectors $\mathbf{n} = \mathbf{k} \times \mathbf{T}$ and $\mathbf{n} = \mathbf{T} \times \mathbf{k}$: Why?

The following diagram is from chapter 16.2 of Thomas's Calculus, 14th Edition:

I've studied and understand the concept of the cross-product, but I don't understand why $\mathbf{n} = \mathbf{k} \times \mathbf{T}$ is the outward unit normal vector if the curve $C$ is parameterised clockwise and $\mathbf{n} = \mathbf{T} \times \mathbf{k}$ is the outward unit normal vector if the curve $C$ is parameterised anti-clockwise?

At the moment, I have to either memorise this fact or refer to my textbook for the information. However, in my experience with mathematics, this necessity is usually a sign of a lack of understanding of why something is the way it is. Therefore, I'm assuming that if I understood why this is the case, I would be able to remember which equation to use ($\mathbf{n} = \mathbf{k} \times \mathbf{T}$ or $\mathbf{n} = \mathbf{T} \times \mathbf{k}$) without needing to rely on rote memorisation or my textbook.

I would greatly appreciate it if people could please take the time to clearly explain why $\mathbf{n} = \mathbf{k} \times \mathbf{T}$ is the outward unit normal vector if the curve $C$ is parameterised clockwise and $\mathbf{n} = \mathbf{T} \times \mathbf{k}$ is the outward unit normal vector if the curve $C$ is parameterised anti-clockwise, so that I can mentally derive these facts, rather then needing to rely on rote memorisation or my textbook.

Vector cross products follow the right-hand rule, by convention. This is the reason $\mathbf{k} \times \mathbf{T}$ points outward for a clockwise parametrised circle. The right-hand rule is arbitrary (there is no objective mathematical reason to use that instead of the opposite, left-hand rule), but it's a choice that the mathematical community at large has come to agree on, as not choosing one way as standard is worse than having an arbitrary standard.
Using that rule (whatever incarnation you have learned), you can tell which way $\mathbf{k} \times \mathbf{T}$ points for your circle. If it points inwards, reverse the product to get a vector pointing in the opposite direction.
It is simply a consequence of the convention of right-handed orientation for the three unit vectors $\mathbf {i,j,k}$ in the 3-D space where, by convention, we have: $\mathbf k= \mathbf i \times \mathbf j$