I Understanding "Hat e" Notation

[iDimension]

Can someone tell me what this notation is

\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}

I saw it here https://socratic.org/questions/how-...t-is-perpendicular-to-both-2i-j-3k-and-i-j-2k

Thanks

 

[PeroK]

Can someone tell me what this notation is

\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}

I saw it here https://socratic.org/questions/how-...t-is-perpendicular-to-both-2i-j-3k-and-i-j-2k

Thanks

Those should be vector cross products not dot products. What don't you understand about it?

 

[Mark44]

Can someone tell me what this notation is

\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}

I saw it here https://socratic.org/questions/how-...t-is-perpendicular-to-both-2i-j-3k-and-i-j-2k

Thanks

As PeroK noted already, the multiplications are cross products, not dot products. In the page you linked to, they they actually wrote was this:
$\hat e =\frac{\overrightarrow{u}\times \overrightarrow{v}}{|| \overrightarrow{u}\times \overrightarrow{v}||} = -\frac{\overrightarrow{v}\times \overrightarrow{u}}{|| \overrightarrow{u} \times \overrightarrow{v}||}$

 

[iDimension]

Ah right, so it's cross product not dot product?

And what is the name of this definition or the name of this generalisation because looking on the wiki page I cannot see this format anywhere.

 

[Mark44]

Ah right, so it's cross product not dot product?

And what is the name of this definition or the name of this generalisation because looking on the wiki page I cannot see this format anywhere.

All they are doing is finding a unit vector ($\hat e$) that is perpendicular to both $\vec u$ and $\vec v$. The equation itself uses the concept that $\vec u \times \vec v = -(\vec v \times \vec u)$. These are very basic properties of vectors and the cross product.