Understanding "Hat e" Notation

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Can someone tell me what this notation is

[itex]\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}[/itex]

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

Thanks
 
on Phys.org
iDimension said:
Can someone tell me what this notation is

[itex]\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}[/itex]

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?
 
iDimension said:
Can someone tell me what this notation is

[itex]\hat e =\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{|| \overrightarrow{u}\cdot \overrightarrow{v}||} = -\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|| \overrightarrow{u} \cdot \overrightarrow{v}||}[/itex]

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}||}##
 
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.
 
iDimension said:
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.
 

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