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2

I don't know what you mean by derive here, but by definition the two products give different types of geometric objects. The exterior product of two vectors (in $\mathbf{R}^n$, for any $n$) is a bivector, whereas the cross product of two vectors (in $\mathbf{R}^3$ only) is another vector. For $n=3$, there happens to be an isomorphism between bivectors and ...

2

I believe the exterior products (wedge products) of two vectors $u$ and $v$ are denoted by $u\wedge v$. Note: the $LaTeX$ command for wedge products is also \wedge. I have borrowed this picture from Wikipedia's page on exterior algebra as it explains the difference very clearly. $a\times b$ is the cross product of the two vectors $a$ and $b$. The cross ...

2

You are asking simply how to find $x$ such that $$x(\vec a\times\vec u) = \vec c$$ where $\vec a$, $\vec u$, $\vec c$ are given vectors (and $\vec u$ has unit length, although this is not relevant). Such $x$ exists if and only if $\vec a\times \vec u$ is a multiple of $\vec c$ and can be calculated simply as $$x = \frac{|\vec c|}{|\vec a\times\vec u|}.$$ In ...

1

The angle between two nonzero vectors $u,v$ is defined to be the number $\theta \in [0,\pi]$ such that $$\cos(\theta)= \frac{u\cdot v}{\|u\|\|v\|}$$ Plugging in your vectors yields $$\cos(\theta) = \frac{-16}{65}$$ The negative sign implies that $\theta \in (\frac {\pi}2,\pi]$. Drawing a triangle and using the Pythagorean theorem we find that the length ...

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The first equation states that the vector field $$\vec{E} + \frac{\partial\vec{A}}{\partial t}$$ has no rotation ($\text{curl}$ of that thing is $\vec{0}$). For vector-fields with $\text{curl}\; \vec{V} = \vec{0}$ have a scalar potential function $f$ for which $\vec{V} = \nabla f$, because for every gradient field $\text{curl}\; \nabla f = \vec{0}$. (https:...

0

This is a consequence of Helmholtz' Theorem: Any sufficiently nice vector field can be decomposed like $$F=\nabla\Phi+\nabla\times B$$ In your case, $F=E+\partial_t A$ and the rotation component is zero.

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