Linear Algebra Confusion You have 2 vectors $\vec{a}$ and $\vec{b}$, with an angle of $\theta$ between them. You get:
cos $\theta = \dfrac{adj}{||\vec{a}||}$ -> $||\vec{a}|| cos\theta$ = adj. But then, as I noted in my notebook:
$\vec{a} . \vec{b}$ = $||\vec{b}|| . adj $
Sadly, I forgot how my professor came to that conclusion, can someone enlighten me?
Also:
You AGAIN have 2 vectors $\vec{a}$ and $\vec{b}$, with an angle of $\theta$ between them. You can make a parallelogram with Area = bh. 
Area = $\vec{b} . ||\vec{a}|| sin \theta $ = $||\vec{a}$ x $\vec{b} || $ 
I also don't understand this at all, so, if again someone could enlighten me.
 A: It's a bit hard to answer your question, because your notation refers to a diagram that your professor presumably drew on the board.  Your second formula expresses the fact that the dot product $\vec{a} \cdot \vec{b}$ is equal to the length of one of the vectors ($\vec{b}$) times the length of the projection of the other vector ($\vec{a}$) onto the first vector ($\vec{b}$).  The length of the projection is equal to $\text{adj}$.  (This is hopefully clear from the diagram.)
For your second formula, it should say that the area = $||\vec{b}|| ||\vec{a}|| \sin \theta$.  You can obtain it as follows.  Draw the vector $\vec{b}$ horizontally, pointing to the right on your paper.  Draw the vector $\vec{a}$ in a SW-NE direction so that it makes an angle $\theta$ with $\vec{b}$.  Extend another copy of $\vec{b}$ horizontally from the head of $\vec{a}$ and another copy of $\vec{a}$ in the same SW-NE direction from the head of $\vec{b}$ to make a parallelogram.  The area of the parallelogram is the length of the base times the height.  The length of the parallelogram is $||\vec{b}||$ and its height is seen to be $||\vec{a}|| \sin \theta$ by looking at a well-chosen right triangle that can be made in the diagram.
The definition of $\vec{a} \times \vec{b}$ is the vector whose direction is perpendicular to both $\vec{a}$ and $\vec{b}$ with its direction determiend by the so-called "right-hand rule" and whose magnitude is equal to the area of the parallelogram described in the previous paragraph.
A: The correct equation for the angle $\phi$ between two vectors $a$ and $b$ is
$$\cos\phi=\frac{a\cdot b}{\|a\|\cdot\|b\|}.$$
The area of the parallelogram which is spanned by the two vectors $a$ and $b$ is given by the norm of the cross product:
$$\|a\times b\|=\|a\|\|b\|\sin\phi$$
for $\phi\in[0,\pi]$.
