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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.

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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.

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    You're very welcome - I edited to address your second question as well.2012-10-20
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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 $\phi\in[0,\pi]$.

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    Ah ok, then I'm sorry. To me it just seemed incomplete :-). I see now that its not wrong.2012-10-20