# Proof that cross product = Area of parrelelogram

• Feb 14th 2010, 10:11 AM
Diemo
Proof that cross product = Area of parrelelogram
Not sure if I should be in this thread or in the geometry one, but here goes:

The problem that I was given was:
Find the area of a paralellogram spanned by the vectors a = i + j and b = 4i - j
Now prove the following |a X b| =|a||b|sin(theta) for ANY ( 2 dimensional ) vectors a and b where theta is the angle between the vectors.

Now I know that the area of a parrallogram is the magnitude of the cross product, and I can easily show that the area of the parrallelogram is the same as |a||b|sin(theta), but how do you prove that |axb| = area of parrelelogram?
• Feb 14th 2010, 10:39 AM
HallsofIvy
Actually, I would be inclined to look at it the other way- with the cross product of two vectors [b]defined[\b] by $\displaystyle |\vec{u}\times \vec{v}= |\vec{u}||\vec{v}|sin(\theta)$ and the direction given by the "right hand rule", prove that $\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}\times x\vec{i}+ y\vec{j}+ z\vec{k}= (bz- cy)\vec{i}- (az- cx)\vec{j}+ (ay-bx)\vec{k}$, but, of course, you can go either way.

In order to show your way, define a "new" product, say *, by $\displaystyle |\vec{u}*\vec{v}= |\vec{u}||\vec{v}|sin(\theta)$, and the "right hand rule". We can easily get the rules $\displaystyle \vec{i}*\vec{j}= \vec{k}$, $\displaystyle \vec{j}*\vec{k}= \vec{i}$, and $\displaystyle \vec{k}*\vec{i}= \vec{j}$ as well as $\displaystyle \vec{i}*\vec{i}= \vec{j}*\vec{j}= \vec{k}*\vec{k}= \vec{0}$, and the general $\displaystyle \vec{u}*\vec{v}= -\vec{v}*\vec{u}$.

Now, for $\displaystyle \vec{u}= a\vec{i}+ b\vec{j}+ c\vec{k}$ and $\displaystyle \vec{v}= x\vec{i}+ y\vec{j}+ z\vec{k}$, multiply $\displaystyle \vec{u}*\vec{v}= a\vec{i}+ b\vec{j}+ c\vec{k}*x\vec{i}+ y\vec{j}+ z\vec{k}$ "term by term" and use the basic products above to show that this "new" product is precisely the cross product.
• Feb 14th 2010, 10:51 AM
Diemo
I think I see. How would you do this in two dimensions though?