area of a parallelogram spanned by (a,b) and (c,d) (positive oriented) I know that the area of a parallelogram spanned by (a,b) and (c,d) is |ad-bc|. My teacher proved this formula in class and told us that it would be just ad-bc if (a,b) and (c,d) are positive oriented but he did not explain why. 
I tried to prove that ad>bc when (a,b) and (c,d) are positive oriented but failed.
Can someone please help?
Many many thanks.
 A: Here's a justification:
Express $(a,b)$ and $(c,d)$ in polar form:
$$\begin{align}(a,b)&=(u\cos\theta,u\sin\theta)\\
(c,d)&=(v\cos\phi,v\sin\phi)\text{.}
\end{align}$$
Then
$$\begin{split}ad-bc&=(u\cos\theta)(v\sin\phi)-(u\sin\theta)(v\cos\phi)\\
&=uv(\cos\theta\sin\phi-\sin\theta\cos\phi)\\
&=uv\sin(\phi-\theta)\text{.}
\end{split}$$
Since $uv$ is positive, the sign of $ad-bc$ is determined by $\phi-\theta$:


*

*if $0<\phi-\theta<180^{\circ}$—$(c,d)$ is counterclockwise relative to $(a,b)$—then $ad-bc>0$.

*if $180^{\circ}<\phi-\theta<0$—$(c,d)$ is clockwise relative to $(a,b)$—then $ad-bc<0$.

A: I think a picture actually does this best (at least for me)

ad corresponds to the purple box, taking the width of the box from a and the height of the box from d. On the other hand for the blue box, bc corresponds to taking the width of the box from c and and the height of the box from b.
So then its pretty clear, if they wrap counterclockwise, ad (the purple box) is always going to be bigger than bc (the blue box). In fact. Its pretty easy to prove the determinant formula from these pictures just by chopping up the blue and purple areas (look up any of the geometric proofs to try it yourself)
But I hope that helps :) 
Sorry about my rushed photo (I didn't line the vectors up so well haha)
