I have seen the proof of why the area of the parallelogram created by 2 vectors $u = \left(\begin{matrix} u_1\\ u_2 \end{matrix}\right)$ and $v = \left(\begin{matrix}v_1 \\ v_2 \end{matrix}\right)$ $\in \mathbb{R}$

enter image description here

can be obtained by taking the absolute value of the determinant of the following matrix:

$$A_P = \left|\det{\left(\begin{matrix} u_1 & v_1 \\ u_2 & v_2 \end{matrix}\right)}\right|$$.

The proof was quite long, but it was quite clear. Sal explains it really well.

Now, I need to extend the definition of finding the area for the parellogram to find the area bounded by 3 vectors ($\in \mathbb{R^2}$) as depicted in the following figure:

enter image description here

This figure seems to be a parallelogram plust a rectangle... but I am not seeing how exactly how could I extend the definition above of finding the area to find the area of the figure bouned by these vectors. Any ideas?


Its simply a matter of adding the areas of three parallelograms. To see this draw the vector $w$ from the origin, and connect its end to $u+w$ and $v+w$. So then its just the sum of the three determinants...

Edit: see image: enter image description here

  • $\begingroup$ Can you explaing why? Sincerely, I am not seeing 3 parellelograms (and therefore why we have to sum the area of 3 parallelograms...). I will edit my question with 3 vectors of example and please use them to explain. $\endgroup$ – nbro Apr 26 '15 at 9:03
  • $\begingroup$ By the sum of the 3 determinants, you mean the determinants of the pair of vectors: $(u, u + w)$, $(v + w, u + v + w)$ and $(v, v + w)$? $\endgroup$ – nbro Apr 26 '15 at 11:13
  • $\begingroup$ Well it should be determinants of $(u,v)$, $(u,w)$ and $(v,w)$ $\endgroup$ – Christiaan Hattingh Apr 26 '15 at 11:31

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