http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 25 Jun 2014 12:27:36 -0500http://calc3.askbot.com/question/11/finding-area-of-a-parallelogram/So I've been working on problem 7 in 12.4. Following the book on page 315 about parallelograms. > "Given two vectors, we can put them tail to tail and form a parallelogram...The height of the parallelogram, h, is |**A**|$ \sin(\theta)$ , and the base is |**B**|, so the area of the parallelogram is |**A**||**B**|$ \sin(\theta)$, exactly the magnitude of |**A** x **B**|." The problem states "7. Find the area of the parallelogram with vertices (0, 0), (1, 2), (3, 7), and (2, 5)." When I sketched out the parallelogram with the given points, it looked like the base of the parallelogram would be the vector from the origin to (1,2) (**B**), and the top vector would be (2,5)(**A**). (I'm not sure if this is where my mistake was, or if it was in trying to find $\theta$.) Using those two points as the vectors, I ended up with |**A**| = $\sqrt{29}$ and |**B**| =$ \sqrt{5}. $ I then used the dot product |**A**| * |**B**| = |**A**||**B**|$\cos$($\theta$) to attempt to find theta. Plugging everything in I ended up with $$ \theta = \cos^{-1}(\frac {12} {\sqrt{29} \sqrt{5}}) $$ Plugging that into the area equation from the book, I ended up with.. $$ area = \sqrt{29}\sqrt{5}(\sin(\cos^{-1}(\frac {12} {\sqrt{29}\sqrt{5}})))$$ Which unless I did wrong in my calculator gave me 0. But the back of the book showed that the answer should have been 1. So I'm not exactly sure where I went wrong, or if I did the whole problem wrong. So if anybody could give some input that would be fantastic! Thanks! TiffanyWed, 25 Jun 2014 12:27:36 -0500http://calc3.askbot.com/question/11/