Find the area of the parallelogram with vertices (5,1), (8, 6), (12, 9), and (15, 14)

How would I start this problem?

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- Sep 6th 2008, 11:13 AMAlliearea of parallelogram
Find the area of the parallelogram with vertices (5,1), (8, 6), (12, 9), and (15, 14)

How would I start this problem? - Sep 6th 2008, 11:17 AMskeeter
have a look how it's done ...

Area of Parallelogram from Vector Cross Product - Sep 6th 2008, 11:36 AMAllie
thank you that is very useful but i must be doing something wrong in my calculations the vector i get from those points are 3i+5j+0k for both, when i take the cross product the answer is 0. I know that is not correct. What's wrong with my calculations

- Sep 6th 2008, 12:27 PMILoveMaths07
Allie, a sketch always helps. What you're doing is... You're finding the vector of a diagonal of the parallelogram BC instead of the adjacent side, AC. Calculate the vector product $\displaystyle \vec{AB}\times\vec{AC}$.

I hope that helps. :)

ILoveMaths07. - Sep 6th 2008, 01:05 PMAllie
haha the sketch does not help i don't even see parallelogram in the sketch, it looks like a straight line

- Sep 6th 2008, 01:11 PMskeeter
not a line, but a very "skinny" parallelogram

two adjacent sides ...

(5,1) to (8,6) and (5,1) to (12,9)

first side vector ...**u**= 3**i**+ 5**j**

adjacent side vector ...**v**= 7**i**+ 8**j**now find the cross product.

- Sep 6th 2008, 01:22 PMAllie
ok i got it now thank you

Now would i do the same for this problem?

Find the volume of the parallelepiped defined by the vectors

(5,2,1) (-3,-2,5) (2,5,-2) - Sep 6th 2008, 01:32 PMskeeter
- Sep 6th 2008, 01:50 PMAlliethank you that website is very helpful