Thread: area 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?

have a look how it's done ...

Area of Parallelogram from Vector Cross Product

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

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 .

I hope that helps.

ILoveMaths07.

haha the sketch does not help i don't even see parallelogram in the sketch, it looks like a straight line

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 = 3i + 5j

adjacent side vector ... v = 7i + 8j

now find the cross product.

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)

http://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/Determinant/Determinant/node11.html

thank you that website is very helpful