## burhan101 Group Title Area of a parallelogram a= [1,-4], = [3,-5] one year ago one year ago

1. tomo

is there supposed to be a b = [3,-5]

2. burhan101

yup, typo *

3. amistre64

magnitude of the cross product i think is a way to determine it

4. amistre64

|a||b| sin(t); where t is the angle between a and b

5. amistre64

$sin(t)=\frac{|axb|}{|a||b|}$ so yeah, the magnitude of the cross product

6. burhan101

@amistre64 how can i do the cross product when only x and y are given ?

7. amistre64

plug in z=0 :)

8. burhan101

ohhh! .. facepalm

9. amistre64

a= [1,-4], = [3,-5] x 1 3 y -4 -5 z 0 0 x = 0-0 -y = 0-0 z = -5+12 the magnitude of [0, 0, 7 ] = 7

10. amistre64

of course the cross of R^2 simplifies as the determinant

11. burhan101

Okay thanks ! :)

12. burhan101

wouldnt axb = 0 ? :S

13. burhan101

oh nvm

14. amistre64

my teacher did an R^2 cross a few semesters ago, and I was like, aint it just the determinant? and she was like or just plug in z=0 :)

15. burhan101

yeah i think its easier to plug in the zero :P

16. amistre64

have fun ;)