Question

Area of a parallelogram a= [1,-4], = [3,-5]

Answer 1

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

Answer 2

yup, typo *

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

plug in z=0 :)

Answer 8

ohhh! .. facepalm

Answer 9

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

Answer 10

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

Answer 11

Okay thanks ! :)

Answer 12

wouldnt axb = 0 ? :S

Answer 13

oh nvm

Answer 14

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 :)

Answer 15

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

Answer 16

have fun ;)