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

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is there supposed to be a b = [3,-5]
yup, typo *
magnitude of the cross product i think is a way to determine it

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Other answers:

|a||b| sin(t); where t is the angle between a and b
\[sin(t)=\frac{|axb|}{|a||b|}\] so yeah, the magnitude of the cross product
@amistre64 how can i do the cross product when only x and y are given ?
plug in z=0 :)
ohhh! .. facepalm
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
of course the cross of R^2 simplifies as the determinant
Okay thanks ! :)
wouldnt axb = 0 ? :S
oh nvm
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 :)
yeah i think its easier to plug in the zero :P
have fun ;)

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