Cross and Dot Products

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January 29th 2009, 05:31 PM
veronicak5678

Cross and Dot Products

Find the cross product axb and verify it is orthogonal to both a and b

a = <e^2t, 1, e^-t>
b = <e^t, 2, e^-t>

I calculated the cross product to be <-e^-t, e^t - 1, 2e^2t - e^t> and tried dotting it with a and b. Problem is, I don't get 0 in either case. Isn't that how you show it is orthogonal?
January 29th 2009, 05:45 PM
skeeter

Quote:

Originally Posted by veronicak5678

Find the cross product axb and verify it is orthogonal to both a and b

a = <e^2t, 1, e^-t>
b = <e^t, 2, e^-t>

I calculated the cross product to be <-e^-t, e^t - 1 , 2e^2t - e^t> and tried dotting it with a and b. Problem is, I don't get 0 in either case. Isn't that how you show it is orthogonal?

correction on your j component ...

<-e^-t, 1 - e^t , 2e^2t - e^t>

now find the dot products
January 29th 2009, 05:47 PM
veronicak5678

I see what I did wrong. Thanks for your help!
January 29th 2009, 06:04 PM
skeeter

Quote:

Originally Posted by veronicak5678

Isn't the j component negative? I originally had 1-e^t, then looked at my notes and saw it had to be negative. It didn't dot out to 0 in either case, anyway.

j component ...

$-(e^{2t} \cdot e^{-t} - e^{-t} \cdot e^{t})<br />$

$-(e^t - 1)$

$1 - e^t$

dot product with a

$<e^{2t}, 1, e^{-t}> \bullet <-e^{-t}, 1 - e^t , 2e^{2t} - e^t>$

$-e^t + 1 - e^t + 2e^t - 1 = 0$

dot product with b

$<e^t, 2, e^{-t}> \bullet <-e^{-t}, 1 - e^t , 2e^{2t} - e^t>$

$-1 + 2 - 2e^t + 2e^t - 1 = 0$