Find all the vectorsuthat satisfy the equation <1,1,1> Xu= <-1,-1,2>

(The answer in the back of the book is <u3-1, u3+1, u3> I just don't know how to get to it!)

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- Oct 10th 2012, 09:13 PMjessicadowningCross Product Help
Find all the vectors

**u**that satisfy the equation <1,1,1> X**u**= <-1,-1,2>

(The answer in the back of the book is <u3-1, u3+1, u3> I just don't know how to get to it!) - Oct 10th 2012, 09:35 PMchiroRe: Cross Product Help
Hey jessicadowning.

Do the know the formula for C = A X B where a = (a0,a1,a2), b = (b0,b1,b2) and c = (c0,c1,c2) where you get c0,c1, and c2 in terms of a0,a1,a2,b0,b1 and b2? - Oct 10th 2012, 09:42 PMjessicadowningRe: Cross Product Help
No, I believe the process people are using for this problem is the determinant with the bottom vector being <u1, u2, u3>

- Oct 10th 2012, 09:59 PMchiroRe: Cross Product Help
Yes that is equivalent (and generally how you do the calculation).

So the determinant with u at the bottom and <1,1,1> in the middle gives

i*(1*u3 - 1*u2) - j*(1*u3 - 1*u1) + k*(1*u2 - 1*u1) = <-1,-1,2>

Now equating components gives you:

u3 - u2 = -1

u1 - u3 = -1

u2 - u1 = 2.

Row reducing this gives:

>> A = [0, -1, 1, -1; 1, 0, -1, -1; -1, 1, 0, 2]

A =

0 -1 1 -1

1 0 -1 -1

-1 1 0 2

>> rref(A)

ans =

1 0 -1 -1

0 1 -1 1

0 0 0 0

So this means that u2 - u3 = 1 and u1 - u3 = -1. So we get

u1 = u3 - 1

u2 = u3 + 1

So plug those in to the following:

<u3 - 1, u3 + 1, u3> = U. - Oct 10th 2012, 10:07 PMjessicadowningRe: Cross Product Help
Thanks!