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htk
Sep11-09, 08:03 PM
find u X v and show that it is orthogonal to both u and v.

u= 6k
v=-i + 3j + k

http://s763.photobucket.com/albums/xx275/trinhkieu888/?action=view&current=666.jpg

This is what I got from the picture, but my teacher said that I have one more step to do to show that they are orthogonal, I need to do uxv

uxw = (0*-18)+(0*-6)+(6*0)= ?

and also vxw=?

If both are zero, the vectors are orthogonal.

But my problem is I don't know how to find w. Can any one please explain it to me? Thank you very much!

LCKurtz
Sep11-09, 08:24 PM
find u X v and show that it is orthogonal to both u and v.

u= 6k
v=-i + 3j + k

http://s763.photobucket.com/albums/xx275/trinhkieu888/?action=view&current=666.jpg

This is what I got from the picture, but my teacher said that I have one more step to do to show that they are orthogonal, I need to do uxv

uxw = (0*-18)+(0*-6)+(6*0)= ?

and also vxw=?

If both are zero, the vectors are orthogonal.

But my problem is I don't know how to find w. Can any one please explain it to me? Thank you very much!

uxw is a vector. Where are your i, j, k? I think you have it correct on your paper.

If you want to show uxw is orthogonal to u and v, remember that two nonzero vectors are orthogonal if their dot product is 0.

njama
Sep12-09, 05:42 AM
Call the orthogonal vector w.

If two vectors are orthogonal then the angle between them is 900

You can prove it using the dot product:

\mathbf{u} \cdot \mathbf{w} = 0

and

\mathbf{v} \cdot \mathbf{w} = 0

or finding cos(u,w) and cos(v,w) for the cross product (since you already find w). :smile:

HallsofIvy
Sep12-09, 06:02 AM
Excuse me, but you first say "find u X v and show that it is orthogonal to both u and v" but then start talking about "u x w" and "v x w". Where did "w" come from? If you mean that w= u x v, then, as njama said, you want to look at the dot product of w with u and v, not the cross product. You do NOT, by the way, need to divide by ||u x v|| since the problem says nothing about a unit vector.