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About LinkBacksIf u,v and w are non-zero vectors from R3 and
u x v = u x w
then
v = w
I know this is false, but I don't understand why.
u x v gives a vector. If that vectore is equal to the one given by u x w, how can v and w be different?
April 27th, 2013, 10:43 PM
If u,v and w are non-zero vectors from R3 and
u x v = u x w
then
v = w
I know this is false, but I don't understand why.
u x v gives a vector. If that vectore is equal to the one given by u x w, how can v and w be different?
I'm taking a guess at this, after just skimming over the wiki on vectors.
I did well in highschool geometry but I think this is little bit (in other words- a buttload) higher level than I ever learned.
but the best way I understood it is that if uxv is describing an angle creaged by two rays both starting at point x and ray one passes through point u and ray two passes through point v and the other uxw the same with x being the point of origin. it creates two angles sharing a common ray of xu.
if this is dealing with 3d spaces then a way of looking at it would be if you look at a corner of a room and point x is a corner where the two walls and the floor meet. point u is where the ceiling meets with the same two walls. (the walls ceiling and floor being 2d planes. )
dammit i need to draw a pic don't i.
vectorsmaybe.jpg
ok looking at the pic, what I am trying to say (as a guess) is that while v and w may be the same distance from x and and u.. they are not necessarily the same point. they are on different planes and are therefor different locations.
If I'm wrong, please don't hurt me .. this is just a guess from someone who has no idea what vectors are other than what she attempted to understand from the wiki link on it.
scratch that, Neverfly told me to look up Euclidean space on wiki and that is easier to understand. I don't want to draw another crappy pic so I will just send you the link.
basically. in Euclidean space every point is defined by three coordinates... so uxw and uxv cannot be equal because they are defining two differnent points. uxv=uxw is not two equations it is declaring two points to be one and the same. or at least that's my next best guess.
Last edited by seagypsy; April 28th, 2013 at 12:23 AM.
what isOriginally Posted by Peyo
and
?
u x u = 0. Therefore If v - w = cu, then u x v = u x w.
This looks like a job for.....mathman!!!u x u = 0. Therefore If v - w = cu, then u x v = u x w.
Actually, in u x v , the "x" meant vectorial product, it is an operation you can apply on two vectors and it results in a vector that is perpendicular to vector [u] AND vector [v].basically. in Euclidean space every point is defined by three coordinates... so uxw and uxv cannot be equal because they are defining two differnent points. uxv=uxw is not two equations it is declaring two points to be one and the same. or at least that's my next best guess.
Let's say you have
vector u = [1 2 3]
vector w = [4 5 6]
you get u x w by building a square matrix, with i j k as the first line that goes like this:
i j k
1 2 3
4 5 6
and multiplying i, j and k by the the determinant of their minor (not sure if you call it a minor or a co-factor)
so it goes like this
i[2(6)-3(5)] -j[1(6)-3(4)] k[1(5)-2(4)]
gives us
i[-3) -j[-6] k[-3] = [-3 6 -3]
i don't understandu x u = 0. Therefore If v - w = cu, then u x v = u x w.
u x u = 0
that's alright, we're talking about the same thing.
but why would v - w = cu ??
well fine! do it YOUR way then......Actually, in u x v , the "x" meant vectorial product, it is an operation you can apply on two vectors and it results in a vector that is perpendicular to vector [u] AND vector [v].basically. in Euclidean space every point is defined by three coordinates... so uxw and uxv cannot be equal because they are defining two differnent points. uxv=uxw is not two equations it is declaring two points to be one and the same. or at least that's my next best guess.
Let's say you have
vector u = [1 2 3]
vector w = [4 5 6]
you get u x w by building a square matrix, with i j k as the first line that goes like this:
i j k
1 2 3
4 5 6
and multiplying i, j and k by the the determinant of their minor (not sure if you call it a minor or a co-factor)
so it goes like this
i[2(6)-3(5)] -j[1(6)-3(4)] k[1(5)-2(4)]
gives us
i[-3) -j[-6] k[-3] = [-3 6 -3]
Oh well, did explaining it to me help you any? Not that I understand a word of what you said. (because, like I said, math ain't my thang) but sometimes I remember how to do something when I am teaching it to someone else. I did ok in highschool algebra as an adult but in highschool not so well. But I got good at it while trying to help my daughter with her homework.
No that's iti understand how it works but the answer doesn't make sense to me. I was hoping someone could help out here, i'll ask on yahoo
Someone here will be able to. Just a matter of the right person seeing the thread.No that's it
Let me see if I can put this in a way that appeals to your intuition a bit more by being a bit more pictorial. What we learn from u x v = 0 is that v is parallel to u . To prove that statement, choose a coordinate system in which u is aligned along x. Then the determinant that you used to get the vector product is given byi don't understandu x u = 0. Therefore If v - w = cu, then u x v = u x w.
u x u = 0
that's alright, we're talking about the same thing.
but why would v - w = cu ??
i j k
ux 0 0
vx vy vz
If you now do the expansion of the determinant by minors that you used earlier, you get that
i [0] - j [ux vz ] + k [ux vy ] = u x v = 0 .
Since there is no point in even doing this if u = 0, clearly ux is not zero, and the vector equation above gives
vy = 0, vz = 0
Hence what we know is that the components of v perpendicular to the x axis, hence perpendicular to u, are zero. Hence u is parallel to
v .
But there are a lot of vectors parallel to v , since any multiple of v is parallel to v . Hence u x (v - w )= 0 means that v - w is parallel to u , not that v - w = 0 implying that v = w .
The proof referred to above made use of this fact by defining w' = w + c u , clearly different from w if c is not equal to 0, and showing that
u x w' = 0 if u x w = 0.
i don't understandu x u = 0. Therefore If v - w = cu, then u x v = u x w.
u x u = 0
that's alright, we're talking about the same thing.
but why would v - w = cu ??
You misunderstood my point. You asked if uxv = uxw implies v = w. What I am trying to say is that uxv = uxw implies that v-w is a multiple of u, not necessarily 0.
I guess this is answered by now but let's find a counter example for clarity:
-let u be (1 0 0)
-let v be (3 2 1)
-let w be (6 2 1)
Therefore, v != w
However, uxv = uxw = (0 -1 2)
It's been a while since I went through linear algebra, but this is a property of vector spaces. Cross product between two vectors multiplies components that are orthogonal between them. In other words, if you take the first vector of the cross product as the X axis, only the components perpendicular to that axis (from the second vector) will matter. You can do a change of basis to allign the first vector to the X axis (like I did in my example) and see that the X component of the second vector will be meaningless in the cross product.
If your vectors are uglier than that, there is another way to test it. You can take any of the two vectors (say, w) and add to it any vector linearly dependent to the other (v), and the cross product will be the same. For instance:
For all v, w:
v x w = v x (a*v + w) where a is any real number.
If you take this bracketed expresion as your new "w" you will see there are infinite different vectors that yield the same result.
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