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This is a bonus problem I was given today in Physics C AP. If anyone would please help me out and point me in the right direction to solving this problem, I'd really appreciate it. I've spent an hour and half staring at this problem and I literally have nothing. If this sounds like your cup of tea and you have a moment, please give me a tip. That said
THE PROBLEM:+ Show Spoiler + If vector A= ((Ax)(i-hat))+((Ay)(j hat))+((Az)(k hat)) and vector B= ((Bx)(i hat))+((By)(j hat))+((Bz)(k hat)), prove that A X B (the cross product of A and B) is perpendicular to A vector and B vector. **(i hat) is pure X direction, (j hat) is the pure Y direction, (k hat) is the pure Z direction
WHAT I DON'T UNDERSTAND:+ Show Spoiler + So firstly, I don't even know how to prove that A and B meet at the point (and are thus co-planear) so I'm assuming that they do meet at a vertex. Beyond that, the problem is, to me, very abstract; I'm having an impossibly hard time separating vectors from magnitudes. I simply have no plan or direction for going about proving this, it's unlike any proof I've ever done before (geometric ones). I don't even know what I would need to get in order to prove perpendicular. I'm just lost
DISCLAIMER:+ Show Spoiler + I know homework threads are generally frowned upon but I don't think they are necessarily prohibited. If they are, please close this, I'm very sorry and I won't make another. I know some people on tl are very very good at math and I am very confused.
SUCCESS: Wow, that was prompt and accurate and generally awesome. Thanks you tl and specifically,
Divinek
Luddite
crate
NevilleS
statix
l10f
jonnyp
OhNoes
You guys are awesome, thank you for the help. I'm gonna go do this now; if I get stuck I'll be back. Thanks again : )
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i'm pretty sure they don't have to meet at a point, perpendicular doesn't mean they touch, only that their dot product = 0 (i think, it's been a while, sorry  ) so if you prove that their dot product = 0 you should be good
edit, they are coplanar already since any two vectors define a plane, although this assumes the two vectors aren't parallel
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does this help?
+ Show Spoiler +
Making the requirement of rotational invariance more stringent for the cross product, we need the cross product of two vectors to yield another vector. Consider, for instance, two 3-dimensional vectors u and v in a plane (two non-parallel vectors always define a plane, in the same way that two lines do. If we rotate this plane, the vectors will change direction, but we don't want the cross product w = u×v to change at all. However, if w has any non-zero components in the plane of u and v , those components will necessarily change under rotation (they get rotated just like everything else). The only vectors that won't change at all under a rotation of the u - v plane are those vectors that are perpendicular to the plane. Hence, the cross product of two vectors u and v must give a new vector which is perpendicular to both u and v .
oh and ill throw in the definition of the cross product too
u×v = (u 1 v 2 - u 2 v 1)k + (u 3 v 1 - u 1 v 3)j + (u 2 v 3 - u 3 v 2)i
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That site is great, I've been there before. I use that site to get concepts and I think I understand vectors, specifically dot and cross products pretty well. I just don't know how to prove any thing with them. I really should make an account there and ask, ty.
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you just have to do (A x B) dot A and (A x B) dot B. Both equal 0.
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Yeah, proving that (A x B) dot A = 0 and that (A x B) dot B = 0 shows that (A x B) is mutually perpendicular to A and B.
I like calculating the cross product using the matrix form. Calculate the determinant of the 3x3 matrix:
[i j k]
[Ax Ay Az]
[Bx By Bz]
Where i, j, k, are the unit vectors in those directions, Ax = the x-component of A, etc. That is taking the cross product. It's easier mathematically for most situations imo.
I can't be bothered to bold the vectors in my post; if you have trouble understanding it let me know.
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This proof lies in the definition of the cross product itself. I don't remember the specifics, but it'll just be a linear algebra thing. Just write out axb as an expression of Ax, Ay, Az and Bx, By and Bz and then do both cross-products. You will find (axb)*a = 0 since all x, y and z components will negate eachother, and the same for (axb)*b. I'd prove it for myself, but a) I don't remember the formal definitions for cross product (I believe dot is just sum of x+y+z...) and b) that would ruin the exercise for you...
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EDIT: more of the same.
a dot b = abval (a) * abval (b) cos theta, where theta is the angle between the two vectors.
a dot b = 0, the vectors are orthogonal.
Also, by it's definition, the cross product of a X b is perpendicular to a and b.
You can prove this by doing (a X b) dot a. If you get zero, hooray!
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+ Show Spoiler +C = ((Ay*Bz)-(By*Az))i+((Az*Bx)-(Bz*Ax))j+((Ax*By)-(Bx*Ay))k
so the dot product of the vectors A . C = Ax*Ay*Bz-AxBy*Az+Ay*Az*Bx-Ay*Bz*Ax+Az*Ax*By-Az*Bx*Ay = 0 thus they are perpendicular
hope it helps, i proved that the first vector was perpendicular to the cross product, you can do the rest 
hope there aren't any mistakes lol
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l10f
United States3241 Posts
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whatever you do don't ask micronesia for advice.
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On September 15 2009 12:04 Divinek wrote:does this help? + Show Spoiler +
Making the requirement of rotational invariance more stringent for the cross product, we need the cross product of two vectors to yield another vector. Consider, for instance, two 3-dimensional vectors u and v in a plane (two non-parallel vectors always define a plane, in the same way that two lines do. If we rotate this plane, the vectors will change direction, but we don't want the cross product w = u×v to change at all. However, if w has any non-zero components in the plane of u and v , those components will necessarily change under rotation (they get rotated just like everything else). The only vectors that won't change at all under a rotation of the u - v plane are those vectors that are perpendicular to the plane. Hence, the cross product of two vectors u and v must give a new vector which is perpendicular to both u and v .
oh and ill throw in the definition of the cross product too
u×v = (u 1 v 2 - u 2 v 1)k + (u 3 v 1 - u 1 v 3)j + (u 2 v 3 - u 3 v 2)i
I really want to understand what you said but it's not clicking unfortunately. I'm going to continue re-reading your post though, thank you very much. I don't think I understand "However, if w has any non-zero components in the plane of u and v , those components will necessarily change under rotation (they get rotated just like everything else". I know the cross product makes a perpendicular line to the plane of A and B (U and V) but aren't the components for vector C (W) in the perpendicular plane of C (and thus none of it's components exist in plane u of v)?
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On September 15 2009 12:17 l10f wrote:+ Show Spoiler +Sorry, I'm not that good with physics. Good luck though!
LOL, thats awesome dude, your a math prodigy
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On September 15 2009 12:08 Luddite wrote:
you just have to do (A x B) dot A and (A x B) dot B. Both equal 0.
Oh... that makes more and more sense the more I think about it... You're amazing, ty.
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l10f
United States3241 Posts
On September 15 2009 12:23 n.DieJokes wrote:LOL, thats awesome dude, your a math prodigy
Just give me a pm if you need solutions for any other Math-related questions 
+ Show Spoiler +I've taken AP Phys B if that means something
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hw threads aren't allowed / you're lucky. or mods just hate me
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It's really easy. Perpendicular means dot product is 0 by definition.
We'll pick vector U = ai + bj + ck and V = di + ej + fk
Just take the cross product W = UxV and show that the dot product of U and W as well as V and W are both 0. I leave the algebra to you.
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oh. ok. I got beaten to the punch.
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On September 15 2009 12:14 OhNoes wrote:
EDIT: more of the same.
a dot b = abval (a) * abval (b) cos theta, where theta is the angle between the two vectors.
a dot b = 0, the vectors are orthogonal.
Also, by it's definition, the cross product of a X b is perpendicular to a and b.
You can prove this by doing (a X b) dot a. If you get zero, hooray!
Just do this but instead of
a dot b = abval (a) * abval (b) cos theta
do:
a dot (axb) = abval(a) x abval(axb)cos(theta)
if you solve [a dot axb] then you will get 0 / (abval(a) x abval(axb)) = Cos theta
then Cos theta will be 0 and that proves that theta might be 90º or 270º doesn't matter because it still proves both vectores are orthogonal
you can do the same with b dot axb
cheers.
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