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Calculating a vector from a known angle and a known vector

  1. Jun 21, 2011 #1
    I've been wrestling with this problem for several days. I really wanted to solve it myself, but at this point I just keep barking up the same wrong trees. Any help would be greatly appreciated. Here is the problem -

    I am trying to determine Vector N', given the following:

    Vector A = [5,3,1]
    Vector N = [3,2,-21]
    Vector A' = [5,3,2] <-- Vector A has moved by one unit in Z

    Furthermore,

    N is normal to A,
    N' is normal to A',
    and N' is to A' what N is to A

    So far I've determined the following:

    theta(A,A') = 9.172 deg = theta(N,N')
    AdotA' = 36
    ScalarA = 5.916
    ScalarN = 21.307
    ScalarA' = 6.164

    I really want to figure this out on my own, but I am convinced there is some simple relationship that I am missing. If someone could point me in the right direction I'd be grateful.
     
  2. jcsd
  3. Jun 21, 2011 #2

    Philip Wood

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    Would you please explain what is meant by "N' is to A' what N is to A".
     
  4. Jun 21, 2011 #3
    Philip,

    Thanks for looking at the post. I was afraid that statement would be unclear. What I meant was that as vector A moves to become vector A', vector N experiences the same motion, thus becoming N'.

    If you make an "L" with your thumb and your index finger. Think of your thumb as Vector A and your index finger as Vector N. Now rotate (assume no translation) your hand to a new position, your thumb would now be A' and your index finger would be N'.

    Does that make sense?
     
  5. Jun 21, 2011 #4

    Philip Wood

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    I'm envisaging your vectors A and A' as position vectors, i.e. as giving displacements of points from the origin. In that way I can make sense of your talking about the vector A 'moving' by one z unit, as I suppose you to mean that the point moves by one z unit. Vector A' is slightly longer than vector A and its direction is also different.

    A and A' define a plane, P. It's quite easy to picture this plane, as it stands 'upright' on the x-y plane. I find that the vector normal to this plane P is (3a, -5a, 0), in which a is 1/sqrt(34), if the normal is to be of unit length.

    Now I think that when you change A to become A' and move N to become N' in the way you describe, then N' will be at the same angle to the normal to the plane as N was. Am I right?

    If so, I think I know what to do, but I'm not at all sure that I've interpreted the question properly.
     
  6. Jun 21, 2011 #5
    One way to solve it is to define a vector V to be the cross product of N and A (N x A = V). You can say that N' x A' = V as well, so then given an N, A and A', you can solve for the N'.

    In this example, N x A = [-65,108,1] = V. N' = [n1,n2,n3] and N' x A' = [ 2*n2 - 3*n3, 5*n3 - 2*n1, 3*n1 - 5*n2]; If N' x A' = V, then N' = [109, 65.3,65.2];

    I'm not sure this gives you what you want, but it's a solution. I think there might be other solutions to the system of equations that would get N' close to the original direction of N.
     
  7. Jun 21, 2011 #6
    Philip,

    I think you are correct, but I haven't been able to fully convince myself of it (sorry, that third dimension is giving me fits). Intuitively it seems to make sense, though and I haven't been able to disprove it.

    There are few easy checks we can do to verify the results (for one A' dot N' should equal 0).

    Thanks again for the help.

    Regards
     
  8. Jun 21, 2011 #7
    Tim,

    Thanks for the reply. I'm not sure the assumption that N' X A' = V is correct. For one, since N' is normal to A', N' dot A' = 0. Also, if V is normal to the plane created by N and A, and N and A move (becoming N' and A') you would expect V to move as well becoming V'. I can imagine a special case where V = V', but that's only for very specific circumstances (I think). Finally, If A' is similar to A, I would expect (perhaps incorrectly) N' to be somewhat similar to N.

    I appreciate the help. If I've missed something please let me know.

    Regards
     
  9. Jun 22, 2011 #8

    Philip Wood

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    Michael,
    Using my interpretation of "N' is to A' what N is to A", as explained above, I find N' to be (5.84, 3.70, -20.2). This assumes that N' has the same magnitude (length) as N. My method uses two equations based on dot products, and I'd happily explain it, but...

    What makes this problem difficult is interpreting "N' is to A' what N is to A", making me ask: who set this problem? It is, in my opinion, very poorly worded. If I'd been set it, I'd ask the teacher to re-word so it makes proper sense. If it came out of a book, I'd wonder about the competence of the author. If I'd set it to test myself, I'd try and make the wording precise, a useful exercise in itself because it will help you to clarify your ideas, but an exercise that should be abandoned if you're not getting anywhere - there are plenty of tightly-worded problems out there, I'm sure.
     
  10. Jun 22, 2011 #9
    Philip,

    Regarding the wording of the sentence - this problem is part of a bigger exercise posed to the class by the professor (briefly, we were asked to determine the impact on alignment if the incoming beam is stationary but the reflective surface moves. I've been able to quantify every other aspect of the problem, this is the last piece to the puzzle). The "N' is to A' what N is to A" portion came out of side discussions with the teaching assistant and probably made more sense in the context of the discussion.

    In any event, your solution seems to pass the litmus tests. For one, theta(N,N') = 9.172°. Also, A' dot N' is about zero (Actually, I calculated -0.1, but this could just be a rounding error somewhere) I set up I'd be curious to hear what your approach was...

    Regards
     
  11. Jun 22, 2011 #10

    Philip Wood

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    Michael. Thank you for explaining how the problem arose. Here's my method.

    Notation: N' = (p, q, r).

    (1) I assume that N' is to have the same magnitude (length) as N', in which case
    p2 + q2 + p2 = 32 + 22 + (-21)2 = 454

    (2) N' is perpendicular to A', so N'.A' = 0.
    So 5p + 3q + 2r = 0

    (3) A unit vector normal to the plane containing A and A' is U = (3a, -5a, 0), in which a = 1/sqrt(34).

    If N' and N are both at the same angle to U, then, because N' and N have the same magnitude, N'.U = N.U. So
    3ap - 5aq = 9a - 10a, that is: 3p - 5q = -1 .

    I used the nice simple linear equations from (1) and (2) to express q and r in terms of p. Then I substituted these expressions for q and r into the equation from (1). This gave me a quadratic equation for p. I chose the most plausible root!

    Best wishes.
     
  12. Jun 22, 2011 #11
    IMHO, the problem that I see is that in 3-space (R3), there are an infinite amount of solutions to the equation N'.A'=0. The constraints you choose allow you to get a solution. In Philip's solution, he chose the constraints that 1) U = A x A' and 2) N'.U = N.U and 3) |N|=|N'|. Even with those 3 constraints, Philip ended up with 2 solutions. As long as you're fine with those constraints, then this is a great solution. However, you might want to check that these constraints can be used in the bigger exercise that this is a part of. I didn't think the problem all the way through when I posted my method, which doesn't even give you a good solution (I posted way too quickly).
     
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