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Dumb question about this linear algebra equation

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that any vector V in a plane can be written as a linear combination of two non-parallel vectors A and B in the plane; that is, find a and b so that V = aA + bB.

    Take components perpendicular to the plane to show that a = (B x V)[itex]\bullet[/itex]n / (B x A)[itex]\bullet[/itex]n

    2. Relevant equations



    3. The attempt at a solution

    (Upper case letters are vectors, it gets tiring to bold everything)

    BxV = B x (aA + bB) = a(BxA) + b(BxB) = a(BxA)

    BxV = a(BxA)

    Now, if I compare this to the solution given for a, I see that they took the dot product with n on both side, so

    (BxV)[itex]\bullet[/itex]n = a(BxA)[itex]\bullet[/itex]n

    It is then trivial to isolate a and find the right answer. However, I was wondering if I could take the dot product with any vector, since technically I'd be applying the same operation of both sides of the equation. If not, why does this only work with the unit vector perpendicular to the plane?
     
  2. jcsd
  3. Jul 21, 2012 #2

    Simon Bridge

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    What does it mean, geometrically, to take a dot product with n, as opposed to any other vector?
     
  4. Jul 21, 2012 #3
    since cos90 = 1 and |n| = 1, it basically gives the length of the cross products.
     
  5. Jul 21, 2012 #4

    vela

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    You can use any vector for n as long as it doesn't lie in the plane.
     
  6. Jul 21, 2012 #5

    Simon Bridge

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    geometrically, the dot product tells you the extent of one vector in the direction of the unit vector doesn't it?

    The cross product of the two vectors in the plane will give you a vector perpendicular to the plane. n is also perpendicular to the plane so the result will be how far above or below the plane the cross-product reaches. As you point out - that is just ||AxB|| etc. (Though I feel I should point out that cos(90) = 0, not 1.)

    What happens, then, if you use another vector, that is not perpendicular to the plane instead of n?

    But your purpose is to find "a"... is there a reason to prefer to use the unit normal compared with any arbitrary vector? Any vector not actually in the plane will get you a relation for a after all? (Note: you see why you cannot substitute a vector actually in the plane?)
     
  7. Jul 22, 2012 #6
    Thanks everyone. So, any vector outside of the plane can be taken, but using n guarantees that the equation is true. If the vectors lays in the plane, the dot product would be zero since all the vectors resulting from the cross products are perpendicular to the plane. (which is why it's a good idea to use n)

    Also, any vector outside of the plane can be used because all the vectors resulting from the cross products are parallel. Is that right?
     
  8. Jul 22, 2012 #7

    Simon Bridge

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    VxB || BxA sure.

    An arbitrary vector U will have length |U| and form an angle θ to the normal.
    If you used U instead of n in the equation, you get:

    |VxB||U|cosθ = a|BxA||U|cosθ
    ... you can solve this for a since you know all the values - but you picked U yourself so the question arises: is there a "best choice" for U?

    If |U|=1 and θ=0 then U=n and you get:

    |VxB| = a|BxA|

    which is simple and easy.
    So, per your original question, you can use any vector in the place of n and still achieve your goal ... but n is the simplest choice. The point is just to prove that a=|VxB|/|BxA| makes sense.

    In general, it is useful for understanding vectors to think about what is going on geometrically rather than as abstract algebra.
     
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