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Direction angles - Proof

  1. Feb 25, 2009 #1
    Hello everyone,

    Thank you in advance for your help!

    1. The problem statement, all variables and given/known data

    10. A vector [itex] \vec{u} [/itex] with direction angles A1, B1, and Y1, is perpendicular to a vector [itex] \vec{v} [/itex] with direction angles A2, B2, and Y2. Prove that:
    [itex] \cos A1 \cos B2 + \cos B1 \cos B2 + \cos Y1 \cos Y2 = 0[/itex].

    3. The attempt at a solution

    I let [itex] \vec{u} = [a, b, c], \vec{v} = [x, y, z] [/itex].

    Since these are perpendicular, therefore:

    [itex] \vec{u} \bullet \vec{v} = ax + by + cz = 0 [/itex].

    Also, [itex] a, b, c, x, y, z [/itex] would all correspond to their direction cosines.

    However, I do not understand how I can prove the above statement with these facts. For example, would [itex] \cos A1 \cos A2 = 0 [/itex] simply because they are the components of two vectors which are parallel to each other?
  2. jcsd
  3. Feb 25, 2009 #2


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    what is a direction angle? i assume if the vector is length r then a = r.cosA1 etc....?

    then just put them in your dot product & you're pretty much there
  4. Feb 26, 2009 #3


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    The whole point of "direction cosines" is that if a vector [itex]\vec{v}[/itex] has direction cosines A1, B1, Y1, (I imagine that in your book those are [itex]\Alpha[/itex], [itex]\Beta[/itex], and [itex]\Gamma[/itex] and that you are told that they are the cosines of the angles the vector makes with the x, y, and z axes, respectively) Then [itex]\vec{v}= A1\vec{i}+ B1\vec{j}+ Y1\vec{k}[/itex]. That makes this problem simple.
  5. Feb 26, 2009 #4
    Thank you for your response.

    Would you mind elaborating on the proof?

    I thought that the direction cosines themselves were the unit vectors, so how would [itex] \cos A1 \cos A2 = 0 [/itex]? Shouldn't the dot product of these direction cosines = 0?
  6. Jun 8, 2009 #5
    Could anyone please offer an explanation for how to prove the above?

    Thank you!
  7. Jun 8, 2009 #6


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    The direction cosines aren't unit vectors. They are the coefficients of the unit vectors. In terms of the direction angles, u=|u|*(cos(A1)*i+cos(B1)*j+cos(Y1)*k) and v=|v|*(cos(A2)*i+cos(B2)*j+cos(Y2)*k). And, yes, u.v=0. Substitute the expressions for the vectors into the dot product.
  8. Jun 8, 2009 #7
    Thanks for your reply, Dick.

    I have:

    [tex] \vec{u} = |\vec{u}| \cos A1 \hat{i} + |\vec{u}| \cos B1 \hat{j} + |\vec{u}| \cos Y1 \hat{k} [/tex]

    [tex] \vec{v} = |\vec{v}| \cos A2 \hat{i} + |\vec{v}| \cos B2 \hat{j} + |\vec{v}| \cos Y2 \hat{k} [/tex]

    If I dot these two expressions on the RS, I do not see how I would get 0 as the final expression.

    Could you please explain more?
  9. Jun 8, 2009 #8


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    Your are GIVEN that the two vectors are perpendicular. If you dot them you will get an expression involving the cosines and |u| and |v|. You can set that equal to zero because you are GIVEN u.v=0.
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