1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    lanedance

    User Avatar
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook