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Math Help - direction cosines/vector help

  1. #1
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    Exclamation direction cosines/vector help

    Hi everyone! I'm working on a vector question involving finding and proving direction cosines/sines/tangents of vector (a,b,c), u = magnitude

    I know:
    cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares). But I don't understand the second part. I know it must be pretty simple and related to unit vectors, and I'm just missing something, so any help would be very much appreciated! Also, any help on the sine/tangent part would be useful, too.

    Thanks so much for any and all help!
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  2. #2
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    Quote Originally Posted by JennK View Post
    Hi everyone! I'm working on a vector question involving finding and proving direction cosines/sines/tangents of vector (a,b,c), u = magnitude

    I know:
    cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares). But I don't understand the second part. ...
    Probably you mean by second part how to calculate the magnitude of the vector(?). If so:

    u = \sqrt{a^2+b^2+c^2}
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  3. #3
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    Hi! Yes, I mean the magnitude. And I understand the formula you posted...but if u = square root of a2 + b2 + c2, why is the formula cos2[alpha] + cos2[beta] + cos2[gamma] = 1, and not the square root of (cos2[alpha] + cos2[beta] + cos2[gamma]) = 1?

    Please help, and thank you!
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  4. #4
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    Quote Originally Posted by JennK View Post
    Hi! Yes, I mean the magnitude. And I understand the formula you posted...but if u = square root of a2 + b2 + c2, why is the formula cos2[alpha] + cos2[beta] + cos2[gamma] = 1, and not the square root of (cos2[alpha] + cos2[beta] + cos2[gamma]) = 1?

    Please help, and thank you!
    To answer this question first: 1 = \sqrt{1}

    I know:
    cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares).
    If \cos(\alpha)=\frac au ~\implies~ a =u \cdot \cos(\alpha) and

    b =u \cdot \cos(\beta)

    c =u \cdot \cos(\gamma)

    Now consider the term:

    a^2+b^2+c^2 = (u \cdot \cos(\beta))^2 + (b =u \cdot \cos(\alpha))^2 + (u \cdot \cos(\gamma))^2 = u^2 \cdot (\cos(\alpha))^2 + u^2 \cdot (\cos(\beta))^2 + u^2 \cdot (\cos(\gamma))^2 = u^2((\cos(\alpha))^2 + (\cos(\beta))^2 + (\cos(\gamma))^2) = u^2 \cdot 1

    and therefore: u = \sqrt{a^2+b^2+c^2}
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