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Math Help - Vector Arithmetic Problem

  1. #1
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    Vector Arithmetic Problem

    Let \vec{u} be an arbitrary fixed unit vector and show that any vector \vec{b} satisfies b^2=(\vec{u}\cdot\vec{b})^2+ (\vec{u}\times\vec{b})^2

    Explain this result in words, with the help of a picture.


    So I got to the point where I can see the end in sight: b_1(u_1^2+u_2^2+u_3^2)+b_2(u_1^2+u_2^2+u_3^2)+b_3(  u_1^2+u_2^2+u_3^2)-2(u_2 u_3 b_2 b_3+u_1 u_3 b_1 b_3 +u_1 u_2 b_1 b_2)

    Obviously the (u_1^2+u_2^2+u_3^2)'s will all go away since u is a unit vector. My only problem is that I have some leftover terms from the dotting of the cross products which I don't see how to get rid of since u and b are so generalized. Since the first part seems to be right, I have to assume that I either did something wrong in the cross product arithmetic or else I'm missing some property of dot products that will make that last term go away. Any thoughts?
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  2. #2
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    Quote Originally Posted by davesface View Post
    Let \vec{u} be an arbitrary fixed unit vector and show that any vector \vec{b} satisfies b^2=(\vec{u}\cdot\vec{b})^2+ (\vec{u}\times\vec{b})^2

    Explain this result in words, with the help of a picture.
    Here's a very brief explanation in words, without the help of a picture. The quantity \vec{u}\cdot\vec{b} represents the magnitude of the component of \vec{b} in the direction of \vec{u}. The quantity \vec{u}\times\vec{b} represents the component of \vec{b} perpendicular to \vec{u}. The result then follows from Pythagoras' theorem.
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  3. #3
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    The picture isn't so much the issue, since the geometric interpretations of dot and cross products is pretty simple. I just included that last sentence as a part of the full problem statement. The real issue I'm having is with that -2... term that's leftover.

    I'll add that, upon starting the problem from scratch again, I see that it is much easier to do using the trigonometric definitions of dot and cross products. I'm convinced that the equation holds true from that, but out of curiosity I still want to know what I did wrong the first time.
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