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Math Help - [proof] Angles Between Two Vectors and Their Projections Are Equal

  1. #1
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    [proof] Angles Between Two Vectors and Their Projections Are Equal

    Q: Assume \vec{x}, \vec{y} \in\mathbb{R}^n such that \vec{x}\cdot\vec{y}\neq0, prove that the angle between \vec{x} and \vec{y} is equal to the angle between proj_{\vec{x}}\vec{y} and proj_{\vec{y}}\vec{x}

    I've tried various things but I have yet to discover a good (formal) way of tackling this. I've written on the first line:

    \cos\theta=\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\  |\vec{y}\|}=\frac{proj_{\vec{x}}\vec{y}\cdot proj_{\vec{y}}\vec{x}}{\|proj_{\vec{x}}\vec{y}\|\|  proj_{\vec{y}}\vec{y}\|}

    Now I've tried expanding this but I don't think it leads me anywhere so next I write the definition of projection vector:

    proj_{\vec{x}}\vec{y} = (\frac{\vec{x}\cdot\vec{y}}{\vec{x}\cdot\vec{x}})\  vec{x}

    Here it occurs to me that \vec{x}\cdot\vec{y} can be either positive or negative, therefore the projections will be either in the same direction or the opposite direction of \vec{x} and \vec{y}. So this leads me to believe that the projections do have the same angle by the congruence law of intersecting lines(?) Is this somewhat correct? Does it hold for \mathbb{R}^n, n>2? How do I format this into a formal proof?
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  2. #2
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    The projection of \vec{x} on \vec{y} is, by definition, a vector in the same direction as \vec{y} and the projection of \vec{y} on \vec{x} is a vector in the same direction as \vec{x}, so of course, the angle between them is the same.
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  3. #3
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    You should know that \dfrac{\vec{x}\cdot\vec{y} }{\vec{x}\cdot\vec{x}}~\&~\dfrac{\vec{x}\cdot\vec{  y} }{\vec{y}\cdot\vec{y}} must have the same sign.

    Also  \left\|\text{proj}_{\vec{x}}\vec{y}\right\|=\left| \frac{\vec{x}\cdot\vec{y} }{\vec{x}\cdot\vec{x}} \right|\left\|\vec{x}\right\|.
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