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Thread: Projection Vectors Problem

  1. #1
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    Projection Vectors Problem

    The angle between two vectors a and b is ∅, where ∅≠90. Under what conditions will |Projab|2 + |Projba|2 = 1?

    Don't really know where to start on this one.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Projection Vectors Problem

    Quote Originally Posted by anonymoususer20 View Post
    The angle between two vectors a and b is ∅, where ∅≠90. Under what conditions will |Projab|2 + |Projba|2 = 1?

    Don't really know where to start on this one.
    Hi anonymoususer20!

    I'm not entirely sure what you mean with Projab.
    Presumably it is:

    $\displaystyle \text{Proj}_{ab} = \mathbf a \cdot \hat{\mathbf b}$

    where $\displaystyle \hat{\mathbf b}$ is the unit vector in the direction of $\displaystyle \mathbf b$.

    If that is the case you can substitute $\displaystyle \mathbf a \cdot \hat{\mathbf b} = a \cos \phi$.
    And you can substitute something similar for the other projection.

    Have you tried that?
    If so, what did you get?
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    Re: Projection Vectors Problem

    Projab is a projection vector.

    |Pab|= |a.b|/|b|.
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    Super Member ILikeSerena's Avatar
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    Re: Projection Vectors Problem

    Quote Originally Posted by anonymoususer20 View Post
    Projab is a projection vector.

    |Pab|= |a.b|/|b|.
    Aha! Then it is exactly what I suspected, since $\displaystyle \mathbf{\hat b} = \frac{\mathbf b}{|\mathbf b|} = \frac{\mathbf b}{b}$.
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    Re: Projection Vectors Problem

    Quote Originally Posted by anonymoususer20 View Post
    Projab is a projection vector.

    |Pab|= |a.b|/|b|.

    Quote Originally Posted by ILikeSerena View Post
    Aha! Then it is exactly what I suspected, since $\displaystyle \mathbf{\hat b} = \frac{\mathbf b}{|\mathbf b|} = \frac{\mathbf b}{b}$.

    There seems to be somewhat of confusion on notation here.
    I have never seen the notation $\displaystyle \text{Proj}_{ab}$.
    In North American text books $\displaystyle \text{Proj}_{\vec{b}}\;\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}{$

    If we use that definition then $\displaystyle \left\|\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}\right\|^2=| \vec{a}\cdot\vec{b}|^2$

    Now that makes a really interesting question.
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    Super Member ILikeSerena's Avatar
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    Re: Projection Vectors Problem

    Quote Originally Posted by Plato View Post
    There seems to be somewhat of confusion on notation here.
    I have never seen the notation $\displaystyle \text{Proj}_{ab}$.
    In North American text books $\displaystyle \text{Proj}_{\vec{b}}\;\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}{$

    If we use that definition then $\displaystyle \left\|\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}\right\|^2=| \vec{a}\cdot\vec{b}|^2$

    Now that makes a really interesting question.
    That does make more sense and is probably what it means.
    I didn't think to include the direction.

    However, $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$
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    Re: Projection Vectors Problem

    Quote Originally Posted by ILikeSerena View Post
    That does make more sense and is probably what it means.
    I didn't think to include the direction.
    However, $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$
    Actually no, it equals $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 =|{\vec{a}\cdot \vec{b}|^2$.

    So we get $\displaystyle |{\vec{a}\cdot \vec{b}|^2=\frac{1}{2}$
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  8. #8
    Super Member ILikeSerena's Avatar
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    Re: Projection Vectors Problem

    $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec a \cdot \vec b|^2} {|\vec{b}\cdot\vec{b}|^2} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^4} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^2}$
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