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Math Help - Orthogonality

  1. #1
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    Orthogonality

    Let \vec{x} and \vec{y} be linearly independent vectors in R^2. If || \vec{x}|| = 2 and ||[ \vec{y}|| = 3, what, if anything, can we conclude about the possible values of | \vec{x}^{T}\vec{y}|?

    So this is what I have so far...

    2 = \sqrt{x_{1}^{2}+x_{2}^{2}}
    3 = \sqrt{y_{1}^{2}+y_{2}^{2}}

    After some algebra I got...

    \vec{x}^{T}\vec{y}=(\sqrt{4-x_{2}^{2}}\sqrt{4-x_{1}^{2}})(\sqrt{9-y_{2}^{2}}\sqrt{9-y_{1}^{2}})^{T}

    This is where I'm stuck, what can I determine from this? Thanks in advance!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pakman View Post
    Let \vec{x} and \vec{y} be linearly independent vectors in R^2. If || \vec{x}|| = 2 and ||[ \vec{y}|| = 3, what, if anything, can we conclude about the possible values of | \vec{x}^{T}\vec{y}|?

    So this is what I have so far...

    2 = \sqrt{x_{1}^{2}+x_{2}^{2}}
    3 = \sqrt{y_{1}^{2}+y_{2}^{2}}

    After some algebra I got...

    \vec{x}^{T}\vec{y}=(\sqrt{4-x_{2}^{2}}\sqrt{4-x_{1}^{2}})(\sqrt{9-y_{2}^{2}}\sqrt{9-y_{1}^{2}})^{T}

    This is where I'm stuck, what can I determine from this? Thanks in advance!
    If I'm not mistaken this looks like it's an Euclidean dot product, so
    |\vec{x}^T \vec{y}| = ||\vec{x}||~||\vec{y}||~|cos(\theta)| which ranges from 0 to 2 * 3 = 6.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    If I'm not mistaken this looks like it's an Euclidean dot product, so
    |\vec{x}^T \vec{y}| = ||\vec{x}||~||\vec{y}||~|cos(\theta)| which ranges from 0 to 2 * 3 = 6.

    -Dan
    How do you figure that it ranges from 0 to 6? I am a bit confused...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pakman View Post
    How do you figure that it ranges from 0 to 6? I am a bit confused...
    ||\vec{x}|| and ||\vec{y}|| are constants. And |cos(\theta)| ranges from 0 to 1.

    -Dan
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    Quote Originally Posted by topsquark View Post
    ||\vec{x}|| and ||\vec{y}|| are constants. And |cos(\theta)| ranges from 0 to 1.

    -Dan
    Ahh okay makes sense now thanks!
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