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Thread: Finding coordinate vector of point

  1. #1
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    Finding coordinate vector of point

    A point P in $\displaystyle \mathbb{R}^n$ has coordinate vector $\displaystyle \mathbf{p}$. Find the coordinate vector of the point Q which is the reflection of P in the line $\displaystyle l$ which passes through the point $\displaystyle \mathbf{a}$ parallel to the direction $\displaystyle \mathbf{d}$
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  2. #2
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    Quote Originally Posted by acevipa View Post
    A point P in $\displaystyle \mathbb{R}^n$ has coordinate vector $\displaystyle \mathbf{p}$. Find the coordinate vector of the point Q which is the reflection of P in the line $\displaystyle l$ which passes through the point $\displaystyle \mathbf{a}$ parallel to the direction $\displaystyle \mathbf{d}$
    I'm not certain if this will help:

    1. Draw a sketch.

    2. Since Q is the image of P by refelection about the line l the midpoint of $\displaystyle \overline{PQ}$ belongs to the line l and the vector $\displaystyle \vec d$ is perpendicular to $\displaystyle \overrightarrow{PQ}$. That means:

    $\displaystyle \frac12(\vec p + \vec q)=\vec a + r \cdot \vec d$
    and
    $\displaystyle \vec d \cdot \overrightarrow{PQ} = 0$

    3. Solve for $\displaystyle \vec q$.

    I've got:

    $\displaystyle \vec q = 2\vec a + 2r \cdot \vec d - \vec p$
    Attached Thumbnails Attached Thumbnails Finding coordinate vector of point-pktspieglangerade.png  
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  3. #3
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    Quote Originally Posted by earboth View Post
    I'm not certain if this will help:

    1. Draw a sketch.

    2. Since Q is the image of P by refelection about the line l the midpoint of $\displaystyle \overline{PQ}$ belongs to the line l and the vector $\displaystyle \vec d$ is perpendicular to $\displaystyle \overrightarrow{PQ}$. That means:

    $\displaystyle \frac12(\vec p + \vec q)=\vec a + r \cdot \vec d$
    and
    $\displaystyle \vec d \cdot \overrightarrow{PQ} = 0$

    3. Solve for $\displaystyle \vec q$.

    I've got:

    $\displaystyle \vec q = 2\vec a + 2r \cdot \vec d - \vec p$
    Thank you so much. That was really helpful
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