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Math Help - closest point

  1. #1
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    closest point

    I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.
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  2. #2
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    Quote Originally Posted by chaddy View Post
    I need to find the point in the plane x + 2y + 3z = 6 that is nearest the origin.
    To do this, find the normal vector to the plane that passes through the origin.
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  3. #3
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    How would I find the normal vector?
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  4. #4
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    Does anyone know?
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  5. #5
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    Quote Originally Posted by chaddy View Post
    How would I find the normal vector?
    For the plane ax + by + cz = d, a normal vector is ai + bj + ck.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    For the plane ax + by + cz = d, a normal vector is ai + bj + ck.
    So the parametric equations of the line passing through the origin and in the direction of this normal vector will be:

    x = at
    y = bt
    z = ct

    The point where this line cuts the plane will be the point that's closest to the origin:

    a(at) + b(bt) + c(ct) = d \Rightarrow t = \frac{d}{a^2 + b^2 + c^2}.

    Therefore the coordinates of the point in the plane closest to the origin are:

    x = at = \frac{ad}{a^2 + b^2 + c^2}.

    y = bt = \frac{bd}{a^2 + b^2 + c^2}.

    x = at = \frac{cd}{a^2 + b^2 + c^2}.

    And as a bonus, the answer to what the closest distance is equal to is ..........
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  7. #7
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    <br />
x = \frac{6}{14}<br />

    <br />
y = \frac{12}{ 14}<br />

    <br />
z = \frac{18}{ 14}<br />

    So the point closest to the origin would be <  \frac{3}{7} ,  \frac{6}{7} ,  \frac{9}{7} >
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  8. #8
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    Quote Originally Posted by chaddy View Post
    <br />
x = \frac{6}{14}<br />

    <br />
y = \frac{12}{ 14}<br />

    <br />
z = \frac{18}{ 14}<br />

    So the point closest to the origin would be <  \frac{3}{7} ,  \frac{6}{7} ,  \frac{9}{7} >
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