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Points On A Plane

  1. Jul 3, 2008 #1

    danago

    User Avatar
    Gold Member

    The three points (1,1,-1), (3,3,2) and (3,-1,-2) determine a plane. Find:
    1. A vector normal to the plane
    2. The equation of the plane
    3. The distance of the plane from the origin


    For part 1 i just said let:
    [tex]
    \begin{array}{l}
    \overrightarrow {OA} = < 1,1, - 1 > \\
    \overrightarrow {OB} = < 3,3,2 > \\
    \overrightarrow {OC} = < 3, - 1, - 2 > \\
    \end{array}
    [/tex]

    then:

    [tex]
    \begin{array}{l}
    \overrightarrow {AB} = < 2,2,3 > \\
    \overrightarrow {AC} = < 2, - 2, - 1 > \\
    \end{array}
    [/tex]

    Since the vectors AB and AC both lie on the plane, i just need to find a vector normal to both.

    [tex]
    \overrightarrow {AB} \times \overrightarrow {AC} = < 4,8, - 8 >
    [/tex]


    For the equation of the plane i just said

    [tex]
    \left( {\vec{p} - \left( {\begin{array}{*{20}c}
    1 \\
    1 \\
    { - 1} \\
    \end{array}} \right)} \right) \cdot \left( {\begin{array}{*{20}c}
    4 \\
    8 \\
    { - 8} \\
    \end{array}} \right) = 0
    [/tex]

    Where vector p is some point on the plane.


    For the final part, i think i did it correctly, but not sure if it is the most efficient way of doing it. I assumed that the question was supposed to say the perpendicular distance.

    What i did was find the projection of OA (since A lies on the plane) onto a vector which is normal to the plane, n, and then calculate its magnitude.

    [tex]
    \begin{array}{l}
    {\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\overrightarrow n .\overrightarrow n }}\overrightarrow n = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}}\widehat{n} \\
    \therefore \left| {{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} } \right| = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = 5/3 \\
    \end{array}
    [/tex]

    Does that look ok? Its mainly the last part i was a bit iffy about; i think ive done parts 1 and 2 correctly, unless ive made errors in my calculations.

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. Jul 3, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    I got the same answer for your first two parts. For the last part you have to find the distance of the perpendicular line from O to the plane. I don't think you need to use projection here. Since the normal is already perpendicular to the plane, you can let [itex]k \vec{n}[/itex] be the vector from the origin to the plane. Clearly the individual directional scalar components of [itex]k\vec{n}[/tex] satisfy the equation of the plane, since [itex]k\vec{n}[/itex] is itself on the plane. You can make use of that to find the length of kn.
     
  4. Jul 4, 2008 #3

    danago

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    Gold Member

    Oh so pretty much form the equation:

    [tex]
    \left( {\begin{array}{*{20}c}
    {4k - 1} \\
    {8k - 1} \\
    { - 8k + 1} \\
    \end{array}} \right) \cdot \left( {\begin{array}{*{20}c}
    4 \\
    8 \\
    { - 8} \\
    \end{array}} \right) = 0
    [/tex]

    And then solve for k? And then find the magnitude of [tex]k\vec{n}[/tex]? I did all that and got the same answer so atleast i know my reasoning in the first method i did was correct :smile:

    Thanks again for your help.
     
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