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Length of vector perpendicular to plane

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\vec{a},\vec{b},\vec{c}[/itex] be three constant vectors drawn from the origin to the points [itex]A,B,C[/itex]. What is the distance from the origin to the plane defined by the points [itex]A,B,C[/itex]? What is the area of the triangle [itex]ABC[/itex]?

    2. Relevant equations

    3. The attempt at a solution
    Starting with the first part:

    The distance is the length of a perpendicular vector from the origin to the plane. If that vector is [itex]\vec{r}[/itex], then
    [tex]\vec{r} \cdot (\vec{a} - \vec{b}) = \vec{r} \cdot (\vec{b} - \vec{c}) = \vec{r} \cdot (\vec{c} - \vec{a}) = \vec{r} \cdot (\vec{a} - \vec{r}) = 0\\ \vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} = r^2[/tex]
    by perpendicularity.

    I can also get
    [tex]\vec{r} \times [(\vec{a} - \vec{b}) \times (\vec{b} - \vec{c})] = 0[/tex]
    by a similar argument, but I don't think it really helps.

    I'm stuck here. Is there another relationship that I'm missing?
     
  2. jcsd
  3. Jun 27, 2012 #2
    There's a simple formula that if you can write the equation of the plane in the form of [itex]ax + by + cz + d = 0[/itex], then the distance from that plane to the origin is [itex]d/\sqrt{a^2 + b^2 + c^2}[/itex]. To be honest, I'm not sure how to prove this using just regular 3d vector geometry. This would be an ideal situation to use homogeneous coordinates, but I expect that's a bit more advanced.
     
  4. Jun 27, 2012 #3

    vela

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    You said ##\vec{r}\cdot\vec{a} = r^2##. If you used the unit vector ##\hat{r}=\vec{r}/||\vec{r}||## instead, you'd have ##\hat{r}\cdot\vec{a} = r##. Can you think of a way to find the unit normal to the plane in terms of ##\vec{a}##, ##\vec{b}##, and ##\vec{c}##.
     
  5. Jun 27, 2012 #4
    The cross product of two vectors between points in the plane is normal to the plane, so we have
    [tex]\hat{r} = \frac{(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})}{\|(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})\|}[/tex]

    [tex](\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \vec{a} \times \vec{c} - \vec{a} \times \vec{b} - \vec{b} \times \vec{c} + \vec{b} \times \vec{b} = \vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}[/tex]

    [tex]r = \vec{a} \cdot \hat{r} = \frac{\vec{a} \cdot (\vec{a} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{b})}{\|\vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}\|} = \frac{\vec{a} \cdot (\vec{c} \times \vec{b})}{\|\vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b}\|}[/tex]

    Thanks for the help.
     
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