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Math Help - Equation of a plane through 3 points

  1. #1
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    Equation of a plane through 3 points

    There is a method of finding the equation by using this equation: r.(di+ej.fk)=1 and substituting the position vectors in turn for r. Then the simultaneous equations are solved for d, e and f. I don't know how it works. Why is the scalar product 1?
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  2. #2
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    Re: Equation of a plane through 3 points

    You did not mention one important point: di+ ej+ fk is a normal vector, perpendicular to the plane. Now, if r is a vector lying in the plane, in particular if r= <x- a, y- b, z- a> where (a, b, c) is given as a specific point in the plane and (x, y, z) represents a variable point in the plane, then r is perpendicular to di+ ej+ fk and so their dot product is 0.

    If you have three points in the plane, say, (a, b, c), (d, e, f), (g, h, i), then (a- d, b- e, c- f) and (a- g, b- h, c- i) are vectors lying in the plane and so you can take di+ ej+ fk to be the cross product of those two vectors.
    Last edited by HallsofIvy; June 6th 2013 at 01:23 PM.
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  3. #3
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    Re: Equation of a plane through 3 points

    Suppose that \hat{n} is a unit vector perpendicular to the plane and that \underline{a} is the position vector of some fixed point within the plane.

    If now \underline{r} is the position vector of another point within the plane, then (\underline{r}-\underline{a}) will be a vector in the plane and so perpendicular to \hat{n}.

    In that case, (\underline{r}-\underline{a}).\hat{n}=0,
    so
    \underline{r}.\hat{n}=\underline{a}.\hat{n},
    or
    \underline{r}.\left(\frac{\hat{n}}{\underline{a}. \hat {n}}\right)=1.

    This can now be written as
    \underline{r}.(d\hat{i}+e\hat{j}+f\hat{k})=1.

    Three different position vectors \underline{r} for three different points within the plane produce three equations in which the unknowns are d, e \text{ and } f.
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