# Thread: Equation of a plane through 3 points

1. ## 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?

2. ## 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.

3. ## Re: Equation of a plane through 3 points

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

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

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

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

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