Equation of a Plane: determine plane with three points?

[Trenters4325]

If the equation of a plane is given by,

ax+by+cz+d=0,

then how do you determine a plane with three points since you end up with only three equations and four unknowns when you plug them in.

 

[galactus]

Do you mean you have three points like, say, \(\displaystyle P_{1}(1,2,-1)\\P_{2}(2,3,1)\\P_{3}(3,-1,2)\)?.

Here's an example using these points.

Since the points \(\displaystyle P_{1},P_{2},P_{3}\) lie in the plane, the vectors

\(\displaystyle P_{1}P_{2}=(1,1,2)\ and\ P_{1}P_{3}=(2,-3,3)\) are parallel to the plane.

Therefore, \(\displaystyle P_{1}P_{2}XP_{1}P_{3}=\begin{vmatrix}1&1&2\\2&-3&3\end{vmatrix}=9i+j+k\\) is normal to the plane, since it is perpendicular to \(\displaystyle P_{1}P_{2}\ and\ P_{1}P_{3}\)

By using this normal and the point P(1,2,-1) in the plane, we arrive at

9(x-1)+(y-2)+5(z+1)=0

Which can be written as:

9x+y-5z-16=0

Does this example help?. Try it with yours.