# Plane Through Three Points

## Plane Through Three Points

This is done in a series of small steps;

• Select One Point to be the Primary Point(P)
• Create Two vectors from the Primary Point(P) to the remaining Points(A) and(B)
• Create the Plane Vector(N) by creating the Cross Product of Vectors (PA) and (PB)
• Resolve the Plane Equation

From Points (P) 30 ; 200 ; 50 , Point(A) 1 ; 2 ; 3 and Point(B) 100 ; 150 ; 200 create a new plane:

### Vectors PA and PB

\overrightarrow{PA} \\
= Ax - Px ; Ay - Py ; Az - Pz \\
= 1 - 30 ; 2 - 200 ; 3 - 50 \\
\overrightarrow{AB}= -29 ; -198 ; -47 \\
I = -29 ; J = -198 ; K = -47

and

\overrightarrow{PB} \\
= Bx - Px ; By - Py ; Bz - Pz \\
= 100 - 30 ; 150- 200 ; 200 - 50 \\
\overrightarrow{AB}= 70 ; -50 ; 150 \\
I = 70 ; J = -50 ; K = 150

### Vector N by Vector Cross Product

\overrightarrow{N}  = \overrightarrow{PA} × \overrightarrow{PB} \\
=\begin{bmatrix}
i & j & k \\
-29 & -198 & -47 \\
70 & -50 & 150
\end{bmatrix}=
\begin{bmatrix}
-198 & -47 \\
-50 & 150
\end{bmatrix} i-
\begin{bmatrix}
-29 & -47 \\
70 & 150
\end{bmatrix} j-
\begin{bmatrix}
-29 & -198 \\
70 & -50
\end{bmatrix} k \\
i=((-198⋅150)−(-50⋅-47)) \\
=((-29,700)−(2,350)) \\
=\boldsymbol{-32,050i} \\
j=−((-29⋅150)−(70⋅-47)) \\
=−((-4,350)-(-3,290)) \\
=-(-1,060) \\
=\boldsymbol{1,060j} \\
k=+((-29⋅-50)−(70⋅-198)) \\
=+((1,450)-(-13,860)) \\
=\boldsymbol{15,310k} \\


### Vector Magnitude

||{N}|| \\
= \sqrt{I^2 + J^2 + K^2}  \\
= \sqrt{−32,050^2 + 1060^2 + 15310^2}  \\
= \sqrt{1,027,202,500 + 1,123,600 + 234,396,100}  \\
= \sqrt{1,262,722,200}  \\
||{N}||= 35,534.8026  \\

### Unit Vector

\widehat{N} \\
= \overrightarrow{N} / ||{N}|| \\
= −32,050 / 35,534.8026 ; 1060 / 35,534.8026 ; 15310 / 35,534.8026 ; \\
= -0.9019 ; 0.02982 ; 0.43084 \\
\hat{I} = -0.9019 ;  \hat{J} = 0.02982 ; \hat{K} = 0.43084

Now we have the Unit Vector we can calculate the last component of the plane equation (D), to do this well plug one of the three points back into the equation, in this case Point(A) 1 ; 2 ; 3:

=−0.9019x+0.02982y+0.43084z−D=0\\
=(−0.9019 \cdot 1 ) + ( 0.02982 \cdot 2) + ( 0.43084 \cdot 3) −D = 0 \\
=(−0.9019 ) + ( 0.05964 ) + ( 1.29252 ) - D = 0 \\
= 0.45026 - D = 0 \\
D = -0.45026 \\
=−0.9019x+0.02982y+0.43084z−(-0.45026)=0\\


Easy Right?