Intersection of Two Planes
Lets take two planes and determine the vector form for the line of intersection.
Plane(A) = 6x+7y+5z−37=0 \\ Plane(B) = 5x+4y+13z−40=0 \\
Vector Cross Product
Let’s start with the vector cross product of the two planes, this will define the intersection vector, and the first part of the lines vector form:
\overrightarrow{N} = \overrightarrow{PA} × \overrightarrow{PB} \\
=\begin{bmatrix}
i & j & k \\
6 & 7 & 5 \\
5 & 4 & 13
\end{bmatrix}=
\begin{bmatrix}
7 & 5 \\
4 & 13
\end{bmatrix} i-
\begin{bmatrix}
6 & 5 \\
5 & 13
\end{bmatrix} j-
\begin{bmatrix}
6 & 7 \\
5 & 4
\end{bmatrix} k \\
i=((7⋅13)−(4⋅5)) \\
=((91)−(20)) \\
=\boldsymbol{71i} \\
j=−((6⋅13)−(5⋅5)) \\
=−((78)-(25)) \\
=-(53) \\
=\boldsymbol{-53j} \\
k=+((6⋅4)−(5⋅7)) \\
=+((24)-(35)) \\
=+(-11) \\
=\boldsymbol{-11k} \\
Now we know the direction of the intersection line. Next we have to establish a point along the intersection, so we can correctly position the vector in space. Let’s rearrange the equations in terms of X.
Defining a Point Along the Intersection
Plane(A) = 6x+7y+5z−37=0 \\ Plane(B) = 5x+4y+13z−40=0 \\ x = (-7y-5z+37)/6 \\ x = (-4y-13z+40)/5 \\ (-7y-5z+37)/6 = (-4y-13z+40)/5 \\ -1.166y -0.833z + 6.166 = -0.8y-2.6z+8 \\ -0.833z+2.6z = 1.166y-0.8y+8-6.166 \\ 1.767z = 0.366y+2.166 \\ z=0.207y+1.225
Let’s substitute Z back into one of the equations:
x = (-7y-5z+37)/6 \\ x = (-7y-5(0.207y+1.225)+37)/6 \\ x = (-7y-1.035y+6.125+37)/6 \\ x = (-8.035y+43.125)/6 \\ x = -1.339y + 7.1875
Let use Parameter T to equal Y: (Vector(T) correct x,y,z incorrect)
x = 7.1875 -1.339T\\ y = T \\ z=1.225 + 0.207T
In vector form (this is correct, from CAD, i need to rework calcs)
\utilde{R} = \begin{pmatrix}
71 \\
-53 \\
-11
\end{pmatrix}
+ \lambda
\begin{pmatrix}
-12 \\
5 \\
\end{pmatrix}