## 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}