Line of Intersection of 2 Planes:
P1: 5x – 3y – 2z = 7, and P2: 2x – y + 3z = 4.
The line of intersection, L, lies in both planes, and hence is perpendicular to the normal vectors of both planes. The cross product, n1 ´ n2, is also perpendicular to n1 and n2, so it can be used as a reference vector in the direction of the line.
From the equations, we can determine the normal vectors:
n1 = 5i – 3j – 2k, and
n2 = 2i – j + 3k.
Now we form the cross product:
= –11i – 19j + k
Now we need a point A on the line. Let's set x to 0 in each equation and solve for yA and zA.
From P1: 0 – 3yA –2zA = 7
From P2: 0 – yA + 3zA = 4
Solving, we get
, and .
We can use this point in setting up the equations of the line:
x = 0 –11t, y = –29/11 – 19t, z = 5/11 + t
or in symmetric form,