Find the parametric equations for the line in which the plan

Find the parametric equations for the line in which the planes x + 4y-2z = 2 and 3x-y + 2z = 4 intersect.

solution

The planes have normal vectors a = (1, 4, -2) and b = (3, -1, 2), respectively.

Let L denote the line of intersection. Then v = a × b = (6, -8,-13) is parallel to L. We only need to find a point P on L

To find P, solved the system of equations of the planes

suppose z=0

x+4y=2=x=2-4y

3x-y=4=3(2-4y)-y=4

6-12y-y=4

y= 2/13

x= 18/13

18/13+8/13-2z=2

2-2z=2

2(1-z)=2

1-z=1

z=0

Hence we get P(18/13, 2/13, 0), and so the equations of the line are

(x, y, z) = (18/13, 2/13, 0) + t(6,-8,-13)