Question

1. Determine the intersection, if any, of the planes with equations x + y – z + 12 =0 and 2x + 4y - 3z + 8 = 0

Answer 1

x+y-z=-12, normal vector=<1,1,-1>

2x+4y-3z=-8, normal vector=<2,4,-3>.

These normals are nor scalar multiples of one another, so the planes are not parallel and must therefore intersect. The intersection will be a line (not a point).

Let x=t, then y-z=-12-t so y=z-12-t. Substitute for x and y in the other equation:

4(z-12-t)-3z=-8, 4z-48-4t-3z=-8, z=40+4t and y=40-12-t=28-t. We now have the equation of the line in parametric form:

x=t, y=28-t, z=40+4t.