Find the equation of a plane that passes through the intersection of the two planes.?

Question

How would one find the equation for a plane that passes through the line of intersection of the two planes:
-4z=0
-18x-30y+24z=0

I can find the vector parallel to the line, which is <-5,3,0>. However, I can't find a common point between these two planes.

Answer 1

I assume you trying to find a third plane that contains the line of intersection of the two given planes. If so there are an infinite number of such planes.

The given planes are:

-4z = 0
-18x - 30y + 24z = 0

Let's simplify the equations. Multiplying thru by any non-zero constant does not change the plane. Divide the first equation by -4 and the second equation by -6.

z = 0
3x + 5y - 4z = 0

By inspection it is immediately obvious that a point in common between the two planes is the origin O(0,0,0).

The directional vector v, of the line of intersection between the two planes is normal to the normal vectors of the two planes. Take the cross product of the normal vectors of the two planes.

v = <0, 0, 1> X <3, 5, -4> = <5, -3, 0>

With a point O, on the line, and the directional vector v, of the line, we can write the equation of the line of intersection.

L(t) = O + tv = <0, 0, 0> + t<5, -3, 0>
where t is a scalar ranging over the real numbers

To find a third plane containing the line L, pick another point that is not on either of the first two planes. There are an infinite number of such points. Let's pick P(1, 1, 1).

The directional vector of the line v, is one directional vector of the desired plane. To create a second directional vector of the plane form the vector:

u = OP = <1, 1, 1>

The normal vector to the third plane n, will be orthogonal to both of the directional vectors u and v. Take the cross product.

n = u X v = <1, 1, 1> X <5, -3, 0> = <3, 5, -8>

With the normal vector n, of the plane, and a point P, on the plane, we can write the equation of the plane.

3(x - 1) + 5(y - 1) - 8(z - 1) = 0
3x - 3 + 5y - 5 - 8z + 8 = 0
3x + 5y - 8z = 0