Points on a plane

Question

Q) Find a normal and a few points on the plane 3x - y + 4z = 9

I know a normal is (3, -1, 4) or any parallel vector

but am not sure how to find the points.

Thank you

Answer 1

The plane equation is 3x - y + 4z = 9.

Write plane equation in parametric form x = x₀ + sd +te, where x₀ = any point on the plane d and e are nonparallel vectors parallel to the plane.

Here one equation and three variables, so the number of parameters = 3 - 1 = 2.

Let y = s, z = t then x = (9 + s - 4t)/3 = 3 + s/3 - 4t/3.

Therefore (x, y, z) = (3 + s/3 - 4t/3, s, t) = (3, 0, 0) + s(1/3, 1, 0) + t(-4/3, 0, 1).

Let P = x₀ = point on the plane = (3, 0, 0).

Let Q and R are two remaining on the plane.

d = PQ = OQ - OP

(1/3, 1, 0) = (x, y, z) - (3, 0, 0)

(1/3, 1, 0) = (x - 3, y - 0, z - 0)

x = 10/3, y = 1 and z = 0.

Therefore, the point Q = (10/3, 1, 0).

e = PR = OR - OP

(-4/3, 0, 1) = (x, y, z) - (3, 0, 0)

(-4/3, 0, 1) = (x - 3, y - 0, z - 0)

x = 5/3, y = 0 and z = 1.

Therefore, the point R = (5/3, 0, 1).

The three points are (3, 0, 0), (10/3, 1, 0) and (5/3, 0, 1).