a plane pases trhough the points (1,0,-1),,(3,3,2) and(4,5,-1).?

Question

1)with the parametric equations find the general equation of the plane.
2)explain why this plane is not parallel to a plane having a normal vector n = (5,-2,-1)

Answer 1

(1)
Calling the given points A, B, C in order, the origin O and P the point (x, y, z), the parametric equation is:
OP = sAB + tAC
(x, y, z) = (1, 0, -1) + s(2, 3, 3) + t(3, 5, 0)

Hence:
x = 1 + 2s + 3t ...(1)
y = 3s + 5t ........(2)
z = 3s - 1 ...........(3)

(2) - (3):
y - z = 5t + 1
2*(2) - 3*(1):
2y - 3x = t - 3

Eliminating t between these two equations:
y - z = 5(3 - 3x + 2y) + 1
15x - 9y - z - 16 = 0.

(2)
The normal vector to this plane is (15, - 9, -1), which is clearly not parallel to (5, - 2, -1) as the ratios 15/5, 9/2 , 1/1 are all unequal. The planes are therefore not parallel.

Answer 2

Find the equation of the plane containing the points
P(1,0,-1); Q(3,3,2); and R(4,5,-1).

Create the directional vectors PQ and PR.

PQ = = <3-1, 3-0, 2+1> = <2, 3, 3>
PR = = <4-1, 5-0, -1+1> = <3, 5, 0>

The normal vector v, of the plane is orthogonal to both directional vectors. Take the cross product.

v = PQ X PR = <2, 3, 3> X <3, 5, 0> = <-15, 9, 1>

With the normal vector v and a point in the plane we can write the equation of the plane. Let's choose P(1, 0, -1).

-15(x - 1) + 9(y - 0) + 1(z + 1) = 0
-15x + 15 + 9y + z + 1 = 0
-15x + 9y + z + 16 = 0
______________________

This plane is NOT parallel to a plane with the normal vector
n = <5, -2, -1> because n does not equal v nor is it a non-zero multiple of v.