equation of the plane?

Question

The plane kx - y + jz + 2 = 0 where k and j are constants intersect the point (2,0,3) and is parallell with the vector (1,-2,2). Determine the equation of the plane.

Answer 1

kx - y + jz + 2 = 0
as(2,0,3) lie on given plane
2k - 0 +3j + 2 = 0
=> 2k +3j = -2 --------(1)
As given plane is parallel to vector (1 ,-2 ,2)
k + 2 +2j +2 = 0
=>k + 2j = - 4 ---------(2)
solving equations (1) & (2) we get
k = 8 , j = - 6
equation of plane is
8x - y -6z +2= 0

Answer 2

The equation of a plane is given as kx - y + jz + 2 = 0, where k and j are constants. The point P(2, 0, 3) is on the plane and the vector v = <1, -2, 2> is a directional vector of the plane. Find the equation of the plane.

kx - y + jz + 2 = 0

Plug in the point P.
2k - 0 + 3j + 2 = 0

For the directional vector v we have:

k + 2 + 2j = 0

Set the two equations equal.

2k - 0 + 3j + 2 = k + 2 + 2j
k = -j

Plug back into the equation of v.

k + 2 + 2j = 0
-j + 2 + 2j = 0
j = -2

k = -j
k = 2

The equation of the plane is:

kx - y + jz + 2 = 0
2x - y - 2z + 2 = 0

Answer 3

This one's easy. The answer is Giraldo Rivera!

Ah-cha-cha-cha-cha, I've done it again!