Finding the equation of the plane?

Question

I dont know how to get an answer for this, can anyone help out please.

[x,y,z]^T = [ -2,3,-3 ]^T + t [ 2,2,2 ]^T

and is parallel to the line:

[x,y,z]^T = [ 4,-1,-4 ]^T + t [ 1,-3,2 ]^T.

My answer should be in the form Ax + By + Cz = D. Adjust your coefficients so that A, B, C and D are in lowest terms (have 1 as the greatest common factor), and D is positive.

A=?
B=?
C=?
D=?

Answer 1

I assume that you meant to say that [x,y,z]^T = [ -2,3,-3 ]^T + t [ 2,2,2 ]^T lies in the plane.

Then [x,y,z]^T = [ -2,3,-3 ]^T + t [ 1,-3,2 ]^T also lies in the plane. The cross product of [ 2,2,2 ]^T and [ 1,-3,2 ]^T then gives a vector perpendicular to the plane.

If [ A,B,C ]^T is perpendicular to the plane and [ p,q,r ]^T is a point in the plane, then for all [ x,y,z ]^T in the plane, the direction [ x-p, y-q, z-r ]^T, being in the plane, is also perpendicular to the normal [ A,B,C ]^ and ***** therefore their dot product is zero *****

A(x - p) + B(y - q) + C(z - r) = 0
or Ax + By + Cz = Ap + Bq + Cr

Here you get the cross product of [ 2,2,2 ]^T and [ 1,-3,2 ]^T is [ 10,-2,-8 ]^T. Reducing to lowest terms, [ 5,-1,-4] ^T is also perpendicular to the plane. Since [ -2,-3,-3 ]^T is in the plane, by the above you have 5 (x - -2) + -1 (y - 3) + -4 (z - -3) = 0
5x + 10 -y + 3 - 4z - 12 = 0
5x - y - 4z + 1 = 0
5x - y - 4z = -1
-5x + y + 4z = 1

A = -5
B = 1
C = 4
D = 1

You should check that [ -2,3,-3 ]^T satisfies that equation and that adding c1 [ 2,2,2, ]^T + c2 [ 1,-3,2 ]^T to [ x,y,z ]^T adds 0 to the left hand side.

Dan

Answer 2

As stated the question doesn't make any sense. I assume you meant to say that the first line lies in the plane.

Since the directional vectors of both lines lie in the plane, the normal vector n, of the plane can be found by taking the cross product of the vectors.

n = [2, 2, 2]^T X [1, -3, 2]^T = [10, -2, -8]^T

Any non-zero multiple of the normal vector will do as well. Divide by -2.

n = [-5, 1, 4]^T

Now pick a point on the plane. Let t = 0 and pick the
point (-2, 3, -3) from the first line.

Now we can write the equation of the plane.

-5(x + 2) + 1(y - 3) + 4(z + 3) = 0
-5x - 10 + y - 3 + 4z + 12 = 0
-5x + y + 4z = 1

A = -5
B = 1
C = 4
D = 1