vector line and plane question, help!!!!!?

Question

Let P1 be the plane defined by the points:

P(0,3,2), Q(2,1,−1) and R(3,4,2).

Find the equation of the plane perpendicular to P1 and containing the line:

[x,y,z]T = [ −1,3,2 ]T + t [ 1,1,0 ]T.

Answer 1

(Q - P) cross (R - P) will be perpendicular to plane P1.
(Q - P) x (R - P) = [ 2,-2,-3 ]^T x [ 3,1,0 ]^T = [ 3,-9,8 ]^T is perpendicular to plane P1, and thus is a direction in the unknown plane, as is [ 1,1,0 ]^T. So their cross product is perpendicular to the unknown plane.

[ 3,-9,8 ]^T x [ 1,1,0 ]^T = [ -8,8,12 ]^T
[ -2,2,3 ]^T is in the same direction, perpendicular to the plane.

So given both [ x,y,z ]^T and [ -1,3,2 ]^ in the plane and [ -2,2,3 ]^T perpendicular to the plane, the following dot product is zero:

-2 (x - -1) + 2 (y - 3) + 3 (z - 2) = 0
-2x - 2 + 2y - 6 + 3z - 6 = 0
-2x + 2y + 3z -14 = 0
-2x + 2y + 3z =14

Dan

Answer 2

P1 is defined by the three points P(0,3,2), Q(2,1,−1) and
R(3,4,2).

Create two vectors from the points.

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

The normal vector n1, to the plane P1 is the cross product of the two vectors.

n1 = <2, -2, -3> X <3, 1, 0> = <3, -9, 8>

The normal vector of the desired plane contains n1 and the directional vector r of the given line. The normal vector n1, of the desired plane is the cross product of the two vectors.

n2 = n1 X r = <3, -9, 8> X <1, 1, 0> = <-8, 8, 12>

Any non-zero multiple will also be a normal vector to the plane. Divide by -4.

n2 = <2, -2, -3>

Set t = 0 to find a point on the desired plane (-1, 3, 2). With the normal vector n2 and the point we can write an equation of the plane.

2(x + 1) - 2(y - 3) - 3(z - 2) = 0
2x + 2 - 2y + 6 - 3z + 6 = 0
2x - 2y - 3z + 14 = 0

Answer 3

Your first answer is erroneous because of the fact 4 situations 7 is 28 no longer 24. A for the 2d, the line is parallel to u = <-a million, a million, 4>. To get yet another vector parallel to the airplane, take the factor (-5, -2, 4) on the line and connect it to the factor (5, -2, 7). This vector is v = <10, 0, 3>. the conventional to the airplane is the bypass product u x v u x v = <3, 40 3, -10>. The airplane has equation 3(x - 5) + 40 3(y + 2) - 10(z - 7) = 0 ==> 3x + 43y - 10z = -141