Calculus 3 question

Question

Find the equation of the plane containing the points P = (-3, 0, 2), Q = (-1, -1, 2), R = (-5, 3, 4). The result should be in the form ax + by + cz = d.

Answer 1

Try using Geogebra to visualize and solve problems like these. Here is the problem solved on Geogebra:

https://www.geogebra.org/calculator/vxu6xyhm

Answer 2

Vectors PQ = <2, - 1, 0>, PR = <- 2, 3, 2>.

Cross product PQ x PR is normal vector to plane, PQ x PR = det[(i j k) (2 -1 0) (-2 3 2)> = - 2i - 4j + 4k =

<- 2, - 4, 4>.

Equation of plane: - 2(x + 3) - 4(y - 0) + 4(z - 2) = 0; x + 3 + 2y - 2z + 4 = 0; x + 2y - 2z = - 7