Christopher J. answered • 05/28/20

Berkeley Grad Math Tutor (algebra to calculus)

Amanda:

One way to think about this problem is to find three points on the plane and then determine the equation of the plane containing these three points by cross product.

We know (2,0,-3) is a point.

Let t = 0 then (-1,3,1) is also a point.

Let t =1 then (-1+2,3+4,1+5) = (1,7,6) is also a point.

Let A= (2,0,-3) B=(-1,3,1) C=(1,7,6)

vector(AB) = < -1-2, 3-0,1+3> = <-3, 3, 4>

vector(AC) = <1-2,7-0,6+3> = <-1,7,9>

Taking the cross product <-3,3,4> X <-1,7,9> = <-1,23,-18>

So the equation of the plane will be -(x-2)+23*(y-0)-18*(z+3)=0

Or -x + 23y -18z = 52. This is the "standard" equation of the plane.

We can use vector(AB) = <-3,3,4> and point A=(2,0,-3) to determine parametric equations of the plane.

x-2 = -3*t

y-0 = 3*t

z+3 = 4*t

I'll let you figure out the rest.