Question

Find the general form of the equation of the plane witht he given characteristics: Passes through (2,2,1) and (-1,1,-1) and is perpendicular to 2x-3y+z=3

Answer 1

plane vector: (2-(-1),2-1,1-(-1)) = (3,1,2) =a
perpendicular to 2x-3y+z=3 means perpendicular to (2,-3,1) = b and a
c = axb = (7,1,-12) (x stands for cross product)
so Plane equation is: 7x+y-12z = 0

Answer 2

got it?

Answer 3

@LivyLou

Answer 4

Where did you get the (3,1,2) and (7,1,-12) from?

Answer 5

(3,1,2) is a vector that goes from (-1,1,-1) to (2,2,1)
(7,1,-12) is a vector product of a and b

Answer 6

Does using the vector product give the perpendicular vector?

Answer 7

yes

Answer 8

Okay thanks