Question

Find the standard equation of the plane containing the three points. Show All Work.
(-2,1,1),(0,2,3),(1,0,-1)

Answer 1

find two vector from that three points

Answer 2

How do you do that?

Answer 3

I will find one for you

choose any two points

(-2,1,1) -(0,2,3)

<-2,-1,-2>

Answer 4

Say you have... (2, -2, 1), (1, 0, -2), and (1, 1, 0)

then your solution would be...

The equation of a plane is Ax + By + Cz = D, where the normal vector to the plane is <A, B, C>

to find the normal vector, you need two vectors that are in the plane. Taking the cross product of these two gives you the normal vector to the plane.

To find two vectors in the plane, subtract two of these points from the other. So:
(2, -2, 1) - (1, 0, -2) = <1, -2, 3>
(1, 1, 0) - (1, 0, -2) = <0, 1, 2>

Now take the cross product of these two vectors. You should get
<-7, -2, 1>

So your equation for the plane is
-7x + -2y + z = D

to solve for D, plug in one of the three points you're given
-7(2) + -2(-2) + 1(1) = D

so D = -9

Final equation:
-7x - 2y + z = -9

You can multiply everything by a -1 if you want the x coefficient to be 7.

Answer 5

Thanks

Answer 6

How do you find the cross product?

Answer 7

(-2,1,1),(0,2,3),(1,0,-1)

Answer 8

There's an easier way.

Answer 9

Wel, not easier, just doesn't confuse you as much.

Answer 10

Please explain it!

Answer 11

Do you know how to solve a system of linear equations with three unknowns?

Answer 12

nope

Answer 13

Like
ax+by+cz=0
dx+ey+fz=0
gx+hy+iz=0

Answer 14

where a,b,c,d,e,f,g,h, and i are known.

Answer 15

wait I only need to find one equation

Answer 16

wow um idk

Answer 17

nvm