Question

a = (2,-1,3), b = (7,5,1), c = (5,5,0)
Determin an equation of the plane containing the points a, b, and c

Answer 1

well the equation: -2x + 3y + 4z =5

i did it by finding a normal vector to ab , ac
you can do this by the determinant :
ab : (5,6,-2)
ac (3,6,-3)
determinant :
i j k
5 6 -2 -> (ab)
3 6 -3 -> (ac)

the normal vector from this determinant : (-6,9,12)

then i did dot product between (x-2,y+1,z-3) which is any vector through the point a

and the normal (-6,9,12) and the dot product must be equal to zero since its dot product with a normal vector.

(x-2,y+1,z-3)(-6,9,12) ->
-2x + 3y + 4z =5

Answer 2

\[-6x+9y+12z=15\]
If you would like me to explain just ask.

Answer 3

hmm

Answer 4

the negative in front is seen as a fauxpaux; and the unreduced form is bad posture as well

Answer 5

zero out a point to see 2 vectors on the plane

(2,-1,3) (7,5,1) (5,5,0)
-5-50 -5-50 -5-50
--------------------
-3,-6, 3 2,0,1

<1,2,-1> and <2,0,1> are on the plane

Answer 6

cross them to get the normal:

<1,2,-1>
<2, 0, 1>
---------
Nx = 2- 0 = 2
Ny = -2-1 = -3
Nz = 0 - 4 = -4

N = <2,-3,-4> right?

Answer 7

Now pick any given point (Px,Py,Pz) and stick it into the equation

2(x-Px)-3(y-Py)-4(z-Pz) = 0

Answer 8

(5,5,0)

2(x-5)-3(y-5)-4(z-0) = 0

2x-10-3y+15-4z = 0

2x -3y -4z +5 = 0

Answer 9

Ok

Answer 10

I need to digest that a bit, but I should be able to get it

Answer 11

Thanks!