find 2 vectors that pass through 3 points ?

Question

p1=[10,20,5]
p2=[15,10,10]
p3=[25,20,10]

these are three points, find two direction vectors which are parallel to the plane defined by p1,p2 and p3 , and hence a normal vector to the plane.

Answer 1

Find 2 vectors that pass through 3 points.

p1 = [10,20,5]
p2 = [15,10,10]
p3 = [25,20,10]

Let u and v be the vectors.

u = p1p2 = <15-10, 10-20, 10-5> = <5, -10, 5>
v = p1p3 = <25-10, 20-20, 10-5> = <15, 0, 5>

The cross product n of u and v is normal to both of them and is the normal vector to the plane.

n = u X v = <5, -10, 5> X <15, 0, 5>
n = -50i + 50j + 150k = <-50, 50, 150>
Any non-zero multiple will do as well. Divide by 50.
n = <-1, 1, 3>

Answer 2

Two vectors that pass through 3 points: Just take the difference between two points to get each vector.

p2 - p1 = [15, 10, 10] - [10, 20, 5] = [5, -10, 5]
p3 - p1 = [25, 20, 10] - [10, 20, 5] = [15, 0, 5]

So, two vectors passing through the 3 points are:
[5, -10, 5]
[15, 0, 5]

To find a normal vector to the plane, take the cross-product:
(5i - 10j + 5k)(15i + 5k)
= 75i² + 25ik - 150ji - 50jk + 75ki + 25k²
= 75(0) + 25(-j) - 150(-k) - 50(i) + 75(j) + 25(0)
= [-50, 75-25, 150]
= [-50, 50, 150]

You could normalize the vector and make it into a unit vector, but I would just divide through by 50 and leave it as:
[-1, 1, 3]

Answer 3

v1 = [5,-10,-5] = p2 - p1
v2 = [10,10,0] = p3 - p2
cross product of above will give you the perpendicular vector
[50, -100, 0] now we need to divide each comp by a number to make it a normal vector
which I believe is:
(2500 + 10000)^0.5 = N
[50/N, -100/N, 0] is the normal vector I believe.