How would I find a unit vector perpendicular to plane ABC where points are A(3,-1,2), B(1,-1,-3) and C(4,-3,1)

Question

thanks for help

Answer 1

Form two vectors from the three points and then take the cross product.

Given A(3,-1,2), B(1,-1,-3) and C(4,-3,1)

u = AB = <1-3,-1+1,-3-2> = <-2,0,-5>
v = AC = <4-3,-3+1,1-2> = <1,-2,-1>

n = u X v = -10i - 7j + 4k = <-10,-7,4>

Now we will make it a unit vector by dividing it by its magnitude.

| n | = √(10² + (-7)² + 4²) = √(100 + 49 + 16) = √165

Unit n = <-10,-7,4>/√165

Answer 2

The vector from A to B is in the plane.
The vector from A to C is in the plane.

The vector cross product of the two is perpendicular to both of them and therefore is perpendicular to the plane.

So take (B - A) x (C - A) (vector cross product) and divide it by its length to get a unit vector.

Dan