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)

thanks for help

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)
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 = %26lt;1-3,-1+1,-3-2%26gt; = %26lt;-2,0,-5%26gt;


v = AC = %26lt;4-3,-3+1,1-2%26gt; = %26lt;1,-2,-1%26gt;





n = u X v = -10i - 7j + 4k = %26lt;-10,-7,4%26gt;





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 = %26lt;-10,-7,4%26gt;/√165
Reply: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