How do you solve the equation of a plane?

Determine the equation of the plane that lies between the points (-1, 2, 4) and (3, 1, -4) and is equidistant from them.How do you solve the equation of a plane?If you know the vector form of a plane then this is the answer.



r*n = a*n



where n is a vector perpendicular to the plane



a is a point on the plane



midpoint of both points = (-1+3)/2, (2+1)/2, (4-4)/2) = (1, 3/2, 0)



this point lies on the plane



a vector perpendicular to the plane can be found by



(-1 2 4) - (3 1 -4) = (-4 1 8)



so equation of plane is



r*(-4 1 8) = (1 3/2 0)(-4 1 8)



-4x+y+8z = -4 +3/2



-8x+2y+16z = 5