Answer:
The normal vector of the plane is the vector pointing from the midpoint to A.
The equation of the plane is
(-3.5)(x)+(1.5)(y)+(2.5)(z) -(-3.5)(0.5) -(1.5)(2.5) -(2.5)(0.5) = 0
Step-by-step explanation:
The set of all points equidistant from A(-3, 4, 3) and B(4, 1, -2) is a plane. The equation of this plane can be found using the midpoint of the line segment connecting A and B and the vector pointing from the midpoint to A (or B), which is perpendicular to the plane.
The midpoint of the line segment connecting A and B is given by:
((-3 + 4)/2, (4 + 1)/2, (3 + -2)/2) = (0.5, 2.5, 0.5)
The vector pointing from the midpoint to A is given by:
(-3 - 0.5, 4 - 2.5, 3 - 0.5) = (-3.5, 1.5, 2.5)
The normal vector of the plane is the vector pointing from the midpoint to A.
The equation of the plane is
(-3.5)(x)+(1.5)(y)+(2.5)(z) -(-3.5)(0.5) -(1.5)(2.5) -(2.5)(0.5) = 0
The set of all points equidistant from A and B is a plane perpendicular to the vector AB, with the equation above.