Wouldn't that be the plane passing through the midpoint of AB and having as its normal the direction vector of AB ??
midpoint of AB = (5/2, 1, 2)
direction vector AB = (3,-2,-2)
the the plane has equation
3x - 2y - 2z + C = 0
but (5/2,1,2) lies on it, so
15/2 - 2 - 4 + C = 0
C = -3/2
so 3x - 2y - 2z - 3/2 = 0 or
6x - 4y - 4z - 3 = 0
"Find an equation for the set of points P(x, y, z) that are equidistant from the points A(1, 2, 3) and B(4, 0, 1)."
Can some one help?
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