Vectors and the Geometry of Space

Question

Find an equation of the plane consisting of all points that are equidistant from (3, -3, 5) and (3, 0, 1).

Answer 1

Let (x, y, z) is arbitrary point on the plane. Then (x - 3)² + (y + 3)² + (z - 5)² = (x - 3)² + y² + (z - 1)²;

6y + 9 - 10z + 25 = - 2z + 1; 6y - 8z + 33 = 0

Answer 2

A GEOMETRIC APPROACH

The plane equidistant of the two given points P (3, -3, 5 ) and Q (3, 0, 1 ) is the plane that passes

through the midpoint of PQ orthogonally to PQ.

PQ = <0, 3, -4 > , the normal vector for the plane in question.

The midpoint Ω has coordinates Ω ( (3+3)/2, (-3+0)/2, (5+1)/2 ) = ( 3, -3/2, 3 )

Then the equation of the plane

0( x -3 ) + 3( y+3/2) -4( z-3) = 0.

6y -8z +33 =0