Find the point in yz-plane which is equidistant from the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).
The general point on yz plane is D(0, y, z). Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).
∴ AD = BD
Squaring both sides,
=
9+ y2 -4y + 4 + z2 + 2z + 1 = 1+ y2 +2y + 1+ z2
-6y + 2z + 12 = 0 ….(1)
Also, AD = CD
Squaring both sides,
=
9+ y2 -4y + 4 + z2 + 2z + 1 = 4+ y2 - 2y + 1+ z2 – 4z + 4
-2y + 6z + 5 = 0 ….(2)
Simultaneously solving equation (1) and (2) we get
Y= 31/16, z = -3/16
The point which is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2) is (0, 31/16, -3/16).