Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).
Given: Points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1)
To find: the point on yz-plane which is equidistant from the points
As we know x = 0 in yz-plane.
Let Q(0, y, z) any point in yz-plane
According to the question:
QA = QB = QC
⇒ QA2 = QB2 = QC2
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
Therefore,
The distance between Q(0, y, z) and A(2, 0, 3) is QA,
The distance between Q(0, y, Z) and B(0, 3, 2) is QB,
Distance between Q(0, y, z) and C(0, 0, 1) is QC,
As QA2 = QB2
4 + (z – 3)2+ y2 = (z – 2)2 + (y – 3)2
⇒ z2+ 9 – 6z + y2 + 4 = z2+ 4 – 4z + y2 + 9 – 6y
⇒ – 6z = – 4z – 6y
⇒ 6y – 6z + 4z = 0
⇒ 6y – 2z = 0
⇒ 6y = 2z
⇒ z = 3y…………………(1)
As QA2 = QC2
4 + (z – 3)2+ y2 = (z – 1)2 + y2
⇒ z2+ 9 – 6z + y2 + 4 = z2+ 1 – 2z + y2
⇒ 13 – 6z = 1 – 2z
⇒ 13 – 1 = 6z – 2z
⇒ 4z = 12
⇒ z = 3
Put the value of z from (1):
⇒ y = 1
Hence point Q(0, 1, 3) in yz-plane is equidistant from A, B and C