Question

Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3,2) and C(0, 0,1).

Answer 1

The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB 

\[\sqrt{\left( 2 - 0 \right)^2 + \left( 0 - y \right)^2 + \left( 3 - z \right)^2} = \sqrt{\left( 0 - 0 \right)^2 + \left( 3 - y \right)^2 + \left( 2 - z \right)}\]
\[ \Rightarrow 4 + y^2 + 9 - 6z + z^2 = 9 - 6y + y^2 + 4 - 4z + z^2 \]
\[ \Rightarrow - 6z = - 6y - 4z\]
\[ \Rightarrow 3y - z = 0 . . . . . \left( 1 \right)\]

Also, PA = PC

\[\sqrt{\left( 2 - 0 \right)^2 + \left( 0 - y \right)^2 + \left( 3 - z \right)^2} = \sqrt{\left( 0 - 0 \right)^2 + \left( 0 - y \right)^2 + \left( 1 - z \right)^2}\]
\[ \Rightarrow 4 + y^2 + 9 - 6z + z^2 = y^2 + 1 - 2z + z^2 \]
\[ \Rightarrow 13 - 6z = 1 - 2z\]
\[ \Rightarrow - 4z = - 12\]
\[ \Rightarrow z = 3 . . . . . \left( 2 \right)\]

Solving (1) and (2),we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).