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
$$\begin{gathered} \sqrt{{\left(2-0\right)}^2+{\left(0-y\right)}^2+{\left(3-z\right)}^2}=\sqrt{{\left(0-0\right)}^2+{\left(3-y\right)}^2+\left(2-z\right)} \\ \Rightarrow 4+y^2+9-6z+z^2=9-6y+y^2+4-4z+z^2 \\ \Rightarrow -6z=-6y-4z \\ \Rightarrow 3y-z=0.....\left(1\right) \end{gathered}$$
Also, PA = PC
$$\begin{gathered} \sqrt{{\left(2-0\right)}^2+{\left(0-y\right)}^2+{\left(3-z\right)}^2}=\sqrt{{\left(0-0\right)}^2+{\left(0-y\right)}^2+{\left(1-z\right)}^2} \\ \Rightarrow 4+y^2+9-6z+z^2=y^2+1-2z+z^2 \\ \Rightarrow 13-6z=1-2z \\ \Rightarrow -4z=-12 \\ \Rightarrow z=3.....\left(2\right) \end{gathered}$$
Solving (1) and (2),we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).