Question

Find the coordinates of a point equidistant from the origin and points A (a, 0, 0), B (0, b, 0) andC(0, 0, c).

Answer 1

Let the point be P(x, y, z).
Now, PO = PA
$$\begin{gathered} \sqrt{{\left(0-x\right)}^2+{\left(0-y\right)}^2+{\left(0-z\right)}^2}=\sqrt{{\left(a-x\right)}^2+{\left(0-y\right)}^2+\left(0-z\right)} \\ \Rightarrow x^2+y^2+z^2=a^2-2ax+x^2+y^2+z^2 \\ \Rightarrow 0=a^2-2ax \\ \Rightarrow x=\frac{a}{2} \end{gathered}$$
Also, PO = PB
$$\begin{gathered} \sqrt{{\left(0-x\right)}^2+{\left(0-y\right)}^2+{\left(0-z\right)}^2}=\sqrt{{\left(0-x\right)}^2+{\left(b-y\right)}^2+\left(0-z\right)} \\ \Rightarrow x^2+y^2+z^2=x^2+b^2-2by+y^2+z^2 \\ \Rightarrow 0=b^2-2by \\ \Rightarrow y=\frac{b}{2} \end{gathered}$$
Again, PO = PC
$$\begin{gathered} \sqrt{{\left(0-x\right)}^2+{\left(0-y\right)}^2+{\left(0-z\right)}^2}=\sqrt{{\left(0-x\right)}^2+{\left(0-y\right)}^2+\left(c-z\right)} \\ \Rightarrow x^2+y^2+z^2=x^2+y^2+c^2-2cz+z^2 \\ \Rightarrow 0=c^2-2cz \\ \Rightarrow z=\frac{c}{2} \end{gathered}$$

Hence, the coordinates of the point is $\left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right)$