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 

\[\sqrt{\left( 0 - x \right)^2 + \left( 0 - y \right)^2 + \left( 0 - z \right)^2} = \sqrt{\left( a - x \right)^2 + \left( 0 - y \right)^2 + \left( 0 - z \right)}\]
\[ \Rightarrow x^2 + y^2 + z^2 = a^2 - 2ax + x^2 + y^2 + z^2 \]
\[ \Rightarrow 0 = a^2 - 2ax\]
\[ \Rightarrow x = \frac{a}{2}\] 

Also, PO = PB 

\[\sqrt{\left( 0 - x \right)^2 + \left( 0 - y \right)^2 + \left( 0 - z \right)^2} = \sqrt{\left( 0 - x \right)^2 + \left( b - y \right)^2 + \left( 0 - z \right)}\]
\[ \Rightarrow x^2 + y^2 + z^2 = x^2 + b^2 - 2by + y^2 + z^2 \]
\[ \Rightarrow 0 = b^2 - 2by\]
\[ \Rightarrow y = \frac{b}{2}\]

Again, PO = PC

\[\sqrt{\left( 0 - x \right)^2 + \left( 0 - y \right)^2 + \left( 0 - z \right)^2} = \sqrt{\left( 0 - x \right)^2 + \left( 0 - y \right)^2 + \left( c - z \right)}\]
\[ \Rightarrow x^2 + y^2 + z^2 = x^2 + y^2 + c^2 - 2cz + z^2 \]
\[ \Rightarrow 0 = c^2 - 2cz\]
\[ \Rightarrow z = \frac{c}{2}\]

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