First slide

Introduction to 3-D Geometry

Question

The coordinates of a point which is equidistant from the points (0, 0, 0), (o, 0, 0), (0, b, 0), (0,0, c) are given by

Easy

A

$(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$
B

$(- \frac{a}{2}, - \frac{b}{2}, \frac{c}{2})$
C

$(\frac{a}{2}, - \frac{b}{2}, - \frac{c}{2})$
D

$(- \frac{a}{2}, \frac{b}{2}, - \frac{c}{2})$

Solution

Let P be any point in a space. According to the given condition,
$\begin{aligned} \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} = \sqrt{(x - a)^{2} + y^{2} + z^{2}} \\ = \sqrt{x^{2} + (y - b)^{2} + z^{2}} = \sqrt{x^{2} + y^{2} + (z - c)^{2}} \end{aligned}$
On squaring both sides, we get
$\begin{array}{l} x^{2} + y^{2} + z^{2} = (x - a)^{2} + y^{2} + z^{2} = x^{2} + (y - b)^{2} + z^{2} \\ = x^{2} + y^{2} + (z - c)^{2} \\ \Rightarrow x^{2} + y^{2} + z^{2} = x^{2} + a^{2} - 2 ax + y^{2} + z^{2} \\ \Rightarrow a^{2} - 2 ax = 0 \\ \Rightarrow x = \frac{a}{2} \end{array}$
Similarly, we can solve the other values, we get
$y = \frac{b}{2}, z = \frac{c}{2}$

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