First slide

Introduction to 3-D Geometry

Question

If the sum of the squares of the distance of a point from the three coordinate axes be 36, then its distance from the origin is

Easy

A

6
B

$3 \sqrt{2}$
C

$2 \sqrt{3}$
D

None of these

Solution

. Let P(x, y,z)be any point in the plane.
According to the given condition,
(distance from P to X-axis)² + (distance from P to Y-axis)² + (distance from P to Z-axis)² = 36
$\begin{array}{lr} \Rightarrow {(\sqrt{y^{2} + z^{2}})}^{2} + {(\sqrt{x^{2} + z^{2}})}^{2} + {(\sqrt{x^{2} + y^{2}})}^{2} = 36 \\ \Rightarrow (y^{2} + z^{2}) + (x^{2} + z^{2}) + (x^{2} + y^{2}) = 36 \\ \Rightarrow 2 (x^{2} + y^{2} + z^{2}) = 36 \\ \Rightarrow x^{2} + y^{2} + z^{2} = 18 - - - - (i) \end{array}$
The distance from origin to the point (x, y, z) is
$\begin{array}{l} = \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{18} [∵ from Eq. (i)] \\ = 3 \sqrt{2} \end{array}$

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