First slide

Introduction to 3-D Geometry

Question

The Locus of a point for which the sum of the squares of the distances from the coordinate axes is 16 units is

Very Easy

A

$x + y + z = 4$
B

$x^{2} + y^{2} + z^{2} = 8$
C

$x^{2} + y^{2} + z^{2} = 16$
D

$x^{2} + y^{2} + z^{2} = 32$

Solution

Suppose that $P (x, y, z)$ be any point on the locus.
The perpendicular distance from $P (x_{1}, y_{1}, z_{1})$ to the $x -$ axis is $\sqrt{y^{2} + z^{2}}$
The perpendicular distance from $P (x_{1}, y_{1}, z_{1})$ to the $y -$ axis is $\sqrt{x^{2} + z^{2}}$
The perpendicular distance from $P (x_{1}, y_{1}, z_{1})$ to the $z -$ axis is $\sqrt{x^{2} + y^{2}}$
Hence, the sum of the squares of the distances from the point $P (x_{1}, y_{1}, z_{1})$ to the coordinate axes is $2 (x^{2} + y^{2} + z^{2})$
Therefore, the locus a point for which the sum of the squares of the distances from the coordinate axes is 16 units is $x^{2} + y^{2} + z^{2} = 8$

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