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  4. The perpendicular distance of the point (8, 15, 6) from y-axis is

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Q. The perpendicular distance of the point $(8, 15, 6)$ from $y$-axis is

Introduction to Three Dimensional Geometry

Solution:

Perpendicular distance of the point $(8 , 15, 6)$ from $y$-axis is
$ \sqrt{8^{2}+0^{2}+6^{2}}= \sqrt{64+36} = \sqrt{100}$
$= 10$ units.