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Distance

The perpendic...

Question

The perpendicular distance of point $(2, - 3, - 4)$ from z-axis is

5 u n i t

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\sqrt{29} u n i t

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\sqrt{20} u n i t

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\sqrt{13} u n i t

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Solution

The correct option is D $\sqrt{13} u n i t$
Given that $x = 2 y = - 3 z = - 4$
the perpendicular distance from Z axis is $d = \sqrt{x^{2} + y^{2}} = \sqrt{4 + 9} = \sqrt{13} u n i t$

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P (3, 4)

y -

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\frac{x - 3}{2} = \frac{y - 3}{3} = \frac{z + 5}{6}

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