Given: point P(a, b, c)
To find: distance of the point P from the z-axis
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a-m)^2+(b-n)^2+(c-o)^2}\)
As, x and y coordinate on z-axis are zero
Let point D on z-axis is (0, 0, z)
Direction cosines of z-axis are (0, 0, 1)
Direction cosines of PD are (a – 0, b – 0, c – z) = (a, b, c – z)
Let \(\overrightarrow{b_1}\) and \(\overrightarrow{b_2}\)are two vectors as shown in the figure:
The dot product of perpendicular vectors is always zero
Therefore, \(\overrightarrow{b_1}.\overrightarrow{b_2}\) = 0
⇒ a × 0 + b × 0 + (c – z) × 1 = 0
⇒ 0 + 0 + c – z = 0
⇒ z = c
Hence point D(0, 0, c) Distance between point P(a, b, c) and D(0, 0, c) is d
=\(\sqrt{(a-0)^2+(b-0)^2+(c-c)^2}\)
= \(\sqrt{a^2+b^2+0^2}\)
= \(\sqrt{a^2+b^2}\)
Hence, the distance of the point P from z-axis is \(\sqrt{a^2+b^2}\) units.
