Given: point P(3, 4, 5)
To find: length of the perpendicular from the point on the y-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 z coordinate on the y-axis is zero
Let point D is the point of the foot of perpendicular on the y-axis from point P be (0, y, 0)
Direction cosines of y-axis are (0, 1, 0)
Direction cosines of PD are (3 – 0, 4 – y, 5 – 0) = (3, 4 – y, 5)
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
⇒ 3 × 0 + (4 – y) × 1 + 5 × 0 = 0
⇒ 0 + 0 + 4 – y= 0
⇒ y = 4
Hence point D(0, 4, 0)
Distance between point P(3, 4, 5) and D(0, 4, 0) is d
=\(\sqrt{(3-0)^2+(4-4)^2+(5-0)^2}\)
= \(\sqrt{3^2+0^2+5^2}\)
= \(\sqrt{9+0+25}\)
= \(\sqrt{34}\)
Hence, the distance of the point P from y-axis is \(\sqrt{34}\) units
