Given: point P(3, 5, 12)
To find: length of the perpendicular drawn from the point P from the x-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, y and z coordinate on x-axis are zero
Let point D on x-axis is (x, 0, 0)
Direction cosines of z-axis are (1, 0, 0)
Direction cosines of PD are (3 – x, 5 – 0, 12 – 0) = (3 – x, 5, 12)
Let \(\overrightarrow{b_1}\) are \(\overrightarrow{b_2}\) 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 – x) × 1 + 5 × 0 + 12 × 0 = 0
⇒ 3 – x + 0 + 0 = 0
⇒ x = 3
Hence point D(3, 0, 0)
Distance between point P(3, 5, 12) and D(3, 0, 0) is d
= 13
Hence, the distance of the point P from x-axis is 13 units.
