- Distance Between Two Points in 3-D Space
Let P`(x_1, y_1, z_1)` and Q `( x_2, y_2, z_2)` be two points referred to a system of rectangular axes OX, OY and OZ. Through the points P and Q draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ.Fig.
Now, since ∠PAQ is a right angle, it follows that, in triangle PAQ, `PQ^2 = PA^2 + AQ^2 ` ... (1)
Also, triangle ANQ is right angle triangle with ∠ANQ a right angle.
Therefore ` AQ^2 = AN^2 + NQ^2 ` ... (2)
From (1) and (2), we have
`PQ^2 = PA^2 + AN^2 + NQ^2`
Now PA = `y_2 – y_1`, AN = `x_2 – x_1` and NQ = `z_2 – z_1`
Hence `PQ^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2`
Therefore PQ = `sqrt((x_2-x_1)^2 + (y_2-y_1)^2+(z_2-z_1)^2)`
This gives us the distance between two points `(x_1, y_1, z_1)` and `(x_2, y_2, z_2)`.
In particular, if `x_1 = y_1 = z_1` = 0, i.e., point P is origin O, then
OQ = `sqrt(x_2^2+y_2^2+z_2^2)` , which gives the distance between
the origin O and any point Q `(x_2, y_2, z_2)`.