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 = PA + AQ ... ( ) Also, triangle ANQ is right angle triangle with ∠ ANQ a right angle.
Fig . MATHEMATICS Therefore AQ = AN + NQ ... ( ) From ( ) and ( ), we have PQ = PA + AN + NQ Now PA = y – y , AN = x – x and NQ = z – z Hence PQ = ( x – x ) + ( y – y ) + ( z – z ) Therefore PQ = z z This gives us the distance between two points ( x , y , z ) and ( x , y , z ). In particular, if x = y = z = , i.e., point P is origin O, then OQ = z , which gives the distance between the origin O and any point Q ( x , y , z ).
Example Find the distance between the points P( , – , ) and Q (– , , ). Solution The distance PQ between the points P ( ,– , ) and Q (– , , ) is PQ = ) ( ) = = units Example Show that the points P (– , , ), Q ( , , ) and R ( , , – ) are collinear. Solution We know that points are said to be collinear if they lie on a line. Now, PQ = ( QR = ) ) and PR = ) Thus, PQ + QR = PR.
Hence, P, Q and R are collinear.