Find the points on z - axis which are at a distance 21 from the point ( 1 , 2 , 3 ) .

Let the point on z - axis be A ( 0 , 0 , z ) and P ( 1 , 2 , 3 ) . Then, AP = 21 ⇒ ( 0 - 1 ) 2 + ( 0 - 2 ) 2 + ( z - 3 ) 2 = 21 [Using distance formula] ⇒ ( - 1 ) 2 + ( - 2 ) 2 + ( z - 3 ) 2 = 21 [Squaring both sides] ⇒ 1 + 4 + ( z - 3 ) 2 = 21 ⇒ ( z - 3 ) 2 = 21 - 5 ⇒ ( z - 3 ) 2 = 16 ⇒ z - 3 = ± 4 ⇒ z - 3 = 4 or z - 3 = - 4 ⇒ z = 7 or z = - 1 Hence, the coordinates of the required point are ( 0 , 0 , 7 ) and ( 0 , 0 , - 1 ) .

Prove that the triangle formed by joining the three points whose coordinates are ( 1 , 2 , 3 ) ( 2 , 3 , 1 ) and ( 3 , 1 , 2 ) is an equilateral triangle.

Let A ( 1 , 2 , 3 ) , B ( 2 , 3 , 1 ) and C ( 3 , 1 , 2 ) are the coordinates of the triangle ∆ A B C We know that distance formula between two points, Distance = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2 Now, A B = ( 2 - 1 ) 2 + ( 3 - 2 ) 2 + ( 1 - 3 ) 2 = ( 1 ) 2 + ( 1 ) 2 + ( - 2 ) 2 = 1 + 1 + 4 = 6 BC = ( 3 - 2 ) 2 + ( 1 - 3 ) 2 + ( 2 - 1 ) 2 = ( 1 ) 2 + ( - 2 ) 2 + ( 1 ) 2 = 1 + 4 + 1 = 6 A C = ( 3 - 1 ) 2 + ( 1 - 2 ) 2 + ( 2 - 3 ) 2 = ( 2 ) 2 + ( - 1 ) 2 + ( - 1 ) 2 = 4 + 1 + 1 = 6 Clearly, AB = BC = AC = 6 units We know that all sides of a triangle are equal in an equilateral triangle. Here all sides are equal. Hence proved that given points are vertices of equilateral triangle.