Question

Distance between the points (1,3,2) and (2,1,3) is
A. $12$
B. $\sqrt {12} $
C. $\sqrt 6 $
D. $6$

Answer 1

Hint: Use the distance formula to calculate the distance between two points. Distance formula is as follows: Given two points $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ the distance between P and Q is given by the formula: distance = \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]

Formula used: Distance = \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]

Complete step by step solution:
We know that the distance between two points, $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ is given by the formula \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \].
Let (1,3,2) be $P({x_1},{y_1},{z_1})$ and (2,1,3) be $Q({x_2},{y_2},{z_2})$.
Distance between the two points $PQ = \sqrt {{{(2 - 1)}^2} + {{(1 - 3)}^2} + {{(3 - 2)}^2}} $ units
$PQ = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(1)}^2}} $ units
$PQ = \sqrt {1 + 4 + 1} $ units
$PQ = \sqrt 6 $ units

Therefore, the correct option is option C. $\sqrt 6 $

Additional information: Three mathematicians are credited for the development of the distance formula- Euclid, Pythagoras and Descartes. Rene Descartes is said to have played the greatest role in developing the distance formula.

Note: Distance between two points $A(p,q)\,{\text{and}}\,B(r,s)$ in a two-dimensional plane is given by the formula: $AB = \sqrt {{{(p - r)}^2} + {{(q - s)}^2}} $. We shall see the proof for this formula:

Image: Representation of points in 2D coordinate system
We know that in the right triangle, ABC, $A{B^2} = A{C^2} + B{C^2}$
$AC = r - p \Rightarrow A{C^2} = {(r - p)^2} = {(p - r)^2}$
$BC = s - q \Rightarrow B{C^2} = {(s - q)^2} = {(q - s)^2}$
$A{B^2} = A{C^2} + {BC^2} = {(p - r)^2} + {(q - s)^2}$
Therefore, $AB = \sqrt {{{(p - r)}^2} + {{(q - s)}^2}} $
Similarly, we can derive the distance between two points in a three-dimensional plane as well.