Question

Prove that the distance between the points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is given by
$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Answer 1

Let O be the origin, and let $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be the given point. From P and Q, draw perpendiculars PM and QN respectively on the xy-plane.

Then, the coordinates of M and N are

$M(x_1,y_1,0)$ and $N(x_2,y_2,0)$.

The two-dimensional coordinates of M and N referred to OX and OY are

$M(x_1,y_1)$ and $N(x_2,y_2)$.

$\therefore M{N}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$.

Now, from P draw $PR\perp QN$.

Then, PR is parallel and equal to MN.

Now, in right triangle PRQ, we have

$P{Q}^2=P{R}^2+R{Q}^2$

$=M{N}^2+(QN-RN{)}^2$

$=M{N}^2+(QN-PM{)}^2$

$=(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2$ $[\because PM=z_1\text{and}QN=z_2]$

$\therefore PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$.