`\color{red} ✍️` We have studied about the distance between two points in two-dimensional coordinate system. Let us now extend this study to three-dimensional system.

`\color{red} ✍️` 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.`

`\color{red} ✍️` 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 12.4).

Now, since `∠PAQ` is a right angle, it follows that, in triangle `PAQ,`

`color(blue)(PQ^2 = PA^2 + AQ^2)` ................................ (1)

Also, triangle `ANQ` is right angle triangle with `∠ANQ` a right angle.

Therefore `color(blue)(AQ^2 = AN^2 + NQ^2)` ................................. (2)

From `(1)` and `(2),` we have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(green)(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 `color{red}(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).`

`\color{red} ✍️` In particular, if `color(blue)(x_1 = y_1 = z_1 = 0)`, i.e., point `P` is origin `O,`

then `color{orange}(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).`