Rectangular Coordinate System in Space

Rectangular Coordinate System in Space

Let 'O' be any point in space and $\overleftrightarrow{X^{\prime }OX},\overleftrightarrow{Y^{\prime }OY}and\overleftrightarrow{Z^{\prime }OZ}$ be three lines perpendicular to each other. These lines are known as coordinate axes and O is called origin. The planes XY, YZ, ZX are known as the coordinate planes.

Coordinates of a Point in Space:

Consider a point P in space. The position of the point P is given by triad (x, y, z) where x, y, z are perpendicular distance from YZ-plane, ZX-plane and XY-plane respectively.

If We assume $\hat{i},\hat{j},\hat{k}$ unit vectors along OX, OY, OZ respectively, then position vector of point P is $x\hat{i}+y\hat{j}+z\hat{k}$ or simply (x, y, z).

Note:

$\bullet$ x-axis = {( x, y, z) | y = z = 0}

$\bullet$ y-axis = {(x, y, z) | x = z = 0}

$\bullet$ z-axis = {(x, y, z) | x = y = 0}

$\bullet$ xy plane = {(x, y, z) | z = 0}

$\bullet$ yz plane = {(x, y, z) | x = 0}

$\bullet$ zx plane = {(x, y, z) | y = 0}

$\bullet$ OP = $\sqrt{x^2+y^2+z^2}$

Shifting the Origin:

Shifting the origin to another point without changing the directions of the axes is called the translation of axes.

Let the origin O be shifted to another point

O' (x', y', z') without changing the direction of axes. Let the new coordinate frame be O'X'Y'Z'. Let P (x, y, z) be a point with respect to the coordinate frame OXYZ.

Then, coordinate of point P w.r.t. new coordinate frame O'X'Y'Z' is (x₁, y₁, z₁), where

x₁ = x – x', y₁ = y – y', z₁ = z – z'

Illustration 1: If the origin is shifted (1, 2, –3) without changing the directions of the axes then find the new coordinates of the point (0, 4, 5) with respect to new frame.

Solution: x' = x – x₁, where (x₁, y₁, z₁) is the shifted origin

y ' = y – y₁

z' = z – z₁

x' = 0 – 1 = –1

y' = 4 – 2 = 2

z' = 5 + 3 = 8

$\therefore$ The coordinates of the point w.r.t. to new coordinate frame is (-1, 2, 8).

Note:

$\bullet$ Distance between the points P(x₁, y₁, z₁) and Q (x₂, y₂, z₂) is

$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

$\bullet$ The point dividing the line joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in m : n ratio is

$\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$ where m + n $\ne$0 .

$\bullet$ The coordinates of centroid of a triangle having vertices A (x₁, y₁, z₁), B (x₂, y₂, z₂) and C (x₃, y₃, z₃) is G $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$.

Illustration 2: Find the coordinates of the point which divides the line joining points (2, 3, 4) and (3, –4, 7) in ratio 3 : 5.

Solution: Let the coordinates of the required point be (x, y, z), then

x = $\frac{2(3)+3(5)}{3+5}=\frac{21}{8}$

y = $\frac{3(3)-4(5)}{3+5}=-\frac{11}{8}$

z = $\frac{4(3)+7(5)}{3+5}=\frac{47}{8}$

Hence the required point is $\left(\frac{21}{8},\frac{-11}{8},\frac{47}{8}\right)$.