# Rectangular Co-ordinate System in Space , Shifting of origin , Solved Examples

#### Rectangular Co-ordinate System in Space :

Let ‘O’ be any point in space and X’OX , Y’OY and Z’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 whose position 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 unit vectors along OX, OY, OZ respectively, then position vector of point P is or simply (x, y, z).

Note: Any point on –

X-axis = {( x, y, z) | y = z = 0}
Y-axis = {(x, y, z) | x = z = 0}
Z-axis = {(x, y, z) | x = y = 0}
XY plane = {(x, y, z) | z = 0}
YZ plane = {(x, y, z) | x = 0}
ZX plane = {(x, y, z) | y = 0}

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

The three co-ordinate planes divide the whole space in eight compartments which are known as eight octants and since each of the coordinates of a point may be positive or negative, there are 23 (= 8) points whose coordinates have the same numerical values and which lie in eight octants, one in each.

#### Shifting of 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 : If the origin is shifted to (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: x1 = x – x’ , where (x’, y’, z’) is the shifted origin

y1 = y – y’

z1 = z – z’

x1 = 0 – 1 = -1

y1 = 4 – 2 = 2

z1 = 5 + 3 = 8

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

Note:

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

$\displaystyle = \sqrt{(x_1 – x_2)^2 +(y_1 – y_2)^2 + (z_1 – z_2)^2 }$

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

$\displaystyle ( \frac{m x_2 + n x_1}{m+n} , \frac{m y_2 + n y_1}{m+n} , \frac{m z_2 + n z_1}{m+n} )$ where m + n ≠ 0

(iii) The coordinates of the centroid of a triangle having vertices A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is

$\displaystyle G(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} , \frac{z_1 + z_2 + z_3}{3} )$

(iv) Tetrahedron: Tetrahedron is a figure bounded by four planes, not all of which pass through the same point. It has four vertices, each vertex arising as a point of intersection of three of the four planes. It has six faces.
(a) The four lines drawn from the vertices of a tetrahedron to the centroids of the opposite faces meet in a point, known as centroid of the tetrahedron, which is at three fourth of the distance from each vertex to the opposite face.
(b) If the vertices of a tetrahedron are (xi, yi, zi), i = 1, 2, 3, 4 then coordinates of its centroid is

$\displaystyle ( \frac{\Sigma x_i}{4} , \frac{\Sigma y_i}{4} , \frac{\Sigma z_i}{4} )$

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

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

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

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

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

Hence the required point is $\displaystyle \frac{21}{8} , -\frac{11}{8} , \frac{47}{8}$

Illustration : Prove that the three points A (3, -2, 4), B (1, 1, 1) and C (-1, 4, -2) are collinear.

Solution: The general coordinates of a point R which divides the line joining A (3, -2, 4) and B (1, 1, 1) in the ratio μ : 1 are

$\displaystyle ( \frac{\mu + 3}{\mu + 1} , \frac{\mu -2}{\mu + 1} , \frac{\mu + 4}{\mu + 1} )$ ……(1)

If C (-1, 4, -2) lies on the line AB, then for some value of μ the coordinates of the point R will be the same as those of C.

Let x-coordinate of point R = x-coordinate of point C.

Then, $\displaystyle \frac{\mu + 3}{\mu + 1} = -1$

=> μ = -2

Putting μ = -2 in (1) the coordinates of R are (-1, 4, -2) which are also the coordinates of C.

Hence the points A, B, C are collinear.

Exercise :
(i) Find the coordinates of the point equidistant from the four points (a, 0, 0), (0, b, 0), (0, 0, c) and (0, 0, 0).

(ii) The three points A (0, 0, ,0), B (2, – 3, 3), C (- 2, 3, – 3) are collinear. Find the ratio in which each point divides the segment joining the other two.