Rectangular co-ordinate system & Shifting of origin

RECTANGULAR COORDINATE 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}
    • OP = √(x2+y2+z2)

    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:

    1. 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 }

    2. 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

    3. 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} )

    4. 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.
    (i) 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.
    (ii) 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.

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