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Introduction to Three Dimensional Geometry - Coordinates of a Point in Space

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Coordinate Axes: In three-dimensional geometry, we establish three mutually perpendicular lines passing through a fixed point OO (the origin). These lines are the XX, YY, and ZZ axes. Visually, imagine the corner of a room where the floor meets two walls; the lines formed by these three intersections represent the three axes extending in both positive and negative directions.

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Coordinate Planes: The three axes taken in pairs determine three coordinate planes: the XYXY-plane, the YZYZ-plane, and the ZXZX-plane. These planes are like the floor and two adjacent walls of a room. They divide the entire 3D space into eight distinct regions called octants. For example, the XYXY-plane is the flat surface where the zz-coordinate is always zero.

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Coordinates of a Point: Every point PP in space is represented by an ordered triplet (x,y,z)(x, y, z). These values represent the perpendicular distances from the YZYZ, ZXZX, and XYXY planes, respectively. Visually, to locate a point like (3,2,4)(3, 2, 4), you would move 3 units along the XX-axis, 2 units parallel to the YY-axis, and then 4 units upward parallel to the ZZ-axis.

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Octants and Sign Convention: The signs of the coordinates (x,y,z)(x, y, z) determine which of the eight octants a point lies in. In the first octant, all coordinates are positive (+,+,+)(+, +, +). In the fifth octant, xx and yy are positive but zz is negative (+,+,−)(+, +, -). You can visualize the first four octants as the 'upper floor' and octants V through VIII as the 'lower floor' of the 3D grid.

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Points on Axes and Planes: Any point lying exactly on the XX-axis has the form (x,0,0)(x, 0, 0), as it has no displacement toward the YY or ZZ directions. Similarly, any point lying on the XYXY-plane has the form (x,y,0)(x, y, 0). This 'zeroing out' of coordinates is a key visual cue for points restricted to specific axes or surfaces.

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Distance Formula Logic: The distance between two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) is the length of the direct line segment connecting them. Visually, this distance represents the space diagonal of a rectangular box (parallelepiped) whose sides are parallel to the coordinate axes and whose length, width, and height are the differences between the coordinates.

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Section Formula Logic: This concept identifies the coordinates of a point that divides the line segment joining two points in a given ratio m:nm:n. If the point is between the two ends, it is an internal division; if it lies on the extension of the segment, it is an external division. Visually, a 1:11:1 ratio identifies the midpoint, located exactly at the center of the segment.

📐Formulae

Distance between points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2): PQ=sqrt(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

Section Formula (Internal): P=left(fracmx2+nx1m+n,fracmy2+ny1m+n,fracmz2+nz1m+nright)P = \\left( \\frac{mx_2+nx_1}{m+n}, \\frac{my_2+ny_1}{m+n}, \\frac{mz_2+nz_1}{m+n} \\right)

Section Formula (External): P=left(fracmx2−nx1m−n,fracmy2−ny1m−n,fracmz2−nz1m−nright)P = \\left( \\frac{mx_2-nx_1}{m-n}, \\frac{my_2-ny_1}{m-n}, \\frac{mz_2-nz_1}{m-n} \\right)

Midpoint of segment ABAB: M=left(fracx1+x22,fracy1+y22,fracz1+z22right)M = \\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2}, \\frac{z_1+z_2}{2} \\right)

Centroid of a Triangle: G=left(fracx1+x2+x33,fracy1+y2+y33,fracz1+z2+z33right)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)

💡Examples

Problem 1:

Find the distance between the points A(3,−1,2)A(3, -1, 2) and B(−1,2,2)B(-1, 2, 2).

Solution:

  1. Identify coordinates: (x1,y1,z1)=(3,−1,2)(x_1, y_1, z_1) = (3, -1, 2) and (x2,y2,z2)=(−1,2,2)(x_2, y_2, z_2) = (-1, 2, 2).\n2. Apply the distance formula: AB=sqrt(−1−3)2+(2−(−1))2+(2−2)2AB = \\sqrt{(-1-3)^2 + (2-(-1))^2 + (2-2)^2}.\n3. Simplify inside the square root: AB=sqrt(−4)2+(3)2+(0)2AB = \\sqrt{(-4)^2 + (3)^2 + (0)^2}.\n4. Calculate squares: AB=sqrt16+9+0=sqrt25AB = \\sqrt{16 + 9 + 0} = \\sqrt{25}.\n5. Result: AB=5AB = 5 units.

Explanation:

We calculate the difference between each corresponding coordinate, square those differences, sum them up, and finally take the square root. Since the zz-coordinates are the same, the distance calculation effectively becomes a 2D distance calculation in the plane z=2z=2.

Problem 2:

Find the coordinates of the point that divides the line segment joining P(1,−2,3)P(1, -2, 3) and Q(3,4,−5)Q(3, 4, -5) internally in the ratio 2:32:3.

Solution:

  1. Identify values: (x1,y1,z1)=(1,−2,3)(x_1, y_1, z_1) = (1, -2, 3), (x2,y2,z2)=(3,4,−5)(x_2, y_2, z_2) = (3, 4, -5), m=2m=2, n=3n=3.\n2. Calculate the x-coordinate: x=frac2(3)+3(1)2+3=frac6+35=frac95x = \\frac{2(3) + 3(1)}{2+3} = \\frac{6+3}{5} = \\frac{9}{5}.\n3. Calculate the y-coordinate: y=frac2(4)+3(−2)2+3=frac8−65=frac25y = \\frac{2(4) + 3(-2)}{2+3} = \\frac{8-6}{5} = \\frac{2}{5}.\n4. Calculate the z-coordinate: z=frac2(−5)+3(3)2+3=frac−10+95=−frac15z = \\frac{2(-5) + 3(3)}{2+3} = \\frac{-10+9}{5} = -\\frac{1}{5}.\n5. Final Point: Rleft(frac95,frac25,−frac15right)R\\left(\\frac{9}{5}, \\frac{2}{5}, -\\frac{1}{5}\\right).

Explanation:

The internal section formula provides a weighted average of the endpoints' coordinates. Because the ratio is 2:32:3, the point RR is located closer to point PP than to point QQ.