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Introduction to Three-Dimensional Geometry - Coordinates of a Point

Grade 11ICSE

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

🔑Concepts

Coordinate Axes and Origin: In 3D geometry, we use three mutually perpendicular lines called the XX, YY, and ZZ axes that intersect at a single point called the Origin O(0,0,0)O(0, 0, 0). Visually, these axes can be imagined as the lines formed where two walls and the floor of a room meet.

Coordinate Planes: The three axes taken in pairs determine three coordinate planes: the XYXY-plane (where z=0z=0), the YZYZ-plane (where x=0x=0), and the ZXZX-plane (where y=0y=0). These planes are like the flat surfaces of the walls and floor in a 3D space.

Octants: The three coordinate planes divide the entire space into eight regions known as octants. The signs of the coordinates (x,y,z)(x, y, z) of a point vary depending on its octant; for instance, in the first octant, all coordinates are positive (+,+,+)(+, +, +), while in the eighth octant, they are (+,,)(+, -, -).

Coordinates of a Point: A point PP in space is represented by an ordered triple (x,y,z)(x, y, z). The value of xx is the signed perpendicular distance from the YZYZ-plane, yy from the XZXZ-plane, and zz from the XYXY-plane. Visually, moving from the origin to PP involves traveling along the xx-axis, then parallel to the yy-axis, and finally parallel to the zz-axis.

Distance from Axes: The perpendicular distance of a point P(x,y,z)P(x, y, z) from the XX-axis is given by y2+z2\sqrt{y^2 + z^2}, from the YY-axis by x2+z2\sqrt{x^2 + z^2}, and from the ZZ-axis by x2+y2\sqrt{x^2 + y^2}.

Section Formula Concept: This describes the position of a point that divides a line segment joining two points PP and QQ. If the point is between PP and QQ, it is internal division; if it lies on the extension of the line segment, it is external division.

Centroid of a Triangle: The centroid is the point where the medians of a triangle meet in 3D space. It is found by taking the arithmetic mean of the xx, yy, and zz coordinates of the three vertices.

📐Formulae

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): d=(x2x1)2+(y2y1)2+(z2z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Distance from Origin O(0,0,0)O(0, 0, 0) to point P(x,y,z)P(x, y, z): OP=x2+y2+z2OP = \sqrt{x^2 + y^2 + z^2}

Internal Section Formula: R=(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)R = (\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n})

External Section Formula: R=(mx2nx1mn,my2ny1mn,mz2nz1mn)R = (\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n})

Midpoint Formula: M=(x1+x22,y1+y22,z1+z22)M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2})

Centroid of a Triangle: G=(x1+x2+x33,y1+y2+y33,z1+z2+z33)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})

💡Examples

Problem 1:

Find the distance between the points A(3,2,5)A(3, -2, 5) and B(1,4,3)B(-1, 4, -3).

Solution:

  1. Identify the coordinates: (x1,y1,z1)=(3,2,5)(x_1, y_1, z_1) = (3, -2, 5) and (x2,y2,z2)=(1,4,3)(x_2, y_2, z_2) = (-1, 4, -3). \n2. Apply the distance formula: d=(x2x1)2+(y2y1)2+(z2z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \n3. Substitute the values: d=(13)2+(4(2))2+(35)2d = \sqrt{(-1 - 3)^2 + (4 - (-2))^2 + (-3 - 5)^2}. \n4. Simplify: d=(4)2+(6)2+(8)2d = \sqrt{(-4)^2 + (6)^2 + (-8)^2}. \n5. d=16+36+64=116d = \sqrt{16 + 36 + 64} = \sqrt{116}. \n6. Final value: d=229d = 2\sqrt{29} units.

Explanation:

The distance formula calculates the straight-line spatial distance between two points by finding the square root of the sum of the squares of the differences between their respective coordinates.

Problem 2:

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

Solution:

  1. Coordinates: P(1,2,3)P(1, -2, 3), Q(3,4,5)Q(3, 4, -5) and ratio m:n=2:3m:n = 2:3. \n2. Use Internal Section Formula: x=mx2+nx1m+nx = \frac{mx_2 + nx_1}{m+n}, y=my2+ny1m+ny = \frac{my_2 + ny_1}{m+n}, z=mz2+nz1m+nz = \frac{mz_2 + nz_1}{m+n}. \n3. For xx: x=2(3)+3(1)2+3=6+35=95x = \frac{2(3) + 3(1)}{2+3} = \frac{6+3}{5} = \frac{9}{5}. \n4. For yy: y=2(4)+3(2)2+3=865=25y = \frac{2(4) + 3(-2)}{2+3} = \frac{8-6}{5} = \frac{2}{5}. \n5. For zz: z=2(5)+3(3)2+3=10+95=15z = \frac{2(-5) + 3(3)}{2+3} = \frac{-10+9}{5} = -\frac{1}{5}. \n6. The point is R(95,25,15)R(\frac{9}{5}, \frac{2}{5}, -\frac{1}{5}).

Explanation:

The section formula determines the coordinates of a point that partitions a line segment according to a specific ratio. Since it is internal division, we add the products in the numerator and the terms in the denominator.