Distance and Midpoint in 3D
Distance and Midpoint in 3D

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Analytic Geometry in Three Dimensions

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The Three-Dimensional Coordinate System 11.1

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Objectives Plot points in the three-dimensional coordinate system.
Find distances between points in space and find midpoints of line segments joining points in space. Write equations of spheres in standard form and find traces of surfaces in space.

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The Three-Dimensional Coordinate System

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The Three-Dimensional Coordinate System
We know that the Cartesian plane is formed by two perpendicular number lines, the x-axis and the y-axis. These axes determine a two-dimensional coordinate system for identifying points in a plane. To identify a point in space, you must introduce a third dimension to the model. The geometry of this three-dimensional model is solid analytic geometry.

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The Three-Dimensional Coordinate System
You can construct a three-dimensional coordinate system by passing a z-axis perpendicular to both the x- and y-axes at the origin, as shown in the figure below.

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The Three-Dimensional Coordinate System
Taken as pairs, the axes determine three coordinate planes: the xy-plane, the xz-plane, and the yz-plane. These planes separate the three-dimensional coordinate system into eight octants. The first octant is the one in which all three coordinates are positive.

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The Three-Dimensional Coordinate System
A point P in space is determined by an ordered triple (x, y, z), where x, y, and z are as follows. x = directed distance from yz-plane to P y = directed distance from xz-plane to P z = directed distance from xy-plane to P A three-dimensional coordinate system can have either a left-handed or a right-handed orientation.

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The Three-Dimensional Coordinate System
In this text, you will work exclusively with right-handed systems, as illustrated below.

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The Three-Dimensional Coordinate System
In a right-handed system, Octants II, III, and IV are found by rotating counterclockwise around the positive z-axis. Octant V is vertically below Octant I. Octants VI, VII, and VIII are then found by rotating counterclockwise around the negative z-axis. See below. Octant II Octant III Octant IV Octant I Octant V Octant VI Octant VII Octant VIII

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Example 1 – Plotting Points in Space
Plot each point in space. a. (2, –3, 3) b. (–2, 6, 2) c. (1, 4, 0) d. (2, 2, –3)

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Example 1 – Solution To plot the point (2, –3, 3), notice that x = 2, y = –3, and z = 3. To help visualize the point, locate the point (2, –3) in the xy-plane (denoted by a cross in Figure 11.1). The point (2, –3, 3) lies three units above the cross. The other three points are also shown in Figure 11.1. Figure 11.1

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The Distance and Midpoint Formulas

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The Distance and Midpoint Formulas
Many of the formulas established for the two-dimensional coordinate system can be extended to three dimensions. For example, to find the distance between two points in space, you can use the Pythagorean Theorem twice, as shown in Figure 11.2. Figure 11.2

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The Distance and Midpoint Formulas
Note that a = x2 – x1, b = y2 – y1, and c = z2 – z1.

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Example 2 – Finding the Distance Between Two Points in Space
Find the distance between (1, 0, 2) and (2, 4, –3). Solution: Distance Formula in Space Substitute. Simplify. Simplify.

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The Distance and Midpoint Formulas
Notice the similarity between the Distance Formulas in the plane and in space. The Midpoint Formulas in the plane and in space are also similar.

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Example 3 – Using the Midpoint Formula in Space
Find the midpoint of the line segment joining (5, –2, 3) and (0, 4, 4). Solution: Using the Midpoint Formula in Space, the midpoint is as shown at the right.

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The Equation of a Sphere

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The Equation of a Sphere
A sphere with center (h, k, j ) and radius r is defined as the set of all points (x, y, z) such that the distance between (x, y, z) and (h, k, j ) is r, as shown in Figure Using the Distance Formula, this condition can be written as Figure 11.3

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The Equation of a Sphere
By squaring each side of this equation, you obtain the standard equation of a sphere.

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The Equation of a Sphere
Notice the similarity of this formula to the equation of a circle in the plane. (x – h)2 + (y – k)2 + (z – j )2 = r 2 (x – h)2 + (y – k)2 = r 2 As is true with the equation of a circle, the equation of a sphere is simplified when the center lies at the origin. In this case, the equation is x2 + y2 + z2 = r 2. Equation of sphere in space Equation of circle in the plane Sphere with center at origin

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Example 4 – Finding the Equation of a Sphere
Find the standard equation of the sphere with center (2, 4, 3) and radius 3. Does this sphere intersect the xy-plane? Solution: (x – h)2 + (y – k)2 + (z – j )2 = r2 (x – 2)2 + (y – 4)2 + (z – 3)2 = 32 Standard equation Substitute.

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Example 4 – Solution cont’d From the graph shown in Figure 11.4, you can see that the center of the sphere lies three units above the xy-plane. Because the sphere has a radius of 3, you can conclude that it does intersect the xy-plane—at the point (2, 4, 0). Figure 11.4

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The Equation of a Sphere
Note that the points satisfying the equation of the sphere are “surface points,” not “interior points.” In general, the collection of points satisfying an equation involving x, y, and z is called a surface in space. Finding the intersection of a surface with one of the three coordinate planes (or with a plane parallel to one of the three coordinate planes) helps one visualize the surface. Such an intersection is called a trace of the surface.

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The Equation of a Sphere
For example, the xy–trace of a surface consists of all points that are common to both the surface and the xy-plane. Similarly, the xz-trace of a surface consists of all points that are common to both the surface and the xz-plane.

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Example 6 – Finding a Trace of a Surface
Sketch the xy-trace of the sphere given by (x – 3)2 + (y – 2)2 + (z + 4)2 = 52. Solution: To find the xy-trace of this surface, use the fact that every point in the xy-plane has a z-coordinate of zero. By substituting z = 0 into the original equation, the resulting equation will represent the intersection of the surface with the xy-plane. (x – 3)2 + (y – 2)2 + (z + 4)2 = 52 Write original equation.

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Example 6 – Solution cont’d (x – 3)2 + (y – 2)2 + (0 + 4)2 = 52 (x – 3)2 + (y – 2) = 25 (x – 3)2 + (y – 2)2 = 9 (x – 3)2 + (y – 2)2 = 32 Substitute 0 for z. Simplify. Subtract 16 from each side. Equation of circle

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Example 6 – Solution cont’d You can see that the xy-trace is a circle of radius 3, as shown below.

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