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The Three-Dimensional Coordinate System 11.1 JMerrill, 2010.

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Presentation on theme: "The Three-Dimensional Coordinate System 11.1 JMerrill, 2010."— Presentation transcript:

1 The Three-Dimensional Coordinate System 11.1 JMerrill, 2010

2 Solid Analytic Geometry The Cartesian plane (rectangular coordinate system) is determined by 2 perpendicular number line (x- and y- axis) and their point of intersection (the origin). To identify a point in space, we need a third dimension. The geometry of this three-dimensional model is called solid analytic geometry. The 3-D coordinate system is formed by passing a z-axis perpendicular to both the x- and y-axes at the origin.

3 Coordinate Planes Notice we draw the x- and y-axes in the opposite direction X = directed distance from yz- plane to some point P Y= directed distance from xz- plane to some point P Z= directed distance from xy- plane to some point P (x,y,z) So, to plot points you go out, over, up/down

4 Octants The 3-D system can have either a right-handed or a left-handed orientation. We’re only using the right-handed orientation meaning that the octants (quadrants) are numbered by rotating counterclockwise around the positive z-axis. There are 8 octants.

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6 Plotting Points in Space Plot the points: (2,-3,3) (-2,6,2) (1,4,0) (2,2,-3) Draw a sideways x, then put a perpendicular line through the origin.

7 Formulas You can use many of the same formulas that you already know because right triangles are still formed.

8 The Distance Formula It looks the same in space as it did before except with a third coordinate:

9 Example Find the distance between (1, 0, 2) and (2, 4,-3)

10 Midpoint Formula The midpoint formula is What is the midpoint if you make a 100 on a test and an 80 on a test? So the midpoint is just the average of the x’s, y’s, and z’s.

11 Midpoint You Do Find the midpoint of the line segment joining (5, -2, 3) and (0, 4, 4)

12 Equation of a Sphere The equation of a circle is x 2 + y 2 = r 2 If the center is not at the origin, then the equation is (x-h) 2 + (y-k) 2 = r 2 The equation of a sphere whose center is at (h,k,j) with radius r is (x-h) 2 + (y-k) 2 + (z–j) 2 = r 2

13 Finding the Equation of a Sphere Find the standard equation of a sphere with center (2,4,3) and radius 3 (x-h) 2 + (y-k) 2 + (z–j) 2 = r 2 (x-2) 2 + (y-4) 2 + (z–3) 2 = 3 2 Does the sphere intersect the plane? Yes. The center of the sphere is 3 units above the y-axis and has a radius of 3. It intersects at (2,4,0).

14 Finding the Center and Radius of a Sphere Find the center and radius of the sphere given by x 2 + y 2 + z 2 – 2x + 4y – 6z +8 = 0 This works the same way as it did in 2-D space. In order to find the center, we must put the equation into standard form, which means completing the square.

15 Finding the Center and Radius of a Sphere x 2 + y 2 + z 2 – 2x + 4y – 6z +8 = 0 (x-1) 2 + (y+2) 2 + (z-3) 2 = 6 The center is (1,-2,3) and the radius is √6.


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