Presentation on theme: "The Three-Dimensional Coordinate System 11.1"— Presentation transcript:
1The Three-Dimensional Coordinate System 11.1 JMerrill, 2010
2Solid 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.
3Coordinate PlanesNotice we draw the x- and y-axes in the opposite directionX = directed distance from yz-plane to some point PY= directed distance from xz-plane to some point PZ= directed distance from xy-plane to some point P(x,y,z)So, to plot points you go out, over, up/down
4OctantsThe 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.
6Plotting 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.
7FormulasYou can use many of the same formulas that you already know because right triangles are still formed.
8The Distance FormulaIt looks the same in space as it did before except with a third coordinate:
9ExampleFind the distance between (1, 0, 2) and (2, 4,-3)
10Midpoint 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.
11Midpoint You DoFind the midpoint of the line segment joining (5, -2, 3) and (0, 4, 4)
12Equation of a Sphere The equation of a circle is x2 + y2 = r2 If the center is not at the origin, then the equation is (x-h)2 + (y-k)2 = r2The equation of a sphere whose center is at (h,k,j) with radius r is (x-h)2 + (y-k)2 + (z–j)2= r2
13Finding 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= r2(x-2)2 + (y-4)2 + (z–3)2 = 32Does 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).
14Finding the Center and Radius of a Sphere Find the center and radius of the sphere given by x2 + y2 + z2 – 2x + 4y – 6z +8 = 0This 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.
15Finding the Center and Radius of a Sphere x2 + y2 + z2 – 2x + 4y – 6z +8 = 0(x-1)2 + (y+2)2 + (z-3)2 = 6The center is (1,-2,3) and the radius is √6.