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The standard-form equation of a plane in xyz space looks like an extrapolation of the standardform equation of a straight line in the xy plane This can confuse some people Don t let it baffle you! An equation of the form ax + by + cz + d = 0 where a, b, c, and d are constants represents a plane, not a line In Chap 18, you ll learn how to describe straight lines in Cartesian xyz space
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Draw a graph of the plane represented by the following equation: 2x 4y + 3z 12 = 0
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Let s work out the graph by finding the coordinate-axis intercepts The x-intercept, or the point where the plane intersects the x axis, can be found by setting y = 0 and z = 0, and then solving the resultant equation for x Let s call this point P We have 2x 4 0 + 3 0 12 = 0 Solving step-by-step, we get 2x 12 = 0 2x = 12 x = 12/( 2) = 6 Therefore P = ( 6,0,0)
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The y-intercept, or the point where the plane intersects the y axis, can be found by setting x = 0 and z = 0, and then solving the resultant equation for y Let s call this point Q We have 2 0 4y + 3 0 12 = 0 Solving, we get 4y 12 = 0 4y = 12 y = 12/( 4) = 3 Therefore Q = (0, 3,0) The z-intercept, or the point where the plane intersects the z axis, can be found by setting x = 0 and y = 0, and then solving the resultant equation for z Let s call this point R We have 2 0 4 0 + 3z 12 = 0 Solving, we get 3z 12 = 0 3z = 12 z = 12/3 = 4 Therefore R = (0,0,4) These three points are shown in Fig 17-2 We can now envision the plane because, as we recall from our courses in spatial geometry, a plane in three dimensions can be uniquely defined on the basis of three points
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+y P = ( 6, 0, 0) Each axis division is 1 unit
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+x Q = (0, 3, 0)
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R = (0, 0, 4) y
Figure 17-2
Here s the graph of a plane, based on the locations of the three axis intercept points P, Q, and R
Surfaces in Three-Space
Spheres
A spherical surface is defined as the set of all points that lie at a fixed distance from a known central point in three dimensions When we recall the formula for the distance between a point and the origin, it s easy to work out equations for spheres in Cartesian xyz space
Center at the origin Imagine a sphere whose center lies at the origin (0,0,0), as shown in Fig 17-3 Any point on the sphere s surface is at the same distance from the origin as any other point on the sphere s surface Suppose that P is one such point whose coordinates are given by
P = (xp,yp,zp) In Chap 7, we learned that the distance r of the point P from the origin in Cartesian xyz space is r = (xp2 + yp2 + zp2)1/2 We can square both sides of the above equation to get r2 = xp2 + yp2 + zp2
+y Center of sphere is at (0, 0, 0)
Radius = r
Figure 17-3
A sphere of radius r in Cartesian xyz space, centered at the origin All points on the sphere s surface are at distance r from the center point (0,0,0)
Spheres
Transposing the left- and right-hand sides, we have xp2 + yp2 + zp2 = r2 Every point on the sphere s surface is the same distance from the origin as P, so we can generalize the above equation to get x2 + y2 + z2 = r2 which defines the set of all points in three dimensions that lie at a fixed distance r from the origin That s all there is to it! We ve found the standard-form equation for a sphere of radius r, centered at the origin in Cartesian xyz space
Center away from the origin Consider a sphere whose center is somewhere other than the origin in Cartesian xyz space Suppose that the coordinates of the center point are (x0,y0,z0), as shown in Fig 17-4 Whatever point P that we choose on the sphere s surface, the distance between P and the center is equal to the sphere s radius r Adapting the distance-between-points formula for Cartesian xyz space from Chap 7, we get
r = [(xp x0)2 + (yp y0)2 + (zp z0)2]1/2
Center of sphere is at (x0, y0, z0)
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