Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

112 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
(b) First we find the distances between points:
I DEl = J(l- 0) 2 + [- 2- ( - 5)]2 + (4 - 5) 2 ::::: VIT
IEFI = )(3 - 1) 2 + [4 -
( - 2)]2 + (2 - 4) 2 = v'44 = 2 VIT
IDFI = )(3 - 0) 2 + [4- (-5)]2 + (2 - 5) 2 = v'99 = 3Vll
Since I DEl + IEFI = ID F I, the three points lie on a straight line.
11. An equation of the sphere with center (-3, 2, 5) and radius 4 is [x - (- 3)] 2 + (y - 2) 2 + (z - 5) 2 = 4 2 or
(x + 3) 2 + {y - 2) 2 + (z- 5) 2 = 16. The intersection of this sphere with the yz-plane is the set of points on the spher~
whose x-coordinate is 0. Putting x = 0 into the equation, we have 9 + (y - 2) 2 + (z- 5) 2 = 16, x = 0 or
(y- 2) 2 + (z- 5) 2 = 7, x = 0, which represents a circle in the yz-plane with center (0, 2, 5) and radius -/7.
13. The radiusofthesphere is the distance between (4,3, - 1) and (3,8, 1): r = )(3 - 4)2 + (8 - 3)2 + [1 - (-1)]2 = V3o.
Thus, an equation of the sphere is (x - 3) 2 + (y - 8) 2 + (z- 1) 2 = 30.
15. Completing squares in the equation x 2 + y 2 + z 2 - 2x - 4y + 8z = 15 gives
recognize as an equation of a sphere with center (1, 2, - 4) and radius 6.
17. Completing squares in the equation 2x 2 - 8x + 2y 2 + 2z 2 + 24z = 1 gives
2(x 2 - 4x + 4) + 2y 2 + 2~z 2 + 12z + 36) = 1 + 8 + 72 :::} 2(x - 2) 2 + 2y 2 + 2(z + 6? = 81 :::}
(x - 2) 2 + y 2 + (z + 6) 2 = 8 , which we recognize as an equation of a sphere with center (2, 0, -6) and
21
radius j¥ = 9/ v'2.
19. (a) lfthe midpointofthe line segment from P 1(x 1 ,y1,z1) to P 2 (x 2 ,y 2 ,z 2 ) is Q = (x 1 ;x 2 , Y 1 ;v 2 , z 1 ; z 2 ) ,
then the distances IPl Q l and IQP21 are equal, and each is half of IPt P21· We verify that this is the case:
IP1QI = v [~ (x1 + X2) - X1] 2 + a