Solução_Calculo_Stewart_6e

F.
106 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ETTX.10
CHAPTER 12
(b) First we find the distances between points:
|DE| = (1 − 0) 2 +[−2 − (−5)] 2 +(4− 5) 2 = √ 11
|EF| = (3 − 1) 2 +[4− (−2)] 2 +(2− 4) 2 = √ 44 = 2 √ 11
|DF| = (3 − 0) 2 +[4− (−5)] 2 +(2− 5) 2 = √ 99 = 3 √ 11
Since |DE| + |EF| = |DF|, the three points lie on a straight line.
11. An equation of the sphere with center (1, −4, 3) and radius 5 is (x − 1) 2 +[y − (−4)] 2 +(z − 3) 2 =5 2 or
(x − 1) 2 +(y +4) 2 +(z − 3) 2 =25. The intersection of this sphere with the xz-plane is the set of points on the sphere
whose y-coordinate is 0. Putting y =0into the equation, we have (x − 1) 2 +4 2 +(z − 3) 2 =25,y=0or
(x − 1) 2 +(z − 3) 2 =9,y=0, which represents a circle in the xz-plane with center (1, 0, 3) and radius 3.
13. The radius of the sphere is the distance between (4, 3, −1) and (3, 8, 1): r = (3 − 4) 2 +(8− 3) 2 +[1− (−1)] 2 = √ 30.
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 − 6x +4y − 2z =11gives
(x 2 − 6x +9)+(y 2 +4y +4)+(z 2 − 2z +1)=11+9+4+1 ⇒ (x − 3) 2 +(y +2) 2 +(z − 1) 2 =25,whichwe
recognize as an equation of a sphere with center (3, −2, 1) and radius 5.
17. Completing squares in the equation 2x 2 − 8x +2y 2 +2z 2 +24z =1gives
2(x 2 − 4x +4)+2y 2 +2(z 2 +12z +36)=1+8+72 ⇒ 2(x − 2) 2 +2y 2 +2(z +6) 2 =81 ⇒
(x − 2) 2 + y 2 +(z +6) 2 = 81 , which we recognize as an equation of a sphere with center (2, 0, −6) and radius
2
81
=9/√ 2.
2
19. (a) If the midpoint of the line segment from P 1(x 1,y 1,z 1) to P 2(x 2,y 2,z 2) is Q =
x1 + x 2
2
, y 1 + y 2
2
, z 1 + z
2
,
2
then the distances |P 1 Q| and |QP 2 | are equal, and each is half of |P 1 P 2 |. We verify that this is the case:
|P 1 P 2 | = (x 2 − x 1 ) 2 +(y 2 − y 1 ) 2 +(z 2 − z 1 ) 2
|P 1Q| = 1
2 (x1 + x2) − 2 x1 + 1
2 (y1 + y2) − 2 y1 + 1
2 (z1 + z2) − 2
z1
= 1
2 x2 − 1 2
2 x1 + 1
2 y2 − 1 2
2 y1 + 1
2 z2 − 1 2
2 z1
= 1
2
(x2 − x
2 1) 2 +(y 2 − y 1) 2 +(z 2 − z 1) 2
= 1 (x
2 2 − x 1) 2 +(y 2 − y 1) 2 +(z 2 − z 1) 2
= 1 |P1P2|
2
x2
|QP 2 | = − 1 (x 2 1 + x 2 ) 2 + y 2 − 1 (y 2 1 + y 2 ) 2 + z 2 − 1 (z 2 1 + z 2 ) 2
= 1 x 2 2 − 1 x 2
2 1 + 1
y 2 2 − 1 y 2
2 1 +
1
z 2 2 − 1 z 2
2 1
= 1
2
(x2 − x
2
1 ) 2 +(y 2 − y 1 ) 2 +(z 2 − z 1 ) 2
(x 2 − x 1 ) 2 +(y 2 − y 1 ) 2 +(z 2 − z 1 ) 2 = 1 |P 2 1P 2 |
= 1 2
So Q is indeed the midpoint of P 1 P 2 .