Three-Dimensional Space; Vectors c c
Exercise Set 11.1
c c
1. (a)c c c (0,c0,c0),c(3,c0,c0),c(3,c5,c0),c(0,c5,c0),c(0,c0,c4),c(3,c0,c4),c(3,c5,c4),c(0,c5,c4).
(b) (0,c 1,c 0),c (4,c 1,c 0),c (4,c 6,c 0),c (0,c 6,c 0),c (0,c 1,c −2),c (4,c 1,c −2),c (4,c 6,c −2),c (0,c 6,c −2).
2. Corners:c (2,c2,c±2),c (2,c−2,c±2),c (−2,c2,c±2),c (−2,c−2,c±2).
z
(–2,c–2,c2) (–2,c2,c2)
(2,c–2,c2) (2,c2,c2) y
(–2,c–2,c–2) (–2,c2,c–2)
(2,c–2,c–2) (2,c2,c–2)
x
3. Corners:c (4,c2,c−2),c (4,2,1),c (4,1,1),c (4,c1,c−2),c (−6,c1,c1),c (−6,c2,c1),c (−6,c2,c−2),c (−6,c1,c−2).
z
(–6,c2,c1)
(–6,c1,c–2) (–6,c2,c–2)
y
(4,c1,c1) (4,c2,c1)
(4,c1,c–2)
x
4. (a)c c (x2,cy1,cz1),c(x2,cy2,cz1),c(x1,cy2,cz1)(x1,cy1,cz2),c(x2,cy1,cz2),c(x1,cy2,cz2).
(b) Thec midpointc ofc thec diagonalc hasc coordinatesc whichc arec thec coordinatesc ofc thec midpointsc ofc thec edges.c The
c c c c
1c
midpointcofc thec edgec (x1c ,cy1c ,cz1c )c andc (xc2,cyc1,czc 1)c is 2(xc +cxc ),cyc ,cz ;c thec midpointc ofc thec edgec (xc ,cyc ,czc )c and
1 2 1 1 2 1 1
c
1 1 .
(xc2,cyc2,czc1)c is x2,c 2c(y1c+cy2),cz1 ;c thec midpointc ofc thec ccedgec (x2c ,cy2c ,cz1c )c andc (x2c ,cy2c ,czc2)c is x2,cy2,c 2c(z1c+cz2)
1 1 1
Thusc thec coordinatesc ofc thec midpointc ofc thec diagonalc are
(x1c+cx2),c (y1c+cy2),c (z1c+cz2)c c .
2c 2c 2c
5. (a)c Ac singlec pointc onc thatc line. (b)c Ac linec inc thatc plane. (c)c Ac planec inc 3−space.
6. (a)c c R(1,c4,c0)c andc Qc liec onc thec samec verticalc line,c andc soc doesc thec sidec ofc thec trianglec whichc connectsc them.
519
,520 Chapterc11
R(1,c4,c0)candc Pc liec inc thec planec zc=c0.c Clearlyc thec twoc sidesc arec perpendicular,c a n√d c thec sumc ofc thec squaresc ofc the
twocsidescisc|RQ|2c +c|RP|2c =c42c+c(22c+c32)c=c29,csocthecdistancecfromcPc tocQcis 29.
z
Qc(1,c4,c4)
y
Rc(1,c4,c0)
Pc(3,c1,c0)
x
(b) Clearly,cSPc iscparallelctocthecy- √ √ √
axis.c S(3,c4,c0)candcQcliecincthecplanec yc=c4,candcsocdoescSQ.c Hencecthectwocsidesc|SP| c andc|SQ|carecperpendicula
r,candc|PQ|c= |PS|2c+c|QS|2c = 32c +c(22c +c42)c= 29.
z
Qc(1,c4,c4)
y
Pc(3,c1,c0) Sc(3,c4,c0)
x
(c) Tc(1,c1,c4)candcQclieconcaclinecthroughc(1,c0,c4)candciscthuscparallelctocthecy-
axis,candcTQ c liesconcthiscline.c TcandcPc liecincthecsamecplanecyc=c1cwhichciscperpendicularctocanyclinecwhichciscpar
allelctocthecy-
z
Tc(1,c1,c4)c c Qc(1,c4,c4)
y
Pc(3,c1,c0)
x
axis,cthuscTP c,cwhichc liesc onc suchc ac line,c isc perpendicularc toc TQ.c Thusc |PQ|2c =c|PT| 2 c +c|QT|2c =c(4c+c16)c+c9c=c
29.
7. (a)c Letc thec basec ofc thec boxc havec sidesc ac andc bc andc diagonalc d1.c Thenc a2c+cb2c =c d2,c1andc d1c isc thec basec ofc a
rectangularcofcheightcccandcdiagonalcd,cwithcd2c=cd1 2c+cc2c=ca2c+cb2c+cc2.
(b) Twocunequalc pointsc (x1,cy1,cz1)c andc (x2,cy2,cz2)c formc diagonallyc opp√
cositec cornersc of c ac rectangularc boxc withc
sidescx1c−cx2,cy1c−cy2,cz1c−cz2,candcbycPartc(a)cthecdiagonalchasclength (x1c −cx2)2c +c(y1c −cy2)2c +c(z1c −cz2)2.
1
8. (a)c Thecverticalc planec thatc passesc throughc(c ,c0,c0)c andc isc perpendicularc toc thec x-axis.
2
1c 1c
(b) Equidistant:c (xc−c )2c+cy2c+cz2c =cx2c+cy2c+cz2,c orc −2xc+c1c=c0c orc xc=c .
2 2
9. Thecdiameterciscdc= √
√ √ √
(1c−c3)2c+c(−2c−c4)2c+c(4c+c12)2c =c 296,c soc thec radiusc isc 296/2c=c 74.c Thec midpointc
(2,c1,c−4)cofcthecendpointscofcthecdiametercisctheccentercofcthecsphere.
√
10. Eachc sidec hasc lengthc 14c soc thec trianglec isc equilateral.
,520 Chapterc11
√
11. ( a √
) cc Thecsideschaveclengthsc7,c14,candc7 5;citciscacrightctrianglecbecausecthecsidescsatisfycthecPythagoreanctheorem,
(7c c 5)2c =c72c+c142.
, ExercisecSetc11.1 521
(b) (2,1,6)c isc thec vertexc ofc thec 90◦c anglec becausec itc isc oppositec thec longestc sidec (thec hypotenuse).
(c) Areac =c (1/2)(altitude)(base)c =c(1/2)(7)(14)c=c49.
√ √ √ √ √ √
12. (a)c 3 (b)c 2 (c)c 5 (d)c (2)2c +c(−3)2c =c 13. (e)c (−5)2c +c(−3)2c =c 34. (f)c (−5)2c +c(2)2c =c 29.
13. (a)c (xc−c7)2c+c(yc−c1)2c+c(zc−c1)2c =c16.
(b) (xc−c1)2c+cy2c+c(zc+c1)2c =c16.
√ √
(c) rc = (−1c−c0)2c +c(3c−c0)2c +c(2c−c0)2c =c 14,c (xc+c1)2c+c(yc−c3)2c+c(zc−c2)2c =c14.
1√ 1c√
(d) rc = (−1c−c0)2c +c(2c−c2)2c +c(1c−c3)2c = 5,ccenterc(−1/2,c2,c2),c(xc+c1/2)2c+c(yc−c2)2c+c(zc−c2)2c=c5/4.
2 2
√ c √ c 2c √
14. rc =c|[distancec betweenc (0,0,0)c andc (3,c−2,c4)]c±c1|c=c 29c±c1,c x2c+cy2c+cz2c =cr2c = 29c±c1 =c30c±c2 29.
15. (xc−c2)2c+c(yc+c1)2c+c(zc+c3)2c =cr2,c so
(a)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 9. (b)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 1. (c)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 4.
1c 1c
16. (a)c Thecsidesc haveclengthc1,csoc thecradiuscisc ;c hencec(xc+c2)2c+c(yc−c1)2c+c(zc−c3)2c=c .
2 4
√ √ 3c
(b) Thecdiagonalchasc lengthc 1c+c1c+c1c=c 3c andc isc ac diameter,c soc (xc+c2)2c+c(yc−c1)2c+c(zc−c3)2c=c .
4
6c+c2c 5c+c9c 4c+c0c
(c) Radius:c(6c−c2)/2c=c2,ccenter:c (c ,c ,c ),c soc(xc−c4)2c+c(yc−c7)2c+c(zc−c2)2c=c4.
2 2 2
√
(d) Centerciscthecsame,cradiuscischalfcthecdiagonal,crc=c2 3,csoc(xc−c4)2c+c(yc−c7)2c+c(zc−c2)2c=c12.
17. Letctheccentercofcthecspherecbec(a,cb,cc).c Thecheightcofctheccentercovercthecx-
ycplaneciscmeasuredcalongcthecradiuscthatciscperpendicularctocthecplane.c Butcthisciscthecradiuscitself,c socheightc=cr
adius,c i.e.c cc=cr.c Similarlycac=crc andc bc=cr.
18. Ifcrciscthecradiuscofcthecsphere,cthencthecc√e nt e2cr cof2cthe cs p h e r e hasccoordinatesc(r,cr,cr)c(seecExercisec17).c Thuscthe
distancecfromctheco r i g i√ n cc toctheccenterc is r +c r c +c r2c = √3r,cfromc whichcc itc followsc thatc thec distancec fromc the
√ √ √ √
originc toc thec sp√hcerec is 3rc√ −c r.cEquate √ that cwithc3c−c 3:c 3rc−crc=c3c−c 3,crc=c 3.c Thecsphereciscgivencbycthe
equationc(xc− 3)2c+c(yc− 3)2c+c(zc− 3)2c =c3.
19. False;c needc bec neitherc rightc norc circular,c seec “extrusion”.
20. False,c itc isc ac rightc circularc cylinder.
21. True;c yc =czc =c0.
22. False,cthec spherec satisfiesc thec equality,c notc thec inequality.
23. (xc+c5)2c+c(yc+c2)2c+c(zc+c1)2c =c49;c sphere,c C(−5,c−2,c−1),c rc=c7.
24. x2c+c(yc−c1/2)2c+cz2c=c1/4;csphere,cC(0,c1/2,c0),crc=c1/2.
√
25. (xc−c1/2)2c+c(yc−c3/4)2c+c(zc+c5/4)2c=c54/16;csphere,cC(1/2,c3/4,c−5/4),crc=c3 6/4.
26. (xc+c1)2c+c(yc−c1)2c+c(zc+c1)2c=c0;c thec pointc (−1,c1,c−1).
Exercise Set 11.1
c c
1. (a)c c c (0,c0,c0),c(3,c0,c0),c(3,c5,c0),c(0,c5,c0),c(0,c0,c4),c(3,c0,c4),c(3,c5,c4),c(0,c5,c4).
(b) (0,c 1,c 0),c (4,c 1,c 0),c (4,c 6,c 0),c (0,c 6,c 0),c (0,c 1,c −2),c (4,c 1,c −2),c (4,c 6,c −2),c (0,c 6,c −2).
2. Corners:c (2,c2,c±2),c (2,c−2,c±2),c (−2,c2,c±2),c (−2,c−2,c±2).
z
(–2,c–2,c2) (–2,c2,c2)
(2,c–2,c2) (2,c2,c2) y
(–2,c–2,c–2) (–2,c2,c–2)
(2,c–2,c–2) (2,c2,c–2)
x
3. Corners:c (4,c2,c−2),c (4,2,1),c (4,1,1),c (4,c1,c−2),c (−6,c1,c1),c (−6,c2,c1),c (−6,c2,c−2),c (−6,c1,c−2).
z
(–6,c2,c1)
(–6,c1,c–2) (–6,c2,c–2)
y
(4,c1,c1) (4,c2,c1)
(4,c1,c–2)
x
4. (a)c c (x2,cy1,cz1),c(x2,cy2,cz1),c(x1,cy2,cz1)(x1,cy1,cz2),c(x2,cy1,cz2),c(x1,cy2,cz2).
(b) Thec midpointc ofc thec diagonalc hasc coordinatesc whichc arec thec coordinatesc ofc thec midpointsc ofc thec edges.c The
c c c c
1c
midpointcofc thec edgec (x1c ,cy1c ,cz1c )c andc (xc2,cyc1,czc 1)c is 2(xc +cxc ),cyc ,cz ;c thec midpointc ofc thec edgec (xc ,cyc ,czc )c and
1 2 1 1 2 1 1
c
1 1 .
(xc2,cyc2,czc1)c is x2,c 2c(y1c+cy2),cz1 ;c thec midpointc ofc thec ccedgec (x2c ,cy2c ,cz1c )c andc (x2c ,cy2c ,czc2)c is x2,cy2,c 2c(z1c+cz2)
1 1 1
Thusc thec coordinatesc ofc thec midpointc ofc thec diagonalc are
(x1c+cx2),c (y1c+cy2),c (z1c+cz2)c c .
2c 2c 2c
5. (a)c Ac singlec pointc onc thatc line. (b)c Ac linec inc thatc plane. (c)c Ac planec inc 3−space.
6. (a)c c R(1,c4,c0)c andc Qc liec onc thec samec verticalc line,c andc soc doesc thec sidec ofc thec trianglec whichc connectsc them.
519
,520 Chapterc11
R(1,c4,c0)candc Pc liec inc thec planec zc=c0.c Clearlyc thec twoc sidesc arec perpendicular,c a n√d c thec sumc ofc thec squaresc ofc the
twocsidescisc|RQ|2c +c|RP|2c =c42c+c(22c+c32)c=c29,csocthecdistancecfromcPc tocQcis 29.
z
Qc(1,c4,c4)
y
Rc(1,c4,c0)
Pc(3,c1,c0)
x
(b) Clearly,cSPc iscparallelctocthecy- √ √ √
axis.c S(3,c4,c0)candcQcliecincthecplanec yc=c4,candcsocdoescSQ.c Hencecthectwocsidesc|SP| c andc|SQ|carecperpendicula
r,candc|PQ|c= |PS|2c+c|QS|2c = 32c +c(22c +c42)c= 29.
z
Qc(1,c4,c4)
y
Pc(3,c1,c0) Sc(3,c4,c0)
x
(c) Tc(1,c1,c4)candcQclieconcaclinecthroughc(1,c0,c4)candciscthuscparallelctocthecy-
axis,candcTQ c liesconcthiscline.c TcandcPc liecincthecsamecplanecyc=c1cwhichciscperpendicularctocanyclinecwhichciscpar
allelctocthecy-
z
Tc(1,c1,c4)c c Qc(1,c4,c4)
y
Pc(3,c1,c0)
x
axis,cthuscTP c,cwhichc liesc onc suchc ac line,c isc perpendicularc toc TQ.c Thusc |PQ|2c =c|PT| 2 c +c|QT|2c =c(4c+c16)c+c9c=c
29.
7. (a)c Letc thec basec ofc thec boxc havec sidesc ac andc bc andc diagonalc d1.c Thenc a2c+cb2c =c d2,c1andc d1c isc thec basec ofc a
rectangularcofcheightcccandcdiagonalcd,cwithcd2c=cd1 2c+cc2c=ca2c+cb2c+cc2.
(b) Twocunequalc pointsc (x1,cy1,cz1)c andc (x2,cy2,cz2)c formc diagonallyc opp√
cositec cornersc of c ac rectangularc boxc withc
sidescx1c−cx2,cy1c−cy2,cz1c−cz2,candcbycPartc(a)cthecdiagonalchasclength (x1c −cx2)2c +c(y1c −cy2)2c +c(z1c −cz2)2.
1
8. (a)c Thecverticalc planec thatc passesc throughc(c ,c0,c0)c andc isc perpendicularc toc thec x-axis.
2
1c 1c
(b) Equidistant:c (xc−c )2c+cy2c+cz2c =cx2c+cy2c+cz2,c orc −2xc+c1c=c0c orc xc=c .
2 2
9. Thecdiameterciscdc= √
√ √ √
(1c−c3)2c+c(−2c−c4)2c+c(4c+c12)2c =c 296,c soc thec radiusc isc 296/2c=c 74.c Thec midpointc
(2,c1,c−4)cofcthecendpointscofcthecdiametercisctheccentercofcthecsphere.
√
10. Eachc sidec hasc lengthc 14c soc thec trianglec isc equilateral.
,520 Chapterc11
√
11. ( a √
) cc Thecsideschaveclengthsc7,c14,candc7 5;citciscacrightctrianglecbecausecthecsidescsatisfycthecPythagoreanctheorem,
(7c c 5)2c =c72c+c142.
, ExercisecSetc11.1 521
(b) (2,1,6)c isc thec vertexc ofc thec 90◦c anglec becausec itc isc oppositec thec longestc sidec (thec hypotenuse).
(c) Areac =c (1/2)(altitude)(base)c =c(1/2)(7)(14)c=c49.
√ √ √ √ √ √
12. (a)c 3 (b)c 2 (c)c 5 (d)c (2)2c +c(−3)2c =c 13. (e)c (−5)2c +c(−3)2c =c 34. (f)c (−5)2c +c(2)2c =c 29.
13. (a)c (xc−c7)2c+c(yc−c1)2c+c(zc−c1)2c =c16.
(b) (xc−c1)2c+cy2c+c(zc+c1)2c =c16.
√ √
(c) rc = (−1c−c0)2c +c(3c−c0)2c +c(2c−c0)2c =c 14,c (xc+c1)2c+c(yc−c3)2c+c(zc−c2)2c =c14.
1√ 1c√
(d) rc = (−1c−c0)2c +c(2c−c2)2c +c(1c−c3)2c = 5,ccenterc(−1/2,c2,c2),c(xc+c1/2)2c+c(yc−c2)2c+c(zc−c2)2c=c5/4.
2 2
√ c √ c 2c √
14. rc =c|[distancec betweenc (0,0,0)c andc (3,c−2,c4)]c±c1|c=c 29c±c1,c x2c+cy2c+cz2c =cr2c = 29c±c1 =c30c±c2 29.
15. (xc−c2)2c+c(yc+c1)2c+c(zc+c3)2c =cr2,c so
(a)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 9. (b)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 1. (c)c (x−2)2c+(yc+1)2c+(zc+3)2c =c 4.
1c 1c
16. (a)c Thecsidesc haveclengthc1,csoc thecradiuscisc ;c hencec(xc+c2)2c+c(yc−c1)2c+c(zc−c3)2c=c .
2 4
√ √ 3c
(b) Thecdiagonalchasc lengthc 1c+c1c+c1c=c 3c andc isc ac diameter,c soc (xc+c2)2c+c(yc−c1)2c+c(zc−c3)2c=c .
4
6c+c2c 5c+c9c 4c+c0c
(c) Radius:c(6c−c2)/2c=c2,ccenter:c (c ,c ,c ),c soc(xc−c4)2c+c(yc−c7)2c+c(zc−c2)2c=c4.
2 2 2
√
(d) Centerciscthecsame,cradiuscischalfcthecdiagonal,crc=c2 3,csoc(xc−c4)2c+c(yc−c7)2c+c(zc−c2)2c=c12.
17. Letctheccentercofcthecspherecbec(a,cb,cc).c Thecheightcofctheccentercovercthecx-
ycplaneciscmeasuredcalongcthecradiuscthatciscperpendicularctocthecplane.c Butcthisciscthecradiuscitself,c socheightc=cr
adius,c i.e.c cc=cr.c Similarlycac=crc andc bc=cr.
18. Ifcrciscthecradiuscofcthecsphere,cthencthecc√e nt e2cr cof2cthe cs p h e r e hasccoordinatesc(r,cr,cr)c(seecExercisec17).c Thuscthe
distancecfromctheco r i g i√ n cc toctheccenterc is r +c r c +c r2c = √3r,cfromc whichcc itc followsc thatc thec distancec fromc the
√ √ √ √
originc toc thec sp√hcerec is 3rc√ −c r.cEquate √ that cwithc3c−c 3:c 3rc−crc=c3c−c 3,crc=c 3.c Thecsphereciscgivencbycthe
equationc(xc− 3)2c+c(yc− 3)2c+c(zc− 3)2c =c3.
19. False;c needc bec neitherc rightc norc circular,c seec “extrusion”.
20. False,c itc isc ac rightc circularc cylinder.
21. True;c yc =czc =c0.
22. False,cthec spherec satisfiesc thec equality,c notc thec inequality.
23. (xc+c5)2c+c(yc+c2)2c+c(zc+c1)2c =c49;c sphere,c C(−5,c−2,c−1),c rc=c7.
24. x2c+c(yc−c1/2)2c+cz2c=c1/4;csphere,cC(0,c1/2,c0),crc=c1/2.
√
25. (xc−c1/2)2c+c(yc−c3/4)2c+c(zc+c5/4)2c=c54/16;csphere,cC(1/2,c3/4,c−5/4),crc=c3 6/4.
26. (xc+c1)2c+c(yc−c1)2c+c(zc+c1)2c=c0;c thec pointc (−1,c1,c−1).
