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BCA 21 / IMCA 21 / Mathematics SVT 43
ANALYTICAL GEOMETRY IN THREE DIMENSIONS ORSOLID GEOMETRY
Co-ordinates of a point in 3-space
Consider three mutually perpendicular lines OX, OY & OZin 3-space with O as origin. Let P be a point in 3-space.Draw rPQ⊥ to the plane XOY.
Draw QA & QB parallel to OY & OX to meet OX at A& OY at B.
zQPyOBxOA === &,Let
Then (x, y, z) are taken as coordinates the point P. Threemutually r⊥ lines OX, OY & OZ divide 3-space into eightequal parts called Octants. Co-ordinates of any point in theplane XOY will be of the form (x, y, o), in XOZ plane (x, o, z)in YOZ plane (o, y, z) co-ordinates of any point on x-axis are(x, o, o) on y-axis, (o, y, o) and z-axis (o, o, z). Co-ordinatesof origin (o, o, o). The position vector of P is
.��� rkzjyixOP by denoted also is ++=
Distance between two points
),,(&),,( 222111 zyxQzyxP
kzjyixOQkzjyixOP ���&���222111 vectors,using ++=++=
kzzjyyixxOPOQPQ �)(�)(�)( 121212 −+−+−=−=∴
212
212
212 )()()( zzyyxxPQ −+−+−=∴
212
212
212222111 by given is points obetween tw Distance )()()(),,(&),,( zzyyxxzyxQzyx −+−+−∴
Section Formula
Let R divide PQ in the ratio m : n
n
m
RQ
RP =then
RQmPRn = ie
RQmPRn =∴
( ) ( )OROQmOPORn −=− ie
ORmOQmOPnORn −=− ie
OPnOQmORnm +=+∴ )(
O
Z
B
X
Y
Q
z
P (x, y, z)
y
x
i�
k�
j�
A
O
Z
R
X
Y
P (x 1, y1
, z1)
i�
k�
j�
n Q (x2, y2, z2)m
44 KSOU Solid Geometry
nm
OPnOQmOR
++=∴
OR represents the position vector of R which divides P & Q in the ratio m : n.
2 then point,-mid theis If
OPOQORR
+=
++
++
++
nm
nzmz
nm
nymy
nm
nxmxR 121212by given are of ordinates-coCartesian ,,
if R lie out segment PQ.
−−
−−
−−
nm
nzmz
nm
nymy
nm
nxmxR 121212 are of ordinates-coThen ,,
Direction cosines of a line
Let P be any point in plane, let the line OP make angles
are then axes ordinate-cowith γβαγβα cos&cos,cos,,
defined as directions cosines and usually denoted as l, m, n.
,. OYPBnml r to rawD1 that prove To 222 ⊥=++
.cos rOPr
y == wherethen β
mrry ==∴ βcos
,&OZOXrs to dropingby Similarly ⊥
nrrzlrrx ==== γα cos&cos that showcan we
Squaring and adding, we get
222222222 rnrmrlzyx ++=++2222 rnml )( ++=
2222but rzyx =++
1222 =++∴ nml
Direction of ratios of a line
If the direction cosines of a line are proportional to a, b, c then a, b, c are called Direction Ratios of the line.
(say)ie kc
n
b
m
a
l ===
cknbkmakl === ,,then
222222222 kckbkanml ++=++∴
1but 222 =++ nml
12222 =++∴ kcba )(
2222222 11
iecba
kcba
k++
=∴++
=
O
Z
B
X
Y
P (x, y, z)rγ
α β
BCA 21 / IMCA 21 / Mathematics SVT 45
∴ If a, b, c are the direction ratios of any line then direction cosines are
222222222 cba
c
cba
b
cba
a
++++++,,
),,(),,( 222111 and If zyxQzyxP are any two points in 3-space. The direction ratios of PQ are given by ,12 xx −
1212 zzyy −− , and hence the direction cosines are
212
212
212
12
212
212
212
12
212
212
212
12
)()()(,
)()()(,
)()()( zzyyxx
zz
zzyyxx
yy
zzyyxx
xx
−+−+−
−
−+−+−
−
−+−+−
−
Note :- Co-ordinates of a unit vector represents the direction cosines of the line of the vector. For any other vectors the co-ordinates represent the direction ratios.
Examples
1. Find the distance between the points (4, 3, –6) & (–2, 1, –3).
),,(),,,( 312634Let :Solution −−=−= QP
222 361324 )()()( +−+−++=PQ
79436326 222 =++=−++= )(
2. Show that the points (–2, 3, 5), (1, 2, 3) & (7, 0, –1) are colinear.
),,(),,,(),,,( 107321532Let :Solution −==−= CBA
14419352312 222 =++=−+−+−−= )()()(AB
1425616436130271 222 ==++=++−+−= )()()(BC
14312636981150372 222 ==++=++−+−−= )()()(AC
ACBCAB ==+=+ 14314214
∴ The points A, B, C are colinear.
Alternate Method
Direction ratios AB are – 2 – 1, 3 – 2, 5 – 3 ie –3, 1, 2
Direction ratios of BC are 1 – 7, 2 – 0, 3 + 1, ie – 6, 2, 4 ie – 3, 1, 2
Direction ratios of AB & BC are same. ∴ A, B & C are colinear.
3. Show that the points (3, 2, 2), (–1, 1, 3), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4).Find also the radius of the sphere.
centre. thebe431Let 212650311223 be pointsgiven Let the :Solution ),,().,,(&),,(),,,(),,,( ===−== CSRQP
3414242331 222 =++=−+−+−= )()()(CP
3144341311 222 =++=−+−++= )()()(CQ
3441463510 222 =++=−+−+−= )()()(CR
3441423112 222 =++=−+−+−= )()()(CS
46 KSOU Solid Geometry
3====∴ CSCRCQCP
3. is radius & sphere aon lie SRQP &,,∴
4. Find the co-ordinates of the point which divide the line joining the points (2, –4, 3) & (–4, 5, –6) in the ratio 2 : 1.
),,(),,,( 654342Let :Solution −−=−= ba
12
2by given issection ofPoint
++ ab
12
34216542
+−+−−= ),,(),,(
3
31241028 ),,( +−−+−=
),,(),,(
3223
966 −−=−−=
5. Find the co-ordinates of the point which divide the line joining (3, 2, 1) & (1, 3, 2) in the ratio – 2 : 1.
),,(),,,( 231123Let :Solution == ba
12
12by given issection ofPoint
+−⋅+− ab
1
12312312
−+−= ),,(),,(
1
123462
−+−−−= ),,(),,(
),,(),,(
3411
341 −=−
−−=
6. Find the ratios in which XY plane divides the join of (– 3, 4, – 8) & (5, – 6, 4).Also obtain the coordinates of the point of section.
),,(),,,( 465843Let :Solution −=−−= ba
Let k : 1 be the ratio.
+−
++−
+−=
++
1
84
1
46
1
35
1
kby given issection ofPoint
k
k
k
k
k
k
k
ab,,
This point lies on XY plane
28401
84 =⇒=⇒=+−∴ kk
k
k
−∴ 0
3
8
3
7 arepoint theof ordinates-co and 1:2 is ratio ,,
7. Direction ratios of a line are 6, 2, 3. Find the direction cosines.
222 326 Now :Solution ++ 79436 =++=
.,,7
3
7
2
7
6 are cosinesdirection ∴
BCA 21 / IMCA 21 / Mathematics SVT 47
8. Find the direction cosines of the line joining the points (1, 4, – 3) & (4, 7, – 6).
Solution : Direction ratios of the line joining the points (1, 4, – 3) & (4, 7, – 6) are 4 – 1, 7 – 4, – 6 + 3
ie 3, 3, – 3.
999
3
999
3
999
3 are cosinesdirection
++−
++++∴ ,,
33
3
33
3
33
3 ie
−,, .,,
3
1
3
1
3
1 ie
−
To find the angle between two lines in 3-space
Let OA & OB be two lines whose direction cosines are .,,&,, 222111 nmlnml
θ=BOA �Let
knjmilOBbknjmilOAa ���,���222111Let ++==++==
212121 nnmmllba ++=⋅
θcosabba =⋅but
22
22
22
21
21
21
212121
nmlnml
nnmmll
ab
ba
++++
++=⋅=∴ θcos
212121 nnmmll ++=
0022
If 212121 =++∴== nnmmllππθ cos,
0is be tolines for twocondition 212121 =++⊥∴ nnmmllr
222111 If cbacba ,,&,, are direction ratios of two lines. Then angle θ between them is given by
22
22
22
21
21
21
212121
cbacba
ccbbaa
++++
++=θcos
.0 iflar perpendicu be willlines 212121 =++ ccbbaa
Note :- Two lines will be parallel if direction cosines of the two lines are same or if the direction ratios are proportional.
The Plane
To find the equation of the plane passing through the point(x1, y1, z1) and having a, b, c as the direction ratios of the normalto the plane.
),,( 111Let zyxA be a point in the plane.
),,( zyxPLet be any point in the plane then direction ratios of AP
are given by 111 zzyyxx −−− ,, as this is perpendicular to the normalto the plane, we have
0111 =−+−+− )()()( zzcyybxxa
0)( ie 111 =++−++ czbyaxczbyax
which is a first degree equation in x, y & z.
O
B
Aθ
l 2, m2
, n2
l1, m1, n1
PlaneA (x1, y1, z1)
P (x, y, z)
Normal tothe Plane
48 KSOU Solid Geometry
Note :- (1) z = 0 represents the equation to the XOY plane.
(2) y = 0 represents the equation of the ZOX plane.
(3) x = 0 represents the equation of YOZ plane.
To prove that ax + by + cy + d = 0 a first degree equation in x, y, z represents a plane.
.),,(&),,( 0 satisfying points any two be Let 222111 =+++ dczbyaxzyxQzyxP
0Then 111 =+++ dczbyax (1)
0222 =+++ dczbyax (2)
multiply (2) by k & add to (1)
0111222 =+++++++ )()( dczbyaxdczbyaxk
01 ie 121212 =+++++++ )()()()( kdzkzcykybxkxa
)( 11by out through divide −≠+ kk
0111
121212 =++++
+++
++
dk
zkzc
k
ykyb
k
xkxa
)()()(
This shows that the point whose coordinates are
++
++
++
111121212
k
zkz
k
yky
k
xkx,,
.0equation esatisfy th =+++ dczbyax
Thus every point on the line joining P & Q lie on the locus. ∴ The equation represents the plane.
To find the length of the perpendicular from (x1, y1, z1) upon the plane ax + by + cy + d = 0.
),,( 111Let zyxP be the given point & PA is the length of the
perpendicular on the plane .0=+++ dczbyax
),,( zyxQLet be any point in the plane, AP is the projection
of QP upon the normal to the plane if n� is the unit normal vectorof the plane then
222 cba
kcjbian
++
++=���
�
( ) nkzzjyyixxnQPAP ��)(�)(�)(� ⋅−+−+−=⋅= 111
222
111
cba
zzcyybxxa
++
−+−+−=
)()()(
222
111
cba
czbyaxczbyax
++
++−++= )(
222
111
cba
dczbyax
++
+++=
dczbyax −=++Q
222
111lar perpendicu theoflength cba
dczbyaxAP
++
+++==∴
Note :- ),,(&),,( 222111 points Two zyxzyx lie on the same side or on opposite sides of the plane 0=+++ dczbyaxdczbyaxdczbyax ++++++ 222111 as according & are of same sign or of opposite signs.
Q (x, y, z)
P (x1, y1, z1)
A
BCA 21 / IMCA 21 / Mathematics SVT 49
Normal form of the equation of the plane
Let p represents length of the r⊥ from the origin upon the plane 0=+++ dczbyax
222then
cba
dp
++=
nmlpnzmylx ,, where=++ are the direction cosines of the normal to the plane, is taken as the normal equation of
the plane.
To find the angle between two planes
00 22221111 =+++=+++ dzcybxadzcybxa &
The two planes intersect along a line, the angle between two planes is nothing but angle between two normals of theplane. ∴ if θ is the angle between two planes
0 if be willplanes&costhen 21212122
22
22
21
21
21
212121 =++⊥++++
++= ccbbaacbacba
ccbbaa rθ
Intercept form of the equation if the plane
cba &,Let be the intercept made by the plane on the co-ordinate axes ; ie the plane passes through the points),,,( ooa ),,(&),,( cooobo
.0 be plane theoequation t Let the =+++ dzyx γβα
a
ddaooa −=∴=+∴ αα 0 plane on the lies),,(
b
ddbobo −=⇒=+∴ ββ 0 plane on the lies),,(
c
ddccoo −=⇒=+∴ γγ 0 plane on the lies),,(
0 is plane theofequation =+−−−∴ dzc
dy
b
dx
a
d
01ie =+−−−c
z
b
y
a
x
form.intercept in the plane theofequation theis1 ie =++c
z
b
y
a
x
Examples
(1) Find the equation of the plane passing through the point (1, 2, 3) and having the vector kjin ��� 32 +−= as normal.
Solution : Direction ratios of the normal to the planes are 2, –1, 3
∴ equation to the plane is
0332112 =−+−−− )()()( zyx
093222 ie =−++−− zyx
0932 ie =−+− zyx
50 KSOU Solid Geometry
(2) Find the equation of the plane through the point (2, 1, 0) and perpendicular to the planes and 52 =−− zyx532 =−+ yyx
0012 is 0) 1, (2,point he through tplane theofEquation :Solution =−+−+− )()()( zcybxa
52 to is plane This =−−⊥ zyxr
02 =−−∴ cba (1)
532 to also is plane the =−+⊥ zyxr
032 =−+∴ xba (2)
Let us eliminate c between (1) & (2)
baba
cba
cba
=⇒=−
=−+×=−−×
055 gSubtractin
032 is1)2(
0336 is3)1(
cacaa =⇒=−− 02 (1), from
cba ==∴
0012 is plane theofEquation =−+−+−∴ )()()( zcybxa
0012 ie =++−+− zyx
.3 ie =++ zyx
(3) Find the equation of the plane through (1, 2, 3) and parallel to the plane .7354 =−+ zyx
7354 toparallel plane theofEquation :Solution =−+ zyx can be taken as kzyx =−+ 354 where k is a constant. The plane
passes through (1, 2, 3)
k=−+∴ )()()( 332514 59104 ie =∴=−+ kk
.5354 is plane required theofEquation =−+∴ zyx
(4) Find the r⊥ distance of the point (3, 2, 1) from the plane passing through the points (1, 1, 0), (3, – 1, 1) & (–1, 0, 2)
),,(),,,(),,,(&),,( 201113011123Let :Solution −=−=== CBAP
kjiAB ��� +−= 22
kjiAC ��� 22 +−−=
212
122
ˆˆˆ
−−−=×∴
kji
ACAB
)(�)(�)(� 422414 −−++−+−= kji kji ��� 663 −−−=
∴ direction ratios of the normal to the plane are –3, –6, –6
0061613by given is plane theofEquation =−−−−−− )()()( zyx
063663 ie =++−−− zyx
0322 ie =−++ zyx
units. 23
6
441
3243 is plane theupon the 1) 2, (3, from theoflength ==
++−++⊥∴ r
BCA 21 / IMCA 21 / Mathematics SVT 51
(5) Show that the four points (0, –1, 0), (2, 1, –1), (1, 1, 1) & (3, 3, 0) are coplanar. Find the equation of the plane.
),,(&),,(),,,(),,,( 033111112010Let :Solution ==−=−= DCBA
Let us find the equation of the plane passing through A, B & C.
The plane passing through (0, –1, 0) is given by 010 =−+++− )()()( ozcybxa
0 ie =+++ bczbyax
iton lies112 ),,( −
02 =+−+∴ bcba
022 ie =−+ cba (1)
(1, 1, 1) lies on it
02 =++∴ cba (2)
242122 (2) & (1) from
−=
−−=
+cba
234 ie
cba =−
=
0021304 is plane theofEquation =−++−−∴ )()()( zyx
.03234 ie =−+− zyx
),,( 033Consider =D
it can be seen that it lies on the plane. ∴ the given points are coplanar.
(6) Find the distance between the parallel planes .& 052440322 =++−=++− zyxzyx
.) 0322on point a is 3 0, (0, Now :Solution =++−− zyx
05244 plane upon the 30,(0, from distance =++−−⊥ zyxr )
41616
532044(0) is
+++−+− )()(
6
1
6
1
6
56 =−=+−=
6
1 is planes parallelbetween distance∴
(7) .032 and 0232 planes ebetween th angle theFind =+++=−−+ zyxzyx
have weplanes, obetween tw angle theis If :Solution θ
22
22
22
21
21
21
212121
cbacba
ccbbaa
++++
++=θcos
212
1
3227
1
614
1
114941
322 =××
==++++
−+=
=∴ −
212
11cosθ
52 KSOU Solid Geometry
(8) Find the equation of the plane passing through the line of intersection of the planes 0432 and 1 =+−+=++ zyxzyx432 plane thelar toperpendicu and =− zy
Solution : Equation of the plane passing through the line of intersection of the given two planes can be taken as
04321 =+−++−++ )()( zyxzyx λ
where λ is a constant.
0411312(1 ie =+−−++++ λλλλ zyx )()()
432 to is plane this =−⊥ zyr
013312210 =−−+++∴ )()()( λλλ
03362 ie =+−+ λλ
9
119 ie =⇒= λλ
04329
11 is plane ofEquation =+−+++++∴ )()( zyxzyx
04329999 ie =+−++−++ zyxzyx
0581211 ie =−++ zyx
is the equation of required plane.
The Straight Line
1. Two planes intersect along a straight line, therefore 0 and0 22221111 =+++=+++ dzcybxadzcybxa taken together
represent a line. These are called General equations of the line.
2. If a line passes through the point nmlzyxA ,,&),,( have 111 as direction cosinesthen
),,( zyxPLet be any point on the line
nmlkzzjyyixxAP ,,�)(�)(�)( are cosinesdirection and 111 −+−+−=
n
zz
m
yy
l
xx 111 −=−=−∴
represent two equations of the line. This is called Symmetric form of the equationof the line.
3. If a, b, c are the direction ratios of the line passing through the point ),,,( 111 zyx then equations of the line are
.c
zz
b
yy
a
xx 111 −=−=−
If the line passes through the points ),,(&),,( 222111 zyxzyx , then the direction ratios of the line are ,12 xx − ,12 yy −
12 zz −
12
1
12
1
12
1 are equations itszz
zz
yy
yy
xx
xx
−−
=−−
=−−
∴
A (x1, y1, z1) P (x, y, z)
O
BCA 21 / IMCA 21 / Mathematics SVT 53
4. .0 plane theand line ebetween th angle thefind To 111 =+++−=−=−dczbyax
n
zz
m
yy
l
xx
Solution : If θ is the normal angle between the line and the plane
θ−°90 is the angle between the line and the normal tothe plane.
22222290
cbanml
ncmbla
++++
++=−°∴ )cos( θ
.sin222 cba
ncmbla
++
++=∴ θ
°= 0 then plane the toparallel is line theIf θ
.0=++∴ ncmbla
c
n
b
m
a
l == then plane thelar toperpendicu is line theIf
Examples
(1) 622 plane the toparallel is 2
13
2
2
3
1 line that theShow =−+−=
−+=−
zyxyx
and find the distance between them.
1 2, 2, are 622 plane the tonormal theof ratiosDirection :Solution −=−+ zyx
Direction ratios of the given line are 3, –2, 2
Now 2(3) + 2(– 2) – 1(2) = 6 – 4 – 2 = 0.
∴ line is r⊥ to the normal to the plane. ∴ line is parallel to the plane.
622 plane upon the 121 of distance line.given on thepoint a is 121 =−+−⊥∴− zyxr ),,(),,(
.)(
33
9
3
742
144
61222(1)is ==−−=
++−−−+
(2) Find the equation of the line through (1, 2, – 1) perpendicular to each of the lines
.543
and 101
zyxzyx ==−
==
Solution : Let a, b, c be the direction ratios of the required line.
.& 05430Then =++=−+ cbacoa
045340 −=
−−=
+∴ cba
484 ie
cba =−
= 121
iecba =
−=
.1
13
2
2
1
1 are line required theof equations
+=−−=−∴ yx
(3) .73 plane theand 6
3
32
1 line ebetween th angle theFind =++−==+
zyxzyx
Solution : If θ is the angle between line and the plane, then
N orm a l
L ine
P lane
90º �
q
q
54 KSOU Solid Geometry
117
15
117
636
1193694
161332 =++=++++
++= )()())((sinθ
117
151−=∴ sinθ
(4) Find the perpendicular distance of the point (1, 1, 1) from the line 12
3
2
2
−=+=− zyx
pointgiven thebe 111 & line on thepoint a is032Let :Solution ),,(),,( PA −
23181161 ==++=∴ AP
kjiOAOPAP ��� ++−=−= 4
144
224line on the of Projection
++−+⋅++−== )���(
)���(kji
kjiAQAP
3
5
3
182 =−+−=
9
137
9
25162
9
2518222 =−=−=−=∴ AQAPPQ
3
137=∴ PQ
Alternate Method
rzyx =
−=+=−
12
3
2
2Let
),,( rrr −−+ 3222 is line on thepoint any
14212 are 1) 1, (1,point the&point thisjoining line theof ratiosdirection −−−+ rrr ,,
011422122 if linegiven the to be willline This =−−−−++⊥ )()()( rrrr
018424 ie =++−++ rrr
9
5 ie059 ie ==− rr
−−
−−+∴
9
5
9
17
9
28 ie
9
53
9
102
9
10 arelar perpendicu theoffoot theof ordinates-oC ,,,,
222
19
51
9
171
9
28 is 111
9
5
9
17
9
28between Distance
−−+
−−+
−
−−
),,(&,,
3
137
9
1379
9
142619 222
=×=++=
(5) Find the image of the point (1, 3, 4) in the plane 032 =++− zyx
Solution : Let P be the given point and Q be its image on the plane 032 =++− zyx
(say)1
4
1
3
2
1 are toEquations r
zyxPQ =−=
−−=−
P (1, 1 , 1 )
A (2, –3 , 0)
Ql ine
BCA 21 / IMCA 21 / Mathematics SVT 55
),,( 4312 are of ordinates-coThen ++−+ rrrQ
bygiven is ofpoint -Mid PQ
++++−++
2
44
2
33
2
112 rrr,,
plane on the liepoint this2
8
2
61 ie
++−+ rrr ,,
032
8
2
612 =+++
+−−+∴ rr
r )(
068644 ie =+++−++ rrr
2126 ie −=⇒−= rr
),,(),,( 253423214 −=+−++−=∴ Q
),,( 253 is planegiven in the 4) 3, (1, of Image −∴
(6)2
2
2
2
2
2
1
1
1
1
1
1 and lines for twocondition thefind Ton
zz
m
yy
l
xx
n
zz
m
yy
l
xx −=
−=
−−=
−=
− to intersect (or to be
coplanar) and to find the equation of the plane.
Solution : The equation of the plane containing the first line is given by 0111 =−+−+− )()()( zzcyybxxa (1)
0 where 111 =++ cnbmal (2)
the second line will lie in the plane if 0222 =++ cnbmal (3)
(1)satisfy and 222 ),,( zyx
0 ie 121212 =−+−+− )()()( zzcyybxxa (4)
eliminating a, b, c from (2), (3) & (4)
0get We
222
111
121212
=−−−
nml
nml
zzyyxx
which is the required condition for coplanarity of two lines.
Equation to the plane is by eliminating a, b, c from (1), (2) & (3)
0 ie
222
111
111
=−−−
nml
nml
zzyyxx
is the required equation of the plane.
Exercise
1. Find the equation of the plane passing through the point (– 2, 2, 2) and containing the line joining the points (1, 1, 1) &(1, – 1, 2).
)( 0863 : Ans =+−− zyx
2. Show that the points (– 6, 3, 2), (3, – 2, 4), (5, 7, 3) and (– 13, 17, – 1) are coplanar.
3. Find the equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 9662 =++ zyx
)( 0543 : Ans =−+ zyx
P (1 , 3 , 4)
Q
56 KSOU Solid Geometry
4. Find the equation of the plane through the points whose position vectors are kjikjikji ������,��� ++−++− and 23
)( 522 : Ans =++ zyx
5. Find the equation of the plane through the points (2, 2, 1), (1, – 2, 3) and parallel to the joining the points (2, 1, – 3),(– 1, 5, – 8)
)( 014161112 : Ans =+−− zyx
6.2
13
1
3
2
1 line thecontains which plane theofequation theFind
−=−=− zyx and is perpendicular to the plane
.3=++ zyx
)( 012 : Ans =+− zx
7. line the toparallel and2
4
2
4
3
1 line he through tplane theofequation theFind
−−=−=− zyx
1
2
4
1
3
1 +=−−=+ zyx
)( 38632 : Ans =++ zyx
8.5
4
4
3
3
2 and
4
3
3
2
2
1 lines that theShow
−=−=−−=−=− zyxzyx are coplanar and find the equation of the plane
containing them.
)( 02 : Ans =+− zyx
9.1
1
5
1
4
1 and
3
7
3
5
2
3 lines that theShow
−+=+=+
−−=+=+ zyxzyx
are coplanar and find the equation of the plane
containing them.
)( 056 : Ans =−− zyx
10. .632 plane in the lies 2
810 line that theShow =++=−=+ zyxz
yx
11. Find the co-ordinates of the foot of the perpendicular drawn from the point (–1, –3, 2) upon the plane 5543 =++ zyx .
−− 3,
5
11,
5
2:Ans
12.8
10
3
1
2
1 and
7
1
4
3
1
4 lines that theProve
+=−+=−+=
−+=− zyxzyx
intersect and find the co-ordinates of the point of
intersection.
[ ]), 67 (5, :Ans −
Sphere
Definition :- A Sphere is the locus of a point which remains at a constant distance from a fixed point in three dimension.
The fixed point is the centre and constant distance is called radius.
Equation of the sphere whose centre is at (x1, y1, z1) and radius r is given by22
12
12
1 rzzyyxx =−+−+− )()()(
1. sphere. a represents0222 that show To 222 =++++++ dwzvyuxzyx
The expression can be written as
0222 222 =++++++ dwzzvyyuxx )()()(
0 ie 222222 =+−++−++−+ dwwzvvyuux )()()(
BCA 21 / IMCA 21 / Mathematics SVT 57
dwvuwzvyux −++=+++++ 222222 ie )()()(
which represents a sphere whose centre is (–u, –v, –w) dwvu −++ 222 is radius and
2. Plane section of a sphere
Plane section of a sphere is always a circle. If the plane passesthrough the centre of the sphere it is called a great circle, otherwise asmall circle.
Let C be the centre of the sphere and A be a point on the sphereand on the plane which intersects the sphere. Let B be the centre of thesmall circle then .ABBC r tois ⊥
222 BCABAC +=∴
sphere) theof radius(& RACpBC ==22222Then pRBCACAB −=−=
22 is circle small theof radius pRAB −=∴
0222Let 222 =++++++ dwzvyuxzyxS: be the equation of
the sphere
plane. thebe0 and =+++ dczbyaxP :
Then the equation of the sphere passing through the circle is constant. a is where0 λλ =+ PS
0222 ie00222 ie0 If 2222222
21111222
1 =++++++==++++++= dzwyvxuzyxSdzwyvxuzyxS &
represents two spheres then the equation of the circle common to 0 is00 2121 =−== SSSS &
.)( 0by given is circle he through tsphere theofequation The 211 =−+∴ SSS λ
3. To find the equation of the sphere having the),,(&),,( 222111 zyxzyx as the extremities
of a diameter.
Let A & B be the extremities of the diameter.
Let P(x, y, z) be a point on the sphere.
.PBAP r tois ⊥
are of ratiosDirection AP
111 zzyyxx −−− ,,
222 are of ratiosDirection zzyyxxBP −−− ,,
,PBAP r tois Since ⊥
.))(())(())(( 0212121 =−−+−−+−− zzzzyyyyxxxx
4. To find the equation of the tangent plane at ),,( 111 zyx on the sphere 0222222 =++++++ dwzvyuxzyx
),,( zyxQLet be any point in the tangent plane then direction ratios of 111 are zzyyxxPQ −−− ,, and direction
ratios of CwzvyuxPC where are 111 +++ ,, is the centre of the sphere.
AB r
CR
B x y z( , , )2 2 2
90 º
( , , ) x y z A1 1 1
P x y z( , , )
58 KSOU Solid Geometry
have wetois As ,PCPQ r⊥
0( 111111 =+−++−++− ))(())(())( wzzzvyyyuxxx
12111
211 ie vyvyyyyuxuxxxx −+−+−+−
01211 =−+−+ wzwzzzz
111111 ie zwwzvyvyuxuxzzyyxx −+−+−+++
021
21
21 =−−− zyx
sphere on thepoint a is Since 111 ),,( zyx
0222 11121
21
21 =++++++ dwzvyuxzyx
dwzvyuxzyx +++=−−−∴ 11121
21
21 222
substituting this in (1) and simplifying, we have
0111111 =+++++++++ dzzwyyvxxuzzyyxx )()()(
is the equation to the tangent plane.
5. To find the condition for two spheres to cut orthogonally (ie. the tangents plane at a point of intersection are at rightangles).
21Let CC & be the centres of two spheres whose radii
are Prr .& 21 is point of intersection.
Let the spheres be
0222: 1111222
1 =++++++ dzwyvxuzyxS
0222: 2222222
1 =++++++ dzwyvxuzyxS
22
21
21121 90ˆ Now rrCCCPC +=∴°=
221
221
221ie )()()( wwvvuu +−++−++−
2
2222
22
2
1211
21
−+++
−++= dwvudwvu
222
22
221
21
21
2121
22
2121
22
2121
22
21 222 ie dwvudwvuwwwwvvvvuuuu −+++−++=−++−++−+
21212121 222 ie ddwwvvuu −−=−−−
21212121 222 ie ddwwvvuu +=++
is the required condition.
Note :- as form standard the toreduced becan it hen equation tgiven thebe 0222 If 222 =++++++ dwzvyuxazayax
0222222 =++++++
a
dz
a
wy
a
vx
a
uzyx
Examples
1. Find the equation of the sphere whose centre is at (2, –3, 4) and radius 3 units.2222 3)4()3()2( is equation required The :Solution =−+++− zyx
091689644 ie 222 =−+−+++++− yzyyxx
020864 ie 222 =+−+−++ zyxzyx
P x y z( , , )1 1 1
C u v w(– , – , – )
C u y w( , , )− − −1 1 1
1 Cu y w( , , )− − −2 2 2
2
r 1 r 2
P
BCA 21 / IMCA 21 / Mathematics SVT 59
2. 04
3346 sphere theof radius and centre theFind 222 =−−+−++ zyxzyx
4
3324262equation standard with Comparing :Solution −=−==−= dwvu ,,,
2
323 −==−=∴ wvu ,,
−=−−−=∴
2
323Centre ,,),,( wvu
.4164
3913
4
3
4
949radius ==++=+++=r
3. Find the equation of the sphere whose diameter is the line joining the points (4, 0, –2) and (0, 3, 1).
0 form theof isequation Required :Solution 212121 =−−+−−+−− ))(())(())(( zzzzyyyyxxxx
0123004 isequation Required =−++−−+−−∴ ))(())(())(( zzyyxx
0234 ie 222 =−++−+− zzyyxx
0234 ie 222 =−+−−++ zyxzyx
4. 3). 2, (1,point theand54329 circle he through tsphere theofequation theFind 222 =++=++ zyxzyx ,
iton lie321054329 isequation required The :Solution 222 ),,(,)( =−+++−++ zyxzyx λ
0512629941 =−+++−++∴ )(λ
3
10155 ie −=⇒=+ λλ )(
)() 54323
19( is sphere theofEquation 222 −++−−++∴ zyxzyx
0543227333 ie 222 =+−−−−++ zyxzyx
022432333 ie 222 =−−−−++ zyxzyx
5. 0742505243 circle thefor which sphere theofequation theFind 222 =++−=−−+−++ zyxzyxzyx , is a great
circle.
0)7425(5243 is sphere theof equation Required :Solution 222 =++−+−−+−++ zyxzyxzyx λ
+−−−−+−− )(),(),( λλλ 42
2
124
2
153
2
1 is sphere thisof Centre
Circle will be a great circle if the centre lies on the plane.
0742224532
5 =++−−−++−−∴ )()()( λλλ
0784242
25
2
15 ie =+−+−+− λλλ
102
45
2
45 ie =⇒=+− λλ
60 KSOU Solid Geometry
074255243 issphererequiredthetoEquation 222 =++−+−−+−++∴ zyxzyxzyx
.02222 ie 222 =++++++ zyxzyx
6. Find the equation of the tangent plane to the sphere 3) 2, (1,point at the 0224323 222 =−−−−++ zyxzyx )(
03
22
3
4
3
2 is sphere theofEquation :Solution 222 =−−−−++ zyxzyx
03
22)3(
3
2)2(
2
1)1(
3
1)3()2()1( is 3) 2, (1,at planeTangent =−+−+−+−++ zyxzyx
03
222
3
21
23
1
3
132 ie =−−−−−−−++ zy
xzyx
multiply through out by 6
044124632218126 =−−−−−−−++ zyxzyx
0641494 ie =−++ zyx
7. angles.right at intersect 02046 and 08146 spheres that theShow 222222 =+++++=+++++ zxzyxzyzyx
21212121 222 is orthogonal be tospheres for twoCondition :Solution ddwwvvuu +=++
8730 spherefirst For the 1111 ==== dwvu ,,,
20203 sphere second For the 2222 ==== dwvu ,,,
282800222 212121 =++=++∴ wwvvuu
2820821 =+=+ dd
21212121 222 ddwwvvuu +=++∴
∴ The two spheres intersect orthogonally.
8. sphere the touches01222 plane that theShow =++− zyx 3242222 =+−−++ zyxzyx and find the point ofcontact.
Solution : Let C be the centre of the sphere, then C = (1, 2, –1).
Cr from theofLength ⊥ upon the plane
.33
9
144
12142 is01222 ==
+++−−=++− zyx
plane. the to is en Contact th ofpoint thebe Let rCPP ⊥
as taken becan oEquation t CP∴
say)1
1
2
2
2
1(r
zyx =+=−−=−
),,( 12212 as taken becan of ordinates-Co −+−+∴ rrrP
0121222122 if plane in the lie willThis =+−+−−+ ))(()( rrr
01214424 ie =+−+−++ rrr
1099 −=⇒=+ rr
),,(),,( 241112212 are of ordinates-Co −−=−−++−∴ P
),,( 241 iscontact ofPoint −−∴
P
C (1 , 2, 1 )−
BCA 21 / IMCA 21 / Mathematics SVT 61
Right Circular Cone
Definition : A right circular cone is a surface genereated by a straight line which passes through a fixed point (called Vertex)and makes a constant angle with a fixed line (called axis).
The constant angle is called semi-vertical angle of the cone.
A BC
V (Ve rtex )
ax is
a
A BC
V (Ve rtex )
ax i s
a
G enerato r
Examples
(1) Find the equation of the right circular cone whose vertex is at the origin, semi-vertical angle is α and having axis of Z asits axis.
generator, on thepoint any be Let :Solution ),,( zyxP OZ is
the axis whose direction cosines are 0, 0, 1.
zyxOP ,, are of ratiosDirection
222222 1
100
zyx
z
zyx
zyx
++=
⋅++
⋅+⋅+⋅=∴ αcos
Squaring both sides
222
22
zyx
z
++=αcos
α22222 seczzyx =++∴
αα 222222 1ie tan)(sec zzyx =−=+
α2222 is cone theofEquation tanzyx =+∴
(2) Find the equation of the right circular cone with semi-vertical angle 30º, vertex at the point (2, 1, –3) and the directionratios of whose axis are 3, 4, –1.
),,(),,( 312 vectex andgenerator on thepoint any be Let :Solution −=VzyxP
312 are of ratiosDirection +−− zyxPV ,,
°=− 302 4, 3, are axis of ratiosDirection α&
1169312
31142330
222 ++++−+−
+−−+−=°∴)()()(
)()()(cos
zyx
zyx
26312
34643
2
3 ie
222 )()()( ++−+−
−−−−+=zyx
zyx
Z
V (0 , 0, 0 )
axis
a
P x y z( , , )
X
Y
