Transcript Document

Mathematics
Session
Three Dimensional Geometry–1(Straight Line)
Session Objectives
 Equation: Passing Through a Fixed Point and Parallel to a Given Vector
 Equation: Passing Through Two Fixed Points
 Co-linearity of Three Points
 Angle Between Two Lines
 Intersection of two lines, Perpendicular distance, Image of a Point
 Shortest Distance Between Two Lines
 Class Exercise
Equation of a Line Passing Through a
Fixed Point and Parallel to a Given Vector
L e t a b e th e p o sitio n v e cto r o f a fix e d p o in t A .
Z
L e t r is th e p o sitio n v e cto r o f a n y a rb itra ry
p o in t P o n th e lin e a n d lin e is p a ra lle l to m
m
i.e . A P =  m , w h e re  is a n y re a l n u m b e r.
a
 r - a = m
 r = a + m
 V ector
form 
O
X
P
A
r
Y
Cartesian Form
ˆ a = x ˆi + y ˆj + z kˆ and m = a iˆ + bjˆ + c kˆ
Let r = x ˆi + yjˆ + zk,
1
1
1

x - x1
a
=
y - y1
b
=
z - z1
c
Example-1
x -5
If carte sian e q u atio n o f a lin e is
3
=
y+4
7
=
z-6
2
, th e n
fin d its v e cto r e q u atio n .
S o lu tio n : L e t
x -5
3
=
y+4
7
=
z-6
2
= ,

is an y re al n u m b e r 
 x = 3  + 5, y = 7  - 4, z = 2  + 6
If r = x ˆi + yjˆ + zkˆ is the position vector of point P(x, y, z) on line.
 r = (3  + 5) ˆi + (7  - 4)jˆ + (2  + 6)kˆ
ˆ
 r = 5 ˆi - 4jˆ + 6kˆ + (3 ˆi + 7jˆ + 2k)
Example –2
F in d th e e q u atio n o f a lin e in b o th v e cto r an d carte sian fo rm ,
w h ich p asse s th ro u g h th e p o in t  5 , - 2 , 4  an d p aralle l to th e
v e cto r 2 ˆi - ˆj + kˆ .
S o lu tio n : W e h av e p o in t :  5 , - 2 , 4  an d p a ralle l v e cto r : 2 ˆi - ˆj + kˆ .
a = 5 ˆi - 2jˆ + 4kˆ and m = 2 iˆ - ˆj + kˆ
 r = a + m
 V ector
form 
ˆ
 r = 5 ˆi - 2 ˆj + 4 kˆ +  (2 ˆi - ˆj + k)
 r = (5 + 2  ) ˆi - (2 +  )jˆ + (4 +  )kˆ
 V e cto r
e q u atio n 
Solution Cont.
L e t P (x , y , z ) lie o n th is lin e a n d r is th e p o sitio n v e cto r o f P .
 x ˆi + yjˆ + zkˆ = (5 + 2  ) ˆi + (-2 -  )jˆ + (4 +  )kˆ
 x = 5 + 2  , y = -2 -  , z = 4 + 

x -5
2
=
y+2
-1
= z-4
(C artesian form )
Passing Through Two Fixed Points
L e t A a n d B b e th e g iv e n p o in ts w ith p o s itio n
v e cto rs a a n d b re sp e ctiv e ly . L e t r b e p o sitio n
v e cto r o f a n y p o in t P o n th e lin e A B .
Z
A P is co llin ear w ith A B
 A P =  A B, w h ere  is an y scalar.

 r-a=  b-a

A
a

 r= a  b-a
O

X
B
b
P
r
Y
Cartesian Form
ˆ a = x ˆi + y ˆj + z kˆ and b = x iˆ + y ˆj + z kˆ
Let r = x ˆi + yjˆ + zk,
1
1
1
2
2
2
x - x1
x 2 - x1
=
y - y1
y 2 - y1
=
z - z1
z 2 - z1
Example –3
Find the vector and the Cartesian equations for the line through
the points A(3, 4, -7) and B(5, 1, 6).
S o lu tio n : L e t a a n d b b e th e p o sitio n v e cto rs o f th e p o in ts
A  3 , 4 , - 7  a n d B 5 , 1 , 6 .
a = 3 ˆi + 4jˆ - 7kˆ an d b = 5 ˆi + ˆj + 6kˆ

 

 b - a = 5 ˆi + ˆj + 6 kˆ - 3 ˆi + 4 ˆj - 7 kˆ = 2 ˆi - 3 ˆj + 1 3 kˆ
Solution Cont.
L e t r b e th e p o sitio n v e cto r o f a n y p o in t o n th e lin e . T h e n ,

r = a+  b - a


 r = 3 ˆi + 4 ˆj - 7 kˆ +  2 ˆi - 3 ˆj + 1 3 kˆ
C artesian form of th e eq u ation is

x-3
5-3
=
y-4
1- 4
=
z+7
6+7

x-3
2
=

x - x1
x 2 - x1
y-4
-3
=
=
y - y1
y 2 - y1
z+7
13
=
z - z1
z 2 - z1
Example –4
Find the coordinates of the points where the line through A(5, 1, 6) and
B(3, 4, 1) crosses the y z- plane.
Solution: The vector equation of the line through the points A and B is
r = 5 ˆi + ˆj + 6 kˆ +    3 - 5  ˆi +  4 - 1  ˆj + 1 - 6  kˆ 



 r = 5 ˆi + ˆj + 6 kˆ +  -2 ˆi + 3 ˆj - 5 kˆ

...  i 
Let P be the point where the line AB crosses the y z-plane. Then,
the position vector of the point P is y ˆj + zkˆ .
Solution Cont.
This point must satisfy (i)

 y ˆj + z kˆ = 5 ˆi + ˆj + 6 kˆ +  -2 ˆi + 3 ˆj - 5 kˆ

 y ˆj + zkˆ =  5 - 2   ˆi + 1 + 3   ˆj +  6 - 5   kˆ
 0 = 5 - 2
...  ii 
y = 1 + 3
...  iii 
z = 6 - 5
...  iv 
Solving (ii), (iii) and (iv), we get y =
17
2
,z=-
13
2
17
13 

 T h e co - o rd in a te s o f th e re q u ire d p o in ts a re  0 ,
,.
2
2


Co-linearity of Three Points
L e t A , B , C b e th re e p o in ts w ith p o sitio n v e cto rs a, b, c re sp e ctiv e ly .
E q u atio n o f lin e p assin g th ro u g h A an d B is

r= a+ b-a

...  i 
A , B , C a re co llin e a r if a n d o n ly if C s a tisfie s  i 

c= a+ b-a

Example -5
ˆ 2 ˆi - ˆj + k,
ˆ 3 ˆi  kˆ are collinear.
Show that points w ith position vectors iˆ - 2jˆ + 3k,
ˆ b = 2 iˆ - ˆj + kˆ and c = 3 ˆi - k.
ˆ
Solution : Let a = ˆi - 2jˆ + 3k,
A s w e k n o w a, b, c w ill b e co llin ear if
c = a +  (b - a) fo r so m e re a l n u m b e r  .
ˆ
 3 ˆi - kˆ = ˆi - 2jˆ + 3kˆ + (2 ˆi - ˆj + kˆ - iˆ + 2jˆ - 3k)
ˆ = 2 ˆi + 2jˆ - 4kˆ
 ( ˆi + ˆj - 2k)
 =2
 a, b, c are co llin ear.
Angle Between Two Lines
L e t r = a1 +  b 1 an d r = a2 +  b 2 b e tw o straig h t li n e s.
T h e se straig h t lin e s are in th e d ire ctio n o f b 1 an d b 2 .
If  b e th e an g le b e tw e e n th e se tw o lin e s , th e n  is also
th e an g le b e tw e e n b 1 an d b 2 .
 b 1 . b 2 = b 1 b 2 co s   co s  =
b1 . b 2
b1 b 2
If lin es are p erp en d icu lar, th en
b1 . b 2 = 0
If lin es are p arallel, th en
b1 =  b 2
Cartesian Form
L e t th e e q u atio n s o f tw o lin e s b e
x - x1
a
=
y - y1
b
=
z - z1
c
an d
x - x2
a'
y - y2
=
b'
a a  + b b  + cc 
co s  =
2
2
a +b +c
2
2
2
2
a + b  + c 
If th e lin es are perpen dicu lar, th en
aa  + bb  + cc  = 0
If th e lin e s are p aralle l, th e n
a
a'
=
b
b'
=
c
c'
=
z - z2
c'
.
Example –6
F in d th e a n g le b e tw e e n th e p a ir o f lin e s

r = 3 ˆi + 2 ˆj - 4 kˆ +  ˆi + 2 ˆj + 2 kˆ

r = 5 ˆi - 2 kˆ +  3 ˆi + 2 ˆj + 6 kˆ


S o lu tio n : T h e g iv e n lin e s is o f th e fo rm
r = a1 +  b 1
r = a2 +  b 2
w h ere b1 = ˆi + 2jˆ + 2kˆ an d b 2 = 3 iˆ + 2jˆ + 6kˆ
Let  b e th e an g le b etw een g iven lin es, t h en  is
also th e an g le b etw een th e d irectio n o f b1 an d b 2 .
Solution Cont.
 co s  =
 co s  =
 co s  =
  = co s
b1 . b 2
b1 b 2
 ˆi + 2ˆj + 2 kˆ  .  3 ˆi + 2 ˆj + 6 kˆ 
1 + 4 + 4 9 + 4 + 36
3 + 4 + 12
3×7
-1
 19 


 21 
Example –7
F in d th e an g le b e tw e e n th e p air o f lin e s :
x -2
3
=
y+3
-2
, z = 5 an d
x +1
1
=
2y - 3
3
=
z-5
2
.
S o lu tio n : T h e g iv e n e q u a tio n s ca n b e w ritte n a s
x -2
3
=
y+3
-2
=
z-5
0
...  i  a n d
x +1
1
y=
3
2 = z-5
3
2
...  ii 
2
Let b 1 an d b 2 b e vecto rs p arallel to i  an d  ii  , th en
3
b1 = 3 ˆi - 2 ˆj + 0 kˆ an d b 2 = ˆi + ˆj + 2 kˆ
2
Solution Cont.
Let  be the angle betw een given lines, then
 co s  =
 co s  =
b1 . b 2
b1 b 2
3-3+0
9 + 4 + 0 1+
 =

2
9
4
=0
+4
Intersection of Two Lines
Example - 8
S h o w th at th e lin e s
x
1
=
y-2
2
=
z+3
3
an d
x-2
2
=
y-6
3
A lso fin d th e p o in t o f in te rse ctio n .
S o lu tio n : L e t
x
1
=
y-2
2
=
z+3
3
=  , th e n
x =  , y = 2  + 2 and z = 3  - 3
A n y g e n e ra l p o in t o n th is lin e is   , 2  + 2 , 3  - 3  .
Let
x-2
2
=
y-6
3
=
z-3
4
=  , th e n
x = 2  + 2, y = 3  + 6 and z = 4   3
=
z-3
4
in te rse c t.
Solution Cont.
A n y g e n e ra l p o in t o n th is lin e is  2  + 2 , 3  + 6 , 4   3  .
If the lines intersect, then they have a common point.
  = 2  + 2, 2  + 2 = 3  + 6 and 3  - 3 = 4   3
   2 = 2
...  i 
2   3 = 4
...  ii 
3  4 = 6
...  iii 
S o lv in g  i  a n d  ii  , w e g e t   2 a n d  = 0
Solution Cont.
  2 a n d  = 0 sa tisfy  iii  .
 T h e g iv e n lin e s in te rse ct.
P u ttin g  = 2 in   , 2  + 2 , 3  - 3  , th e p o in t o f in te rse ctio n is
2 , 6 , 3 .
Perpendicular Distance
Example –9
F in d th e fo o t o f th e p e rp e n d icu la r fro m th e p o in t (1 , 2 , - 3 ) o n th e lin e
x+1
2
=
y-3
-2
=
z
-1
.
A lso fin d th e le n g th o f th e p e rp e n d icu la r.
S olu tion : Let
x +1
2
=
y-3
-2
=
z
-1
=
P(1, 2, -3)
 x = 2  - 1, y = -2  + 3, z = - 
A n y g e n e ra l p o in t o n th is lin e is
2 - 1 , - 2 + 3 , -   .
x +1
2
=
y-3
-2
=
z
-1
L 2 - 1 , - 2  + 3 , -  
Solution Cont.
L e t th e co - o rd in a te s o f L b e  2  - 1 , - 2  + 3 , -  
...  i 
 D ire ctio n ra tio s o f P L a re 2  - 1 - 1 , - 2  + 3 - 2 , -  + 3 .
i.e . 2  - 2 , - 2  + 1 , -  + 3
Direction ratios of the given line are 2, - 2, -1.
PL is perpendicular to the given line.
 2 2  - 2  - 2 - 2  + 1  - - + 3   0   = 1
Solution Cont.
P u ttin g  = 1 in  i  , th e co o rd in a te s o f L a re  1 , 1 , - 1  .
PL =
1 - 1 
2
2
+ 1 - 2  +  -1 + 3 
2
=
5 u n its.
Image of a Point
Example –10
F in d th e im ag e o f th e p o in t (2 , - 1 , 5 ) in th e lin e




r = 1 1 ˆi - 2 ˆj - 8 kˆ +  1 0 ˆi - 4 ˆj - 1 1 kˆ .
P (2, -1, 5)
Solution:
Let Q be the image of the given
point P(2, -1, 5) in the given
A
L
line and let L be the foot of the
perpendicular from P on the given line.
Q
B
Solution Cont.
r in th e e q u atio n o f a lin e g iv e s th e p o si tio n v e cto r o f
a p o in t o n it fo r d iffe re n t v alu e s o f  .
 L e t th e p o sitio n v e cto r o f L b e


1 1 ˆi - 2 ˆj - 8 kˆ +  1 0 ˆi - 4 ˆj - 1 1 kˆ = 1 1 + 1 0   iˆ -  2 + 4   ˆj -  8 + 1 1   kˆ

 P L =   1 1 + 1 0   ˆi -  2 + 4   ˆj -  8 + 1 1    - 2 ˆi - ˆj + 5 kˆ


=  9 + 1 0   ˆi - 1 + 4   ˆj - 1 3 + 1 1   kˆ

...  i 
Solution Cont.
P L is p e rp e n d icu la r to th e g iv e n lin e w h ich is
p a ra lle l to b = 1 0 iˆ - 4 ˆj - 1 1 kˆ


 P L . b = 0    9 + 1 0   ˆi - 1 + 4   ˆj - 1 3 + 1 1   kˆ  . 1 0 ˆi - 4 ˆj - 1 1 kˆ = 0


 1 0  9 + 1 0   + 4 1 + 4   + 1 1 1 3 + 1 1   = 0
  = -1
 P o sitio n v ecto r o f L is 1 1 - 1 0  1 iˆ -  2 - 4  1 ˆj -  8 - 1 1  1 kˆ = ˆi + 2 ˆj + 3kˆ
Solution Cont.
L e t th e co o rd in a te s o f Q b e  x 1 , y 1 , z 1  .
L is th e m id - p o in t o f P Q .

 2 ˆi - ˆj + 5 kˆ  +  x1 ˆi + y1 ˆj + z1kˆ 
2
 2 + x1
 
2

 ˆ  -1 + y 1
i+
2


2 + x1
-1 + y 1

2
= 1,
2
= ˆi + 2 ˆj + 3kˆ
 ˆ  5 + z1
 j+ 

 2
= 2,
5 + z1
2
ˆ
ˆ
ˆ
ˆ
k = i + 2j+ 3k

= 3  x 1 = 0 , y 1 = 5 , z1 = 1
Hence, the image of P(2, -1, 5) in the given line is (0, 5, 1).
Shortest Distance Between Two Lines
Two straight lines in space, which do not intersect and are also not
parallel, are called skew lines.
(Which do not lie in the same plane)
b2
B
D
I2
o
90
o
90
C
Let r = a1 +  b1 an d r = a2 + b 2 b e th e eq u atio n s
o f tw o skew lin es 1 an d 2 resp ectively.
I1
A
b1
Cont.
If A B = d  is th e sh o rte st d ista n ce b e tw e e n 1 a n d 2 . T h e n ,
A B is p a ra lle l to b 1 × b 2 .
If nˆ is a u n it v e cto r a lo n g A B, th e n
nˆ =
b1 × b 2
b1 × b 2
 A B =  d  nˆ
L e t C b e a p o in t o n
D be on
2
1
w ith p o sitio n v e cto r a 1 a n d
w ith p o sitio n v e cto r a 2 .
Cont.
If  is th e an g le b etw een C D an d A B, th en
 w ill b e an g le b etw een C D an d nˆ .
uuur
C D.nˆ
 co s  = uuur
C D nˆ
uur uur
uur uur
a2  a1 . b 1 × b 2
uuur
 C D co s  =
uur uur
b1 × b 2



uuur
uuur
uuur
uuur
D i sta n ce A B = T h e le n g th o f p ro je ctio n o f C D o n A B = C D co s 
uuur
 A B = d  =

uur uur
uur uur
a2 - a1 . b 1 × b 2
uur uur
b1 × b 2


Cont.



Th e tw o lin es w ill in tersect w h en d = 0 i.e. a2 - a1 . b1  b 2 = 0.
D(B) b
I2
If tw o lin e s a re p a ra lle l, th e n
90
o
–

90
d=

b × a2 - a1
o
b
o

C
90
A
I1
b
Example –11
F in d th e sh o rte st d ista n ce b e tw e e n th e lin e s

 

r =  2 ˆi + 4 ˆj + 5 kˆ  +   3 ˆi + 4 ˆj + 5 kˆ  .
r = ˆi + 2 ˆj + 3 kˆ +  2 ˆi + 3 ˆj + 4 kˆ a n d
S o lu tio n : T h e g iv e n lin e s are







r = ˆi + 2 ˆj + 3 kˆ +  2 ˆi + 3 ˆj + 4 kˆ an d r = 2 iˆ + 4 ˆj + 5 kˆ +  3 iˆ + 4 ˆj + 5 kˆ
Th e sh o rtest d istan ce b etw een th e lin es
r = a1 +  b1 an d r = a2 + b 2 is
d=
 a2 - a1  .  b1 × b 2 
b1 × b 2

Solution Cont.
ˆ a = 2 ˆi + 4jˆ + 5k,
ˆ b = 2 iˆ + 3jˆ + 4kˆ and b = 3 ˆi + 4jˆ + 5kˆ
a1 = ˆi + 2jˆ + 3k,
2
1
2

 

a2 - a1 = 2 ˆi + 4 ˆj + 5 kˆ - ˆi + 2 ˆj + 3 kˆ  ˆi + 2 ˆj + 2 kˆ
ˆi
ˆj
kˆ
b1 × b 2 = 2
3
4 = - iˆ + 2 ˆj - kˆ
3
4
5


 


 a2 - a1 . b 1 × b 2 = ˆi + 2 ˆj + 2 kˆ . - ˆi + 2 ˆj - kˆ = -1 + 4 - 2 = 1
an d b1 × b 2 =
1+ 4 +1 =
6
Solution Cont.
d=
 a2 - a1  .  b1 × b 2 
b1 × b 2
=
1
6
=
1
6
Example -12
S h o w th at th e fo llo w in g p air o f lin e s in te rse ct :








r = ˆi + ˆj - kˆ +  3 ˆi - ˆj an d r = 4 ˆi - kˆ +  2 ˆi + 3 kˆ .
Solution: The lines will intersect if shortest distance between them = 0



i.e. a2 - a1 . b1  b 2 = 0
 4 ˆi - kˆ    ˆi + ˆj - kˆ  .   3 ˆi - ˆj    2 ˆi + 3 kˆ  


= 3 ˆi - ˆj . -3 ˆi - 9 ˆj + 2 kˆ

= -9 + 9 = 0
Therefore, the given lines intersect.
Thank you