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Mathematics
Session
Three Dimensional Geometry–1(Straight Line)
Session Objectives
Class Exercise
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
Equation of a Line Passing Through a Fixed Point and Parallel to a Given Vector
a
OY
Z
X
,,,,,,,,,,,,,,m
r
AP,
,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Let a be the position vector of a fixed point A.
Let r is the position vector of any arbitrary
point P on the line and line is parallel to m
i.e. AP = m where is any real number.
,,,,,,,,,,,,, ,r - a= m
,,,,,,,,,,,,, ,r =a+ m Vector form
Cartesian Form
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ,,,,,,,,,,,,, ,
1 1 1Let r = xi+yj+zk, a= x i+y j+z k and m=ai+bj+ck
1 1 1x - x y - y z - z= =
a b c
Example-1
x - 5 y+4 z- 6If cartesian equation of a line is = = , then
3 7 2find its vector equation.
, x - 5 y+4 z - 6
Solution: Let = = = is any real number3 7 2
x =3 +5, y =7 - 4, z =2 +6
ˆˆ ˆ
I f r = xi+yj+zk is the position vector of point P(x, y, z) on line.
ˆˆ ˆ r =(3 +5) i+(7 - 4)j+(2 +6)k
ˆ ˆˆ ˆ ˆ ˆ r =5i - 4j+6k + (3i+7j+2k)
Example –2
ˆˆ ˆ
Find the equation of a line in both vector and cartesian form,
which passes through the point 5, - 2, 4 and parallel to the
vector 2i - j+k.
ˆˆ ˆSolution: We have point : 5, - 2, 4 and parallel vector : 2i - j+k.
ˆ ˆˆ ˆ ˆ ˆ,,,,,,,,,,,,, ,a=5i - 2j+4k and m=2i - j+k
ˆ ˆˆ ˆ ˆ ˆ r =5i - 2j+4k + (2i - j+k)
,,,,,,,,,,,,, ,r =a+ m Vector form
ˆˆ ˆ r =(5+2 ) i - (2+ )j+(4+ )k Vector equation
Solution Cont.
Let P(x, y, z) lie on this line and r is the position vector of P.
ˆ ˆˆ ˆ ˆ ˆ xi+yj+zk =(5+2 ) i+(-2- )j+(4+ )k
x =5+2 , y =-2- , z = 4+
x - 5 y+2
= =z- 4 (Cartesian form)2 -1
Passing Through Two Fixed Points
a
b
OY
Z
X
r
AB P
Let A and B be the given points with position
vectors a and b respectively. Let r be position
vector of any point P on the line AB.
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,AP is collinear with AB
AP = AB, where is any scalar.
r - a= b- a
r =a b- a
Cartesian Form
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ
1 1 1 2 2 2Let r = xi+yj+zk, a= x i+y j+z k and b= x i+y j+z k
1 1 1
2 1 2 1 2 1
x - x y - y z - z= =
x - x y - y z - z
Example –3
Find the vector and the Cartesian equations for the line through
the points A(3, 4, -7) and B(5, 1, 6).
Solution: Let a and b be the position vectors of the points
A 3, 4, - 7 and B 5, 1, 6 .
ˆ ˆˆ ˆ ˆ ˆa=3i+ 4j - 7k and b=5i+ j+6k
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆb- a= 5i+ j+6k - 3i+ 4j - 7k =2i - 3j+13k
Solution Cont.
Let r be the position vector of any point on the line. Then,
r = a+ b- a
ˆ ˆˆ ˆ ˆ ˆ r =3i+ 4j - 7k + 2i - 3j+13k
1 1 1
2 1 2 1 2 1
x - x y - y z - zCartesian form of the equation is = =
x - x y - y z - z
x - 3 y - 4 z+7 x - 3 y - 4 z+7
= = = =5- 3 1- 4 6+7 2 -3 13
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
ˆ ˆˆ ˆ ˆ ˆ r =5i+ j+6k + 3- 5 i+ 4- 1 j+ 1- 6 k
ˆ ˆˆ ˆ ˆ ˆ r =5i+ j+6k + -2i+3j - 5k ... i
Let P be the point where the line AB crosses the y z-plane. Then,
the position vector of the point P is ˆˆyj+zk.
Solution Cont.
This point must satisfy (i)
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ yj+zk =5i+ j+6k + -2i+3j - 5k
ˆ ˆˆ ˆ ˆ yj+zk = 5- 2 i+ 1+3 j+ 6- 5 k
0=5- 2 ... ii
y =1+3 ... iii
z =6- 5 ... iv
Solving (ii), (iii) and (iv), we get17 13
y = , z =-2 2
17 13The co- ordinates of the required points are 0, , - .
2 2
Co-linearity of Three Points
Let A, B, C be three points with position vectors a, b, c respectively.
Equation of line passing through A and B is
r =a + b- a ... i
A, B, C are collinear if and only if C satisfies i
c=a + b- a
Example -5
ˆˆ ˆ ˆ ˆ ˆ ˆ ˆk, 3i k are collinear.Show that points with position vectors i - 2j+3k, 2i - j+
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ
Solution: Let a= i - 2j+3k, b=2i - j+k and c =3i - k.
As we know a, b, c will be collinear if
. c=a+ (b- a) for some real number
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ 3i - k = i - 2j+3k + (2i - j+k - i+2j - 3k)
ˆ ˆˆ ˆ ˆ ˆ ( i+ j - 2k) =2i+2j - 4k
=2
a, b, c are collinear.
Angle Between Two Lines
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 1 2 2
1 2
Let r =a + b and r =a + b be two straight lines.
These straight lines are in the direction of b and b .
I f be the angle between these two lines, then is also
the angle between,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2 b and b .
,,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,
,,,,,,,,,,,,, ,1 2
1 2 1 21 2
b . bb . b = b b cos cos =
b b
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2
I f lines are perpendicular, then
b . b =0
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2
I f lines are parallel, then
b = b
Cartesian Form
1 1 1 2 2 2
Let the equations of two lines be
x - x y - y z - z x - x y - y z - z= = and = = .
a b c a' b' c'
2 2 2 2 2 2
aa +bb +cccos =
a +b +c a +b +c
I f the lines are perpendicular, then
aa +bb +cc =0
If the lines are parallel, then
a b c= =
a' b' c'
Example –6
ˆ ˆˆ ˆ ˆ ˆ
ˆ ˆˆ ˆ ˆ
Find the angle between the pair of lines
r =3i+2j - 4k + i+2j+2k
r =5i - 2k + 3i+2j+6k
,,,,,,,,,,,,,,,,,,,,,,,,,, ,,
,,,,,,,,,,,,,,,,,,,,,,,,,, ,, 1 1
2 2
Solution: The given lines is of the form
r =a + b
r =a + b
ˆ ˆˆ ˆ ˆ ˆ,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2where b = i+2j+2k and b =3i+2j+6k
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2
Let be the angle between given lines, then is
also the angle between the direction of b and b .
Solution Cont.
ˆ ˆˆ ˆ ˆ ˆ
i+2j+2k . 3i+2j+6kcos =
1+4+4 9+4+36
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1 2
1 2
b . bcos =
b b
3+4+12
cos =3×7
-1 19=cos
21
Example –7
.
Find the angle between the pair of lines:
x - 2 y+3 x+1 2y - 3 z - 5= , z =5 and = =
3 -2 1 3 2
Solution: The given equations can be written as
3y -x - 2 y+3 z - 5 x+1 z - 52= = ... i and = = ... ii
33 -2 0 1 22
ˆ ˆˆ ˆ ˆ ˆ
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2
1 2
Let b and b be vectors parallel to i and ii , then
3b =3i - 2j+0k and b = i+ j+2k
2
Solution Cont.
Let be the angle between given lines, then
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1 2
1 2
b . bcos =
b b
3- 3+0
cos = =09
9+4+0 1+ +44
=
2
Intersection of Two Lines Example - 8
x y - 2 z+3 x - 2 y - 6 z - 3Show that the lines = = and = = intersect.
1 2 3 2 3 4Also find the point of intersection.
x y - 2 z+3
Solution: Let = = = , then1 2 3
x = , y =2 +2 and z =3 - 3
. Any general point on this line is , 2 +2 , 3 - 3
x - 2 y - 6 z - 3
Let = = = , then2 3 4
3 x =2 +2, y =3 +6 and z = 4
Solution Cont.
3 . Any general point on this line is 2 +2, 3 +6, 4
If the lines intersect, then they have a common point.
3 =2 +2, 2 +2=3 +6 and 3 - 3= 4
4
2 =2 ... i
2 3 = 4 ... ii
3 =6 ... iii
2 and 0 Solving i and ii , we get =
Solution Cont.
2 0
and = satisfy iii .
The given lines intersect.
Putting =2 in , 2 +2 , 3 - 3 , the point of intersection is
2, 6, 3 .
Perpendicular Distance Example –9
Find the foot of the perpendicular from the point (1, 2, - 3) on the line
x+1 y - 3 z= = .
2 -2 -1Also find the length of the perpendicular.
P(1, 2, -3)
x+1 y - 3 z= =
2 -2 -1 L 2 - 1, - 2 +3, -
x+1 y - 3 z
Solution: Let = = =2 -2 -1
x =2 -1, y =-2 +3, z =-
.
Any general point on this line is
2 - 1, - 2 +3 , -
Solution Cont.
Let the co- ordinates of L be 2 - 1, - 2 +3 , - ... i
. Direction ratios of PL are 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.
0 2 2 - 2 - 2 - 2 +1 - - +3 =1
Solution Cont.
Putting =1 in i , the coordinates of L are 1, 1, - 1 .
2 2 2PL = 1- 1 + 1- 2 + -1+3 = 5 units.
Image of a Point Example –10
ˆ ˆˆ ˆ ˆ ˆFind the image of the point (2, - 1, 5) in the line
r = 11i - 2j - 8k + 10i - 4j - 11k .
Q
A B
P (2, -1, 5)
L
Solution:
Let Q be the image of the given
point P(2, -1, 5) in the given
line and let L be the foot of the
perpendicular from P on the given line.
Solution Cont.
r
.
in the equation of a line gives the position vector of
a point on it for different values of
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ
Let the position vector of L be
11i - 2j - 8k + 10i - 4j - 11k = 11+10 i - 2+4 j - 8+11 k ... i
ˆˆ ˆ ˆ ˆ ,,,,,,,,,,,,,,
PL = 11+10 i - 2+4 j - 8+11 - 2i - j+5k
ˆˆ ˆ = 9+10 i - 1+4 j - 13+11 k
Solution Cont.
ˆˆ ˆ
,,,,,,,,,,,,,,
PL is perpendicular to the given line which is
parallel to b=10i - 4j - 11k
ˆ ˆˆ ˆ ˆ ˆ. ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
PL . b=0 9+10 i - 1+4 j - 13+11 k 10i - 4j - 11k =0
10 9+10 +4 1+4 +11 13+11 =0
=-1
ˆ ˆˆ ˆ ˆ ˆ1 1 1 2 3 Position vector of L is 11- 10 i - 2- 4 j - 8- 11 k = i+ j+ k
Solution Cont.
ˆ ˆˆ ˆ ˆ ˆˆˆ ˆ2 3
1 1 1
L is the mid- point of PQ.
2i - j+5k + x i+y j+z k = i+ j+ k
2
1 1 1Let the coordinates of Q be x , y , z .
ˆ ˆˆ ˆ ˆ ˆ
1 1 12+x -1+y 5+zi+ j+ k = i+2j+3k
2 2 2
1 1 11 1 1
2+x -1+y 5+z=1, =2, =3 x =0, y =5, z =1
2 2 2
Hence, the image of P(2, -1, 5) in the given line is (0, 5, 1).
Shortest Distance Between Two Lines
DB I2
I1C A
90 o
90 o
b 2
b 1
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)
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,
1 1 2 2
1 2
Let r =a + b and r =a + b be the equations
of two skew lines and respectively.
Cont.
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
1 2
1 2
I f AB =d is the shortest distance between and . Then,
AB is parallel to b ×b .
ˆ
ˆ
,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1 2
1 2
I f n is a unit vector along AB, then
b ×bn=
b ×b
ˆ,,,,,,,,,,,,,,AB = d n
1 1
2
a
,,,,,,,,,,,,,,
,,,,,,,,,,,,,, 2
Let C be a point on with position vector and
D be on with position vector a .
Cont.
Di uuur uuur uuur uuur
stance AB = The length of projection of CD on AB = CD cos
ˆ
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,I f is the angle between CD and AB, then
will be angle between CD and n.
ˆ
ˆ
uuur
uuurCD.n
cos =CD n
.
uur uur uur uuruuur
uur uur2 1 1 2
1 2
a a b ×bCD cos =
b ×b
uur uur uur uuruuur
uur uur2 1 1 2
1 2
a - a . b ×bAB =d =
b ×b
Cont.
. ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1 1 2The two lines will intersect when d=0 i.e. a - a b b =0.
b
b
90 o
90–o
D(B)
C
A
I2
I1
90 o
,,,,,,,,,,,,,,,,,,,,,,,,,, ,,
2 1
I f two lines are parallel, then
b× a - ad=
b
Example –11
ˆ ˆˆ ˆ ˆ ˆ
ˆ ˆˆ ˆ ˆ ˆ
Find the shortest distance between the lines
r = i+2j+3k + 2i+3j+4k and
r = 2i+4j+5k + 3i+4j+5k .
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Solution: The given lines are
r = i+2j+3k + 2i+3j+4k and r = 2i+4j+5k + 3i+4j+5k
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 1 2 2
The shortest distance between the lines
r =a + b and r =a + b is
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1 1 2
1 2
a - a . b ×bd=
b ×b
Solution Cont.
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2 1 2a = i+2j+3k, a =2i+4j+5k, b =2i+3j+4k and b =3i+4j+5k
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1a - a = 2i+4j+5k - i+2j+3k i+2j+2k
ˆˆ ˆ
ˆˆ ˆ,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2
i j k
b ×b = 2 3 4 =- i+2j - k
3 4 5
ˆ ˆˆ ˆ ˆ ˆ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1 1 2a - a . b ×b = i+2j+2k . - i+2j - k =-1+4- 2=1
,,,,,,,,,,,,,,,,,,,,,,,,,,,,1 2and b ×b = 1+4+1 = 6
Solution Cont.
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1 1 2
1 2
a - a . b ×b 1 1d= = =
6 6b ×b
Example -12
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ . Show that the following pair of lines intersect :
r = i+ j - k + 3i - j and r = 4i - k + 2i+3k
Solution: The lines will intersect if shortest distance between them = 0
. ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2 1 1 2i.e. a - a b b =0
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ. 4i - k i+ j - k 3i - j 2i+3k
ˆˆ ˆ ˆ ˆ= 3i - j . -3i - 9j+2k
=-9+9=0
Therefore, the given lines intersect.
Thank you
