Transcript Document

Mathematics
Session
Three Dimensional Geometry–2(The Plane)
Session Objectives
Normal Form of the Equation of a Plane, Cartesian Form
Plane Passing Through a Given Point and Perpendicular to
a Given Vector, Cartesian Form
Plane Passing Through a Point and Parallel to Two Given Lines,
Cartesian Form
Plane Containing Two Lines, Cartesian Form
Plane Passing Through Three Points, Cartesian Form
Passing Through the Intersection of Two Planes
Angle Between Two Planes, Angle Between a Line and a Plane
Intercept Form of a Plane
Condition of Co-planarity of Two Lines
Distance of a Point From a Plane
 Class Exercise
Plane
A plane can be defined as a surface such that if any two points A, B
are taken on it, the line segment AB lies on the surface i.e., every
point on the line segment AB lies on the plane.
Normal Form of the Equation of a Plane
ON= dnˆ
 NP  ON


NP . ON = 0

P
N

 r - dnˆ .dnˆ = 0


 r - dnˆ .nˆ = 0
r
dnˆ
d  0
n̂
ˆ ˆ =0
 r . nˆ - dn.n
 r . nˆ = d
 Vector form
O
ˆ ˆ =1
 n.n
Cartesian Form
Let P  x, y, z  be any point on the plane, then its
ˆ
ˆ
position vector r is r = xiˆ + yj+zk.
ˆ then
Let l, m, n be the direction cosines of n,
ˆ
ˆ
nˆ = liˆ +mj+nk
P(x,y,z)
N
r . nˆ = d



r
dnˆ
ˆ . liˆ +mj+nk
ˆ =d
ˆ
ˆ
 xiˆ +yj+zk
n̂
 lx+my+nz = d
Note: If the plane passes through the origin, then
ˆ = 0 and lx+my+nz = 0
r .n
O
Example –1
Find the vector equation of the plane which is at a distance
of 7 units from the origin and which is normal to the vector
ˆ
3iˆ + 5jˆ - 6k.
Solution: Let n = 3iˆ + 5jˆ - 6kˆ
nˆ =
n
n
=
3iˆ + 5jˆ - 6kˆ
9+25+36
=
3iˆ + 5jˆ - 6kˆ
70
The required equation of the plane is
3iˆ + 5jˆ - 6kˆ 

r.
=7
70
Using r . nˆ = d


Example -2
Find the cartesian equation of the plane
r.  s - 2t  ˆi + 3 - t  ˆj + 2s+ t  kˆ  =15.


Solution: Let P  x, y, z  be any point on the plane, then its
ˆ
ˆ
position vector r is r = xiˆ + yj+zk.
ˆ .  s -2t  ˆi + 3- t  ˆj + 2s+t kˆ  =15
ˆ
xiˆ +yj+zk


 s-2t  x+ 3- t  y+ 2s+t  z =15
To Reduce the Cartesian Equation to
Normal Form
ax+by+cz+d= 0
 -ax -by - cz = d
-
ax
a2 +b2 +c2
-
by
a2 +b2 +c2
-
cz
a2 +b2 +c2
=
d
a2 +b2 +c2
Example -3
Write the normal form of the equation of the plane
2x -3y+6z+14 = 0.
Solution: The given equation of the plane is
2x -3y+6z+14 = 0
 -2x +3y - 6z =14
-
2x
4+9+36
+
3y
4+9+36
2
3
6
 - x+ y- z =2
7
7
7
-
6z
4+9+36
=
14
4+9+36
Plane Passing Through a Given Point and
Perpendicular to a Given Vector


AP = r - a
r - a lies in the plane and perpendicular to n which is
normal to the plane.
A( a)
 The equation of the plane is
N
 r - a . n = 0
n
O
P( r)
Cartesian Form
ˆ a= a ˆi + a ˆj + a kˆ and n = n ˆi + n ˆj + n k,
ˆ
If r = xiˆ + yjˆ + zk,
1
2
3
1
2
3 then
r - a . n = 0


  x - a1  ˆi +  y - a2  ˆj +  z - a3 kˆ . n1ˆi + n2ˆj + n3kˆ = 0


  x - a1 n1 +  y - a2 n2 + z - a3 n3 = 0
Example –4
Find the vector and cartesian equation of the plane through the point
ˆ
ˆ 3k.
with position vector 2iˆ - ˆj+kˆ and perpendicular to the vector 4iˆ +2j-
ˆ 3kˆ
Solution: Here a = 2iˆ - ˆj+kˆ and n = 4iˆ +2j-
The vector equation of the plane is
r - a. n = 0  r. n = a. n

 





ˆ = 2iˆ - ˆj+kˆ . 4iˆ +2j-3k
ˆ
ˆ
ˆ
 r . 4iˆ +2j-3k
ˆ = 8-2-3
ˆ
 r . 4iˆ +2j-3k
ˆ =3
ˆ
 r . 4iˆ +2j-3k

Solution Cont.
Let P  x, y, z  be any point on the plane, then its
ˆ
ˆ
position vector r is r = xiˆ + yj+zk.
ˆ . 4iˆ +2j-3k
ˆ
xiˆ +yj+zk
  ˆ ˆ  =3
 4x+2y - 3z = 3
Plane Passing Through a Point
and Parallel to Two Given Lines
Let a be the position vector of the given point A in the plane.
Let the plane be parallel to the lines r = a+ b and r = a'+ b'.
Let r be the position vector of any point P in the plane.
P lies in the plane if AP, b and b' are coplanar, i.e.




AP. b×b' = 0

 r - a . b×b' = 0
Cartesian Form
ˆ a = x ˆi + y ˆj+ z kˆ
ˆ zk,
r = xiˆ + yj+
1
1
1
b = b1ˆi +b2 ˆj+b3kˆ and b' = b'1ˆi +b'2 ˆj+b'3kˆ



 r - a . b×b' = 0
x - x1
y - y1
z - z1
 b1
b2
b3
b'1
b'2
b'3
=0
Example –5
Find the vector and cartesian forms of the equation of the plane
passing through the point 1, 2, - 4  and parallel to the lines


ˆ +  ˆi + ˆj- kˆ .
ˆ
and r = ˆi - 3j+5k


ˆ 4kˆ +  2iˆ +3j+
ˆ 6kˆ
r = ˆi +2j-
ˆ b = 2iˆ +3j+6k
ˆ and b' = ˆi + ˆj-k.
ˆ
ˆ 4k,
ˆ
Solution: Here a= ˆi +2jThe vector equation of the plane is
r - a.b×b' = 0

 
 

ˆ  ˆi + ˆj-kˆ  = 0
ˆ 4kˆ .  2iˆ +3j+6k
ˆ
 r - ˆi +2j
 

Solution Cont.

 

ˆ 4kˆ . -9iˆ +8j-k
ˆ ˆ =0
 r - ˆi +2j



ˆ ˆ =11
 r. -9iˆ +8j-k
The Cartesian equation of the plane is
x -1
y-2
2
3
1
1
z+ 4
6 =0
-1
 -9x +8y -z =11
Plane Containing Two Lines
Let a plane contains the two lines
r = a+ b and r = a'+ b'
Let r be the position vector of any point P in the plane. Then, P lies in


the plane if AP, b and b' are coplanar i.e. AP. b×b' = 0.



 r - a . b×b' = 0
Cartesian Form
ˆ a = x ˆi + y ˆj+ z kˆ
ˆ zk,
r = xiˆ + yj+
1
1
1
b = b1ˆi +b2 ˆj+b3kˆ and b' = b'1ˆi +b'2 ˆj+b'3kˆ



 r - a . b×b' = 0
x - x1
y - y1
z - z1
 b1
b2
b3
b'1
b'2
b'3
=0
Example -6
Find the equation of the plane containing the lines
x- 4 y-3 z-2
x-3 y-2 z
=
=
and
=
= .
1
4
5
1
-4
5
Solution: The cartesian equation of the plane is given by
x-4
y-3
z-2
1
4
5
1
-4
5
=0
  x - 420+20 -  y -35-5 + z -2-4- 4 =0
 40x -160 - 8z +16 = 0
 5x - z -18 = 0
Plane Passing Through Three Points
Let A, B and C be the three points on the plane with position vectors
a, b and c respectively. The vector AB×AC is perpendicular to the plane
containing points A, B and C. The equation of the plane passing through
A and perpendicular to the vector AB  AC is
r - a. AB×AC = 0
  
 

 r - a . b - a × c - a  = 0


Cartesian Form
Let A   x1, y1, z1 
B   x2 , y2 , z2 
C   x3 , y3 , z3 
Let  x, y, z  be the coordinates of any point P with position vector r.
P lies on the plane if AP, AB and AC are coplanar, i.e.


x - x1
y - y1
z - z1
 x2 - x1
y2 - y1
z2 - z1 = 0
x3 - x1
y3 - y1
z3 - z1
AP. AB×AC = 0
Example –7
Find the equation of the plane through the three points
A 1, 1, 1 , B 1, -1, 1 and C -7, - 3, -5.
ˆ b = ˆi - ˆj+kˆ and c = -7iˆ - 3j-5k
ˆ
ˆ
Solution: Let a= ˆi + ˆj+k,
The vector equation of the plane is
r - a. AB×AC  = 0  r - a. b - a× c - a = 0

   

 

ˆ 6kˆ  = 0
 r - ˆi + ˆj+kˆ .  -2jˆ × -8iˆ - 4j
 


 r - ˆi + ˆj+kˆ . 12iˆ -16kˆ = 0


Solution Cont.


 r.-3iˆ +4kˆ  =1
 r. 12iˆ -16kˆ = -4
Cartesian form of the equation is
-3x + 4z =1
Passing Through the intersection of Two Planes
r . n1 = d1 and r . n2 = d2
r . n1 - d1  +  r . n2 - d2  = 0
Vector form
a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0
a1x+b1y+c1z+d1  +  a2x+b2y+c2z+d2  =0 Cartesian form
Example –8
Find the equation of the plane through the point 2iˆ + ˆj - kˆ
and passing through the line of intersection of the planes




r . ˆi + 3jˆ - kˆ = 0 and r . ˆj +2 kˆ = 0.
Solution: The equation of a plane through the intersection of




r . ˆi + 3jˆ - kˆ = 0 and r . ˆj +2 kˆ = 0 is




r . ˆi + 3jˆ - kˆ +  r . ˆj +2 kˆ  = 0


 r . ˆi + 3+   ˆj+ -1+2  kˆ  = 0


...i
Solution Cont.
ˆ then
If i passes through the point 2iˆ + ˆj - k,


 2iˆ + ˆj - kˆ . ˆi + 3+   ˆj+ -1+2  kˆ  = 0


 2+ 3+   - -1+2  =0
 6- =0   =6
Putting  = 6 in i ,we get
ˆ
r . ˆi +9j+11
kˆ  = 0


Example –9
Find the equation of the plane containing the line of intersection
of the plane x + y +z - 6 = 0 and 2x +3y + 4z+5 = 0 and passing
through the point 1, 1, 1.
Solution: The equation of the plane through the line of
intersection of the given planes is
 x+y+z - 6 +   2x+3y+4z+5 = 0
If (i) passes through (1, 1, 1), then
-3+14 = 0   =
3
14
...i
Solution Cont.
 x + y +z - 6  +
3
 2x +3y + 4z+5 = 0
14
 20x+23y+26z - 69 = 0
Angle Between Two Planes
If  be angle between the planes r . n1 = d1 and r . n2 = d2, then
n .n
cos  = 1 2
n1 n2
Perpendicular if n1 . n2 = 0
Parallel if n1 = n2
Cont.
If  be the angle between the planes a1x +b1y + c1z + d1 = 0
and a2x +b2y + c2z + d2 = 0, then
cos =
a1a2  b1b2  c1c 2
a12  a22  a22 b12  b22  b22
Perpendicular if a1a2 +b1b2 +c1c2 = 0
a
b
c
Parallel if 1 = 1 = 1
a2 b2 c2
Example -10
Find the angle between the planes




ˆ 2kˆ = 9.
r. 2iˆ - ˆj+2kˆ = 6 and r. 3iˆ +6j-


Solution: The angle between the planes r. 2iˆ - ˆj+2kˆ = 6


ˆ 2kˆ = 9 is
and r. 3iˆ +6j-
ˆ 2kˆ 
2iˆ - ˆj+2kˆ .3iˆ +6j
cos =
4+1+ 4 9+36 + 4
6-6-4
 cos =
3×7
 -4 
  = cos-1  
 21 


n
.
n
 cos = 1 2 

n1 n2 


Example -11
Find the angle between the planes 2x -3y+4z =1 and - x+y = 4.
Solution: If  is the angle between the planes. Then,
cos =
2× -1 + -3 ×1+ 4×0
4+9+16 1+1+0
 cos =
-5
58
-1 
  = cos
-5 


58



a1a2  b1b2  c1c 2
 cos =

2
2
2
2
2
2
a

a

a
b

b

b
1
2
2
1
2
2





Angle Between a Line and a Plane
Let the line be r = a+ b and the plane be r . n= d.
If  is the angle between the line and the plane then, 90°- 
is the angle between the line and the normal to the plane.
 cos  90° -   =
 sin =
b .n
bn
b .n
b n
n
b
90°– 

Example -12

 

ˆ and
ˆ ˆ +  2iˆ - 3j+2k
ˆ
Find the angle between the line r = 2iˆ +2j+k


ˆ = 4.
ˆ
the plane r . 3 ˆi - 2j+5k
ˆ and n= 3 ˆi - 2j+5k
ˆ
ˆ
ˆ
Solution: Here b =2iˆ - 3j+2k
If  is the angle between the line and the plane, then
ˆ . 3 ˆi - 2j+5k
ˆ
ˆ - 3j+2k
ˆ
ˆ
2i




b.n
6+6+10
sin =
=
=
b n
4+9+ 4 9+ 4+25


22
  = sin-1 

 17 38 
17 38
Intercept Form of a Plane
The equation of a plane intersecting lengths a, b and c with x-axis,
y–axis and z-axis respectively is
x y z
+ + =1
a b c
Example –13
Find the equation of a plane which meets the axes in A, B and C,
given that the centeriod of the triangle ABC is the point  , ,   .
Solution: Let the equation of the required plane be
x y z
+ + =1
a b c
The coordinates of A, B and C are A  a, 0, 0 , B 0, b, 0 and
C 0, 0, c  respectively.
a b c
 The centeriod of the triangle is  , , .
3 3 3
Solution Cont.
But the given centeriod is  , ,  .
a
b
c
   ,   ,    a  3, b  3, c  3
3
3
3
Substituting the values of a, b and c in (i), we get
x
y
z
x y z
+ + =1  + + = 3
3 3 3
  
Condition of Co-planarity of Two Lines
r = a1 + b
...i
r = a2 + b'
...ii
 
The line i passes through B  a2  and is parallel to b'.
The line i passes through A a1 and is parallel to b.
The given lines are coplanar if AB, b and b' are coplanar


i.e. AB. b×b' = 0.



 a2 - a1 . b×b' = 0
Cartesian Form
Let  x1, y1, z1  and  x2 , y2 , z2  be the coordinates of A and B
respectively. Then
AB =  x2 - x1  ˆi +  y2 - y1  ˆj+  z2 - z1 kˆ
b = a1ˆi +b1ˆj+c1kˆ and b' = a2ˆi +b2ˆj+c2kˆ


The given lines are coplanar if AB. b×b' = 0
x2 - x1
y2 - y1
z2 - z1
 a1
b1
c1
a2
b2
b2
=0
Example -14

 

ˆ +  2iˆ +3j+
ˆ
r = 2iˆ +6j+3k
  ˆ 4kˆ  are coplanar.
ˆ and
ˆ 3kˆ +  ˆi +2j+3k
ˆ
Show that the lines r = 2j-
ˆ a = 2iˆ +6j+3k,
ˆ b = ˆi +2j+3k
ˆ
ˆ 3k,
ˆ
ˆ
Solution: Here a1 = 2j2
ˆ
ˆ 4k.
and b' = 2iˆ +3j+



Given lines are coplanar if a2 - a1 . b×b' = 0.
ˆ
ˆ
a2 - a1  = 2iˆ + 4j+6k


 
ˆi ˆj
b×b' = 1 2
2 3

kˆ
ˆ ˆ
3 = -iˆ +2j-k
4

ˆ . -iˆ +2j-k
ˆ
ˆ ˆ =-2+8- 6 = 0
 a2 - a1 . b×b' = 2iˆ +4j+6k
Distance of a Point From a Plane
The equation of AB is r = a+ n.

A a
The line AB intersects the plane at B


 r . n = d  a+ n . n = d
 a . n+ n . n = d
 =

d- a . n
n

2


 d- a . n
 r = a+ 
2

n

 n



B
r .n= d
Solution Cont.


 d- a . n
AB = a+ 
2

n

 AB =

d- p . n
n
2



 n- a


n
a . n - d

 AB =
n
Note: The length of the perpendicular from origin to the plane
r . n= d is
d
n
 a = 0


Cartesian Form
A  x1, y1, z1 
AB =
ax1 +by1 + cz1 + d
a2 +b2 + c2
B
ax+by+cz+d= 0
Example –15
Find the distance of the point 3, 4, 5 from the plane


ˆ =13.
ˆ
r . 2 ˆi -5j+3k
ˆ n = 2 ˆi -5j+3k
ˆ and d=13
ˆ
ˆ
Solution: Here a= 3iˆ +4j+5k,
 The distance of the point from the given plane is
ˆ . 2 ˆi -5j+3k
ˆ
3iˆ + 4j+5k
  ˆ ˆ  -13
ˆ
ˆ
2 ˆi -5j+3k
=
=
6 - 20+15 -13
4+25+9
12
38
Example -16
If the product of distances of the point 1, 1, 1 from the origin and
the plane x - y+z+k = 0 be 5, then find the value of k.
Solution: The distance of the point A 1, 1, 1 from the
origin O  0, 0, 0 is 12 +12 +12 = 3
The distance of the point A 1, 1, 1 from the plane x - y+z+k =0 is
1-1+1+k
12 +12 +12
 3×
1+k
3
=
1+k
3
= 5  1+k = ±5  k = 4 or - 6
Thank you