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POLAR COORDINATES &
VECTORS
rahimahj@ump.edu.my
rahimahj@ump.edu.my
Suppose that a particle moves along a curve C in the
xy-plane in such a way that its x and y coordinates, as
functions of times are
The variable t is called the parameter for the
equations.
)(tfx  )(tgy 
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EXAMPLE 1
Solution:
Form the Cartesian equation by eliminate parameter t from the
following equations
tx 2 14 2
 ty
Given that , thus
Then
tx 2
2
x
t 
1
1
2
4
2
2








x
x
y
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EXAMPLE 2
Solution:
Find the graph of the parametric equations
ttx  2
12  ty
We plug in some values of t .
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EXAMPLE 13
Find the graph of the parametric equations
tx cos ty sin 20  t
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x
y
z
Graph the following ordered triples:
a. (3, 4, 5)
b. (2, -5, -7)
EXAMPLE 1
Distance Formula in
3
R
The distance between and
is
21PP ),,( 1111 zyxP
),,( 2222 zyxP
2
12
2
12
2
1221 )()()( zzyyxxPP 
Find the distance between (10, 20, 10) and
(-12, 6, 12).
EXAMPLE 2
Vectors in
3
R
 A vector in R3 is a directed line segment (“an arrow”)
in space.
 Given:
-initial point
-terminal point
Then the vector PQ has the unique standard
component form
),,( 111 zyxP
),,( 222 zyxQ
 121212 ,, zzyyxxPQ
Standard Representation
of Vectors in the Space
 The unit vector:
points in the directions of the positive x-axis
points in the directions of the positive y-axis
points in the directions of the positive z-axis
 i, j and k are called standard basis vector in R3.
 Any vector PQ can be expressed as a linear combination of i, j and
k (standard representation of PQ)
with magnitude
 0,0,1i
 0,1,0j
 1,0,0k
kjiPQ )()()( 121212 zzyyxx 
2
12
2
12
2
12 )()()( zzyyxx PQ
Find the standard representation of the vector PQ
with initial point P(-1, 2, 2) and terminal point
Q(3, -2, 4).
EXAMPLE 3
rahimahj@ump.edu.my
(1) Circular
Cylinder
922
 zx
three.radiusofcircle
aisgraphtheplane-On thexz
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(2) Ellipsoid
12
2
2
2
2
2

c
z
b
y
a
x
rahimahj@ump.edu.my
(3) Paraboloid
0c,2
2
2
2
 cz
b
y
a
x
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(5) Cone
02
2
2
2
2
2

c
z
b
y
a
x
rahimahj@ump.edu.my
12
2
2
2
2
2

c
z
b
y
a
x
(3) Hyperboloid of One Sheet
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(4) Hyperboloid of Two Sheets
12
2
2
2
2
2

c
z
b
y
a
x
rahimahj@ump.edu.my
0c,2
2
2
2
 cz
b
y
a
x
(7) Hyperbolic Paraboloid
rahimahj@ump.edu.my
22
222
22
2
)1(z(v)
1(iv)
16y(iii)
9z(ii)
1535(i)





yx
zyx
x
y
zy
EXAMPLE 29
Sketch the graph of the following equations in 3-dimensions.
Identify each of the surface.
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Parametric Form of a
Line in
3
R
If L is a line that contains the point and
is parallel to the vector , then L has
parametric form
Conversely, the set of all points that satisfy
such a set of equations is a line that passes
through the point and is parallel to a
vector with direction numbers .
),,( 000 zyx
kjiv cba 
ctzzbtyyatxx  000
),,( zyx
),,( 000 zyx
],,[ cba
Find parametric equations for the line that
contains the point and is parallel to the
vector .
Find where this line passes through the
coordinate planes.
 1, 1, 2 
3 2 5  v i j k
EXAMPLE 18
Solution:
 
0 0 0
The direction numbers are 3, 2, 5 and
1, 1 and z 2, so the 1ine has the
parametric form
1 3 1 2 2 5
x y
x t y t z t

    
       
 
2
5
2 11 9 11 9
, ,
5 5 5 5 5
*This 1ine wi11 intersect the -p1ane when 0;
0 2 5 imp1ies
If , then and . This is the point 0 .
*This 1ine wi11 intersect the -p1ane when 0;
0 1 2 imp1ies
xy z
t t
t x y
xz y
t t
 

   
  

  
 
 
1
2
1 1 9 1 9
2 2 2 2 2
1
3
1 1 11 1 11
3 3 3 3 3
If , then and z . This is the point ,0, .
*This 1ine wi11 intersect the -p1ane when 0;
0 1 3 imp1ies
If , then and z . This is the point 0, , .
t x
yz x
t t
t y

    

    

  

  
  
…continue solution:
Symmetric Form of a Line in
3
R
If L is a line that contains the point and
is parallel to the vector
(A, B, and C are nonzero numbers), then the point
is on L if and only if its coordinates satisfy
kjiv cba 
),,( 000 zyx
),,( zyx
c
zz
b
yy
a
xx 000 




Find symmetric equations for the line L
through the points
and .
Find the point of intersection with the
xy-plane.
 2,4, 3A 
 3, 1,1B 
EXAMPLE 19
 
 
 
 0 0 0
The required 1ine passes through or and
is para11e1 to the vector
3 2, 1 4,1 3 1, 5,4 @ 5 4
Thus, the direction numbers are 1, 5,4 .
Let say we choose as , , .
2 4 3
Then,
1 5 4
The sy
A B
A x y z
x y z
         

  
 


AB i j k
4 3
mmetric equation is 2
5 4
y z
x
 
  
Solution:
 11 1
, ,
4 4
This 1ine wi11 intersect the -p1ane when 0;
3 4 3
2 and
4 5 4
11 1
4 4
The point of intersection of the 1ine with the -p1ane is 0 .
xy z
y
x
x y
xy


  
 
…continue solution:
3
R
 
1. Find the parametric and symmetric equations for the
point 1,0, 1 which is para11e1 to 3 4 .
2. Find the points of intersection of the 1ine
4 3
2 with each of the coordinate p1anes
4 3
x y
z
 
 
  
i j
.
3. Find two unit vectors para11e1 to the 1ine
1 2
5
2 4
x y
z
 
  
Line may Intersect, Parallel or Skew…
 Recall two lines in R2 must intersect if their slopes are
different (cannot be parallel)
 However, two lines in R3 may have different direction
number and still not intersect. In this case, the lines are
said to be skew.
In problems below, tell whether the two lines are
intersect, parallel, or skew . If they intersect, give the
point of intersection.
3 3 , 1 4 , 4 7 ;
2 3 , 5 4 , 3 7
x t y t z t
x t y t z t
      
     
1
2 4 , 1 , 5 ;
2
3 , 2 , 4 2
x t y t z t
x t y t z t
     
     
3 1 4 2 3 2
;
2 1 1 3 1 1
x y z x y z     
   
 
EXAMPLE 20
)(a
)(b
)(c
 
 
 
1 2
1
2
3 1 4 2 5 3
1. Let : and :
3 4 7 3 4 7
has direction numbers 3, 4, 7
and has direction numbers 3, 4, 7 .
Since both 1ines have same direction numbers
(or 3, 4, 7 = 3
x y z x y z
L L
L
L
t
     
   
   
 
 
   
   
 
1 2
, 4, 7 , where 1),
therefore they are para11e1 or coincide.
Obvious1y, has point 3,1, 4 and has point 2,5,3 .
4 7 , with the direction numbers 1,4,7 .
Because there is no ' ' for w
t
L A L B
a
  

    AB i j k
   hich 1,4,7 3, 4, 7 ,
the 1ines are not coincide, but just para11e1.
a   
Solution:
 
 
   
1
2
1 2
1
2
2 1 2 4
2. Let : and :
4 1 5 3 1 2
has direction numbers 4,1,5
and has direction numbers 3, 1, 2 .
Since there is no for which 4,1,5 3, 1, 2 ,
the 1ines are not pa
zx y x y z
L L
L
L
t t
   
   
  

 
   
1
1 1 1 12
2 2 2 2
ra11e1 or coincide, maybe skew or intersect.
Express the 1ines in parametric form
: 2 4 , 1 , 5 ;
: 3 , 2 , 4 2
L x t y t z t
L x t y t z t
     
     
Solution:
1 2 1 2
1 2 1 2
1 7
1 2 1 22 2
1 2
Continue : 2
At an intersection point we must have
2 4 3 4 3 2
1 2 3
5 4 2 5 2
So1ving the first two equations simu1taneous1y,
11 and 14 and since the so1ution is
t t t t
t t t t
t t t t
t t
    
       
     
   not
satisfy the third equation, so the 1ines are skew.
…continue solution:
 
 
   
1 2
1
2
3 1 4 2 3 2
3. Let : and :
2 1 1 3 1 1
has direction numbers 2, 1,1
and has direction numbers 3, 1,1 .
Since there is no for which 2, 1,1 3, 1,1 ,
the 1ines are not para
x y z x y z
L L
L
L
t t
     
   
 


  
1 1 1 1
2 2 2 2
11e1 or coincide, maybe skew or intersect.
Express the 1ines in parametric form
: 3 2 , 1 , 4 ;
: 2 3 , 3 , 2
L x t y t z t
L x t y t z t
     
      
Solution:
1 2
1 2 1 2
1 2
Continue : 3
At an intersection point we must have
3 2 2 3
1 3 1 and 1
4 2
Satisfy a11 of the equation,
then these two 1ines are intersect to each other.
The point of intersectio
t t
t t t t
t t
    

     
   
 
 
 
 
1
1 2 2 2
1
n is
3 2 3 2 1 1
1 1 1 2 or 2 3 , 3 , 2
4 4 1 3
1,2,3
x t
y t x t y t z t
z t
      

            

      

…continue solution:
CLASS ACTIVITY 2 :
In problems below, tell whether the two lines are
intersect, parallel, or skew. If they intersect, give
the point of intersection.
1.
2.
3.
6 , 1 9 , 3 ;
1 2 , 4 3 ,
x t y t z t
x t y t z t
     
    
1 2 , 3 , 2 ;
1 , 4 , 1 3
x t y t z t
x t y t z t
    
      
1 2 3 2 1
;
2 3 4 3 2
y z x y z
x
    
   
 
REMEMBERTHAT…
Theorem:The orthogonal vector theorem
Nonzero vectors v and n are orthogonal
(or perpendicular) if and only if
where n is called the normal vector.
0nv
 
 
     
0 0 0
0 0
Let say, we have a p1ane containing point , , and
is orthogona1 (norma1) to the vector
So1ution:
If we have another any point , , in the p1ane, then
0
Q x y z
A B C
P x y z
Ai Bj Ck x x y y z
  

       
N i j k
N.QP
N.QP . i j   
     
     
 
0
0 0 0
0 0 0
0 0 0 0 0 0
0 @
0, as ,
Then 0
z
A x x B y y C z z
A x x B y y C z z
Ax By Cz Ax By Cz D Ax By Cz
Ax By Cz D
     
     
          
   
k
An equation for the plane with normal
that contains the point has the following forms:
Point-normal form:
Standard form:
Conversely, a normal vector to the plane
is
A B C  N i j k
 0 0 0, ,x y z
     0 0 0 0A x x B y y C z z     
0Ax By Cz D   
0Ax By Cz D   
A B C  N i j k
Find an equation for the plane that contains
the point P and has the normal vector N
given in:
1.
2.
 1,3,5 ; 2 4 3P    N i j k
 1,1, 1 ; 2 3P     N i j k
EXAMPLE 21
Point-Normal form
Standard form
 1,3,5 ; 2 4 3P    N i j k
     2 1 4 3 3 5 0x y z     
     2 1 4 3 3 5 0
2 2 4 12 3 15 0
2 4 3 5 0
x y z
x y z
x y z
     
     
   
1.
Solution :
REMEMBERTHAT..
Theorem: Orthogonality Property ofThe
Cross Product
If v and w are nonzero vectors in that are not
multiples of one another, then v x w is
orthogonal to both v and w
3
R
wvn 
Find the standard form equation of a
plane containing
and
   1,2,1 , 0, 3,2 ,P Q 
 1,1, 4R 
EXAMPLE 22
 0 0 0
Hint :
What we need?
?
Point , , ?x y z
N  N PQ PR
Since, a11 point , and
are points in the p1ane,
so just pick one of them !!
P Q R
 
0 0 0
0 0 0
Hint :
Equation for 1ine; , , ,
so, obvious1y, you just have to find
the va1ue of , and .
and , ,
x x At y y Bt z z Ct
A B C
x y z
     
EXAMPLE 23
Find an equation of the line that passes through the point
Q(2,-1,3) and is orthogonal to the plane 3x-7y+5z+55=0
N = Ai + Bj + Ck
(2, -1, 3)
 
 
1. Find an equation for the p1ane that contains the
point 2,1, 1 and is orthogona1 to the 1ine
3 1
.
3 5 2
2. Find a p1ane that passes through the point 1,2, 1
and is para11e1 to the p1ane 2 3 1.
3. Sh
x y z
x y z

 
 

  
1 1 2
ow that the 1ine
2 3 4
is para11e1 to the p1ane 2 6.
x y z
x y z
  
 
  
 Find the equation of a 1ine passing through 1,2,3
that is para11e1 to the 1ine of intersection of the p1anes
3 2 4 and 2 3 5.x y z x y z

     
Equation of a Line Parallel toThe
Intersection ofTwo Given Planes
EXAMPLE 24
Find the standard-form equation of the p1ane
determined by the intersecting 1ines.
2 5 1 1 16
and
3 2 4 2 1 5
x y z x y z    
   
 
Equation of a Plane Containing
Two Intersecting Lines
EXAMPLE 25
Find the point at which the 1ine with parametric
equations 2 3 , 4 , 5 intersects the
p1ane 4 5 2 18
x t y t z t
x y z
     
  
Point where a Line intersects with
a Plane.
EXAMPLE 26
INTERSECTING PLANE
The acute angle
between the planes :
21
21
cos
nn
nn 

EXAMPLE 27
Find the acute angle of intersection between the
planes 4326and6442  zyxzyx
DISTANCE PROBLEMS INVOLVING
PLANES
The distance D between a point and the
plane is
 0000 ,, zyxP
0 dczbyax
222
000
cba
dczbyax
D



EXAMPLE 28
Find the distance D between the point (1,-4,-3) and the plane
1632  zyx
A polar coordinate system consists of :
A fix point O, called the pole or origin
Polar coordinates where
r : distance from P to the origin
: angle from the polar axis to the ray OP
),( r

),( rP
Polar axisOrigin
O 
rahimahj@ump.edu.my 596
I
sin
cos
tan
II
sin
tan
III
cos
IV
THETRIGONOMETRIC RATIOS
rahimahj@ump.edu.my 607
For any angle θ;
A
CT
S
θ
θ
+ve
-ve
 
 
  


tantan
coscos
sinsin



THETRIGONOMETRIC RATIOS
rahimahj@ump.edu.my 618
Trigonometrical ratios of some special angles;
A
1
2
O
B
30°
60°
3
B
A
1
1
O
45°
45°
2
THETRIGONOMETRIC RATIOS
1/ 2 1/ 2 3 / 2
3 / 2 1/ 2 1/ 2
1/ 3 3
θ 30° 45° 60°
sin θ 0 1
cos θ 1 0
tan θ 0 1 undefined
0 90
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EXAMPLE 4
Plot the points with the following polar coordinates
 0
225,3)(a 






3
,2)(

b
Solution:
(a)
0
225
 0
225,3P
x
O
)(b







3
,2

P
xO
3


Relationship between Polar and
Rectangular Coordinates
O
x
y P
sinry 
cosrx 
r

cosrx  sinry 
x
y
yxr  tan222
Change the polar coordinates to Cartesian coordinates.





3
,2

 3,1isscoordinateCartesianThe
3
3
sin2sin
1
3
cos2cos
then,
3
and2Since










ry
rx
r
EXAMPLE 5
Solution:
rahimahj@ump.edu.my
EXAMPLE 6
Find the rectangular coordinates of the point P whose
polar coordinates are   






3
2
,4,

r
Solution:
2
2
1
4
3
2
cos4 





 x
32
2
3
4
3
2
sin4 







 y
Thus, the rectangular coordinates of P are    32,2, yx
Change the coordinates Cartesian to polar coordinates. 1,1 
   
















4
7
,2and
4
,2arescoordinatepolarpossibleThe
4
7
or
4
,1tan
211
thenpositive,betochooseweIf
2222




x
y
yxr
r
 1,1 
x
y
EXAMPLE 7
rahimahj@ump.edu.my
EXAMPLE 8
Find polar coordinates of the point P whose rectangular
coordinates are  4,3 
Solution:
   
5or5Thus
2543
22222


rr
yxr
Then,
3
4
3
4
tan 



x
y

Therefore, 000
13.23313.53180 
Symmetry Tests
SYMMETRIC CONDITIONS
about the x axis
about the y axis
about the origin
 ,r  ,r
   ,r
 ,r   ,r
  ,r
 ,r   ,r
 ,r


),( r
),( r

 
),( r
),(  r
 0
Symmetry with respect to x axis
Symmetry with respect to y axis

),( r
),( r
Symmetry with respect to the
origin
(c)Given that . Determine the symmetry of the
polar equation and then sketch the graph.
sin33r
(d) Test and sketch the curve for symmetry.2sinr
(a) What curve represented by the polar equation 5r
(b) Given that . Determine the symmetry of the
polar equation and then sketch the graph.
cos2r
EXAMPLE 9
2cosr
(a) Find the area enclosed by one loop of four petals 2cosr
(b) Find the area of the region that lies inside the circle
and outside the cardioid
sin3r
sin1r
drA
b
a
 2
2
1
:
8
Answer

:Answer 
EXAMPLE 10
rahimahj@ump.edu.my
TANGENT LINES TO PARAMETRIC CURVES
dt
dx
dt
dy
dx
dy

If 0and0 
dt
dx
dt
dy Horizontal
If 0and0 
dt
dx
dt
dy Infinite slope
Vertical
If 0and0 
dt
dx
dt
dy Singular points
rahimahj@ump.edu.my
EXAMPLE 14
(a) Find the slope of the tangent line to the unit circle
at the point where
(b) In a disastrous first flight, an experimental paper airplane
follows the trajectory of the particle as
but crashes into a wall at time t = 10.
i) At what times was the airplane flying horizontally?
ii) At what times was it flying vertically?
tx cos ty sin
3

t
ttx sin3 ty cos34
ARC LENGTH OF PARAMETRIC CURVES
 












b
a
dt
dt
dy
dt
dx
L
22
EXAMPLE 15
Find the exact arc length of the curve over the stated interval
2
tx 
3
3
1
ty  10  t)(a
tx 3cos ty 3sin  t0)(b
rahimahj@ump.edu.my
Consider the parametric equations,
a) Sketch the graph.
b) By eliminating t, find the Cartesian equation.
12
3
9, for 3 2x t y t t     
2
R
EXAMPLE 16
12
3
9, for 3 2x t y t t     
2
9 9x y 
)(a
Solution:
)(b
 
 
Sketch the graph of 2 4 , 1 5 ,3
So1ution:
In this form we can see that 2 4 , 1 5 , 3
Notice that this is nothing more than a 1ine, with
a point 2, 1,3 and a vector para11e1 is 4,5,1 .
F t t t t
x t y t z t
    
      
  v
Graph in 3
REXAMPLE 17
rahimahj@ump.edu.my
sCoordinatePolarlCylindrica
cosrx 
sinry 
22
yxr 






 
x
y1
tan
 r0
 20 
 z
rahimahj@ump.edu.my
sCoordinatePolarSpherical
rahimahj@ump.edu.my
Figure 11.8.3 (p. 833)
Figure 11.8.4 (p. 833)
Table 11.8.1 (p. 833)
Table 11.8.2 (p. 835)
rahimahj@ump.edu.my
EXAMPLE 30
(a) Convert from rectangular to cylindrical coordinates
(i) (-5,5,6) (ii) (0,2,0)
(b) Convert from cylindrical to rectangular coordinates
(c) Convert from spherical to rectangular coordinates
(d) Convert from spherical to cylindrical coordinates
 9,7(ii)3,
6
,4)( ,πi 




 












4
,
6
5(ii)
2
,0,7)(
 π
,i
  





3
2
,
4
5(ii)0,0,3)(
π
,i
rahimahj@ump.edu.my
Success = 90% Perspiration + 10% Inspiration

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Chapter1polarcoordinatesandvector 150105021140-conversion-gate02

  • 2. rahimahj@ump.edu.my Suppose that a particle moves along a curve C in the xy-plane in such a way that its x and y coordinates, as functions of times are The variable t is called the parameter for the equations. )(tfx  )(tgy 
  • 3. rahimahj@ump.edu.my EXAMPLE 1 Solution: Form the Cartesian equation by eliminate parameter t from the following equations tx 2 14 2  ty Given that , thus Then tx 2 2 x t  1 1 2 4 2 2         x x y
  • 4. rahimahj@ump.edu.my EXAMPLE 2 Solution: Find the graph of the parametric equations ttx  2 12  ty We plug in some values of t .
  • 5.
  • 6. rahimahj@ump.edu.my EXAMPLE 13 Find the graph of the parametric equations tx cos ty sin 20  t
  • 8. Graph the following ordered triples: a. (3, 4, 5) b. (2, -5, -7) EXAMPLE 1
  • 9. Distance Formula in 3 R The distance between and is 21PP ),,( 1111 zyxP ),,( 2222 zyxP 2 12 2 12 2 1221 )()()( zzyyxxPP  Find the distance between (10, 20, 10) and (-12, 6, 12). EXAMPLE 2
  • 10. Vectors in 3 R  A vector in R3 is a directed line segment (“an arrow”) in space.  Given: -initial point -terminal point Then the vector PQ has the unique standard component form ),,( 111 zyxP ),,( 222 zyxQ  121212 ,, zzyyxxPQ
  • 11. Standard Representation of Vectors in the Space  The unit vector: points in the directions of the positive x-axis points in the directions of the positive y-axis points in the directions of the positive z-axis  i, j and k are called standard basis vector in R3.  Any vector PQ can be expressed as a linear combination of i, j and k (standard representation of PQ) with magnitude  0,0,1i  0,1,0j  1,0,0k kjiPQ )()()( 121212 zzyyxx  2 12 2 12 2 12 )()()( zzyyxx PQ
  • 12.
  • 13. Find the standard representation of the vector PQ with initial point P(-1, 2, 2) and terminal point Q(3, -2, 4). EXAMPLE 3
  • 19. rahimahj@ump.edu.my (4) Hyperboloid of Two Sheets 12 2 2 2 2 2  c z b y a x
  • 22. rahimahj@ump.edu.my Parametric Form of a Line in 3 R If L is a line that contains the point and is parallel to the vector , then L has parametric form Conversely, the set of all points that satisfy such a set of equations is a line that passes through the point and is parallel to a vector with direction numbers . ),,( 000 zyx kjiv cba  ctzzbtyyatxx  000 ),,( zyx ),,( 000 zyx ],,[ cba
  • 23. Find parametric equations for the line that contains the point and is parallel to the vector . Find where this line passes through the coordinate planes.  1, 1, 2  3 2 5  v i j k EXAMPLE 18 Solution:   0 0 0 The direction numbers are 3, 2, 5 and 1, 1 and z 2, so the 1ine has the parametric form 1 3 1 2 2 5 x y x t y t z t              
  • 24.   2 5 2 11 9 11 9 , , 5 5 5 5 5 *This 1ine wi11 intersect the -p1ane when 0; 0 2 5 imp1ies If , then and . This is the point 0 . *This 1ine wi11 intersect the -p1ane when 0; 0 1 2 imp1ies xy z t t t x y xz y t t                   1 2 1 1 9 1 9 2 2 2 2 2 1 3 1 1 11 1 11 3 3 3 3 3 If , then and z . This is the point ,0, . *This 1ine wi11 intersect the -p1ane when 0; 0 1 3 imp1ies If , then and z . This is the point 0, , . t x yz x t t t y                        …continue solution:
  • 25.
  • 26.
  • 27. Symmetric Form of a Line in 3 R If L is a line that contains the point and is parallel to the vector (A, B, and C are nonzero numbers), then the point is on L if and only if its coordinates satisfy kjiv cba  ),,( 000 zyx ),,( zyx c zz b yy a xx 000     
  • 28. Find symmetric equations for the line L through the points and . Find the point of intersection with the xy-plane.  2,4, 3A   3, 1,1B  EXAMPLE 19
  • 29.        0 0 0 The required 1ine passes through or and is para11e1 to the vector 3 2, 1 4,1 3 1, 5,4 @ 5 4 Thus, the direction numbers are 1, 5,4 . Let say we choose as , , . 2 4 3 Then, 1 5 4 The sy A B A x y z x y z                   AB i j k 4 3 mmetric equation is 2 5 4 y z x      Solution:
  • 30.  11 1 , , 4 4 This 1ine wi11 intersect the -p1ane when 0; 3 4 3 2 and 4 5 4 11 1 4 4 The point of intersection of the 1ine with the -p1ane is 0 . xy z y x x y xy        …continue solution:
  • 31. 3 R   1. Find the parametric and symmetric equations for the point 1,0, 1 which is para11e1 to 3 4 . 2. Find the points of intersection of the 1ine 4 3 2 with each of the coordinate p1anes 4 3 x y z        i j . 3. Find two unit vectors para11e1 to the 1ine 1 2 5 2 4 x y z     
  • 32. Line may Intersect, Parallel or Skew…  Recall two lines in R2 must intersect if their slopes are different (cannot be parallel)  However, two lines in R3 may have different direction number and still not intersect. In this case, the lines are said to be skew.
  • 33. In problems below, tell whether the two lines are intersect, parallel, or skew . If they intersect, give the point of intersection. 3 3 , 1 4 , 4 7 ; 2 3 , 5 4 , 3 7 x t y t z t x t y t z t              1 2 4 , 1 , 5 ; 2 3 , 2 , 4 2 x t y t z t x t y t z t             3 1 4 2 3 2 ; 2 1 1 3 1 1 x y z x y z            EXAMPLE 20 )(a )(b )(c
  • 34.       1 2 1 2 3 1 4 2 5 3 1. Let : and : 3 4 7 3 4 7 has direction numbers 3, 4, 7 and has direction numbers 3, 4, 7 . Since both 1ines have same direction numbers (or 3, 4, 7 = 3 x y z x y z L L L L t                             1 2 , 4, 7 , where 1), therefore they are para11e1 or coincide. Obvious1y, has point 3,1, 4 and has point 2,5,3 . 4 7 , with the direction numbers 1,4,7 . Because there is no ' ' for w t L A L B a         AB i j k    hich 1,4,7 3, 4, 7 , the 1ines are not coincide, but just para11e1. a    Solution:
  • 35.         1 2 1 2 1 2 2 1 2 4 2. Let : and : 4 1 5 3 1 2 has direction numbers 4,1,5 and has direction numbers 3, 1, 2 . Since there is no for which 4,1,5 3, 1, 2 , the 1ines are not pa zx y x y z L L L L t t                   1 1 1 1 12 2 2 2 2 ra11e1 or coincide, maybe skew or intersect. Express the 1ines in parametric form : 2 4 , 1 , 5 ; : 3 , 2 , 4 2 L x t y t z t L x t y t z t             Solution:
  • 36. 1 2 1 2 1 2 1 2 1 7 1 2 1 22 2 1 2 Continue : 2 At an intersection point we must have 2 4 3 4 3 2 1 2 3 5 4 2 5 2 So1ving the first two equations simu1taneous1y, 11 and 14 and since the so1ution is t t t t t t t t t t t t t t                       not satisfy the third equation, so the 1ines are skew. …continue solution:
  • 37.         1 2 1 2 3 1 4 2 3 2 3. Let : and : 2 1 1 3 1 1 has direction numbers 2, 1,1 and has direction numbers 3, 1,1 . Since there is no for which 2, 1,1 3, 1,1 , the 1ines are not para x y z x y z L L L L t t                  1 1 1 1 2 2 2 2 11e1 or coincide, maybe skew or intersect. Express the 1ines in parametric form : 3 2 , 1 , 4 ; : 2 3 , 3 , 2 L x t y t z t L x t y t z t              Solution:
  • 38. 1 2 1 2 1 2 1 2 Continue : 3 At an intersection point we must have 3 2 2 3 1 3 1 and 1 4 2 Satisfy a11 of the equation, then these two 1ines are intersect to each other. The point of intersectio t t t t t t t t                         1 1 2 2 2 1 n is 3 2 3 2 1 1 1 1 1 2 or 2 3 , 3 , 2 4 4 1 3 1,2,3 x t y t x t y t z t z t                               …continue solution:
  • 39. CLASS ACTIVITY 2 : In problems below, tell whether the two lines are intersect, parallel, or skew. If they intersect, give the point of intersection. 1. 2. 3. 6 , 1 9 , 3 ; 1 2 , 4 3 , x t y t z t x t y t z t            1 2 , 3 , 2 ; 1 , 4 , 1 3 x t y t z t x t y t z t             1 2 3 2 1 ; 2 3 4 3 2 y z x y z x           
  • 40. REMEMBERTHAT… Theorem:The orthogonal vector theorem Nonzero vectors v and n are orthogonal (or perpendicular) if and only if where n is called the normal vector. 0nv
  • 41.           0 0 0 0 0 Let say, we have a p1ane containing point , , and is orthogona1 (norma1) to the vector So1ution: If we have another any point , , in the p1ane, then 0 Q x y z A B C P x y z Ai Bj Ck x x y y z             N i j k N.QP N.QP . i j                  0 0 0 0 0 0 0 0 0 0 0 0 0 0 @ 0, as , Then 0 z A x x B y y C z z A x x B y y C z z Ax By Cz Ax By Cz D Ax By Cz Ax By Cz D                            k
  • 42. An equation for the plane with normal that contains the point has the following forms: Point-normal form: Standard form: Conversely, a normal vector to the plane is A B C  N i j k  0 0 0, ,x y z      0 0 0 0A x x B y y C z z      0Ax By Cz D    0Ax By Cz D    A B C  N i j k
  • 43. Find an equation for the plane that contains the point P and has the normal vector N given in: 1. 2.  1,3,5 ; 2 4 3P    N i j k  1,1, 1 ; 2 3P     N i j k EXAMPLE 21
  • 44. Point-Normal form Standard form  1,3,5 ; 2 4 3P    N i j k      2 1 4 3 3 5 0x y z           2 1 4 3 3 5 0 2 2 4 12 3 15 0 2 4 3 5 0 x y z x y z x y z                 1. Solution :
  • 45. REMEMBERTHAT.. Theorem: Orthogonality Property ofThe Cross Product If v and w are nonzero vectors in that are not multiples of one another, then v x w is orthogonal to both v and w 3 R wvn 
  • 46. Find the standard form equation of a plane containing and    1,2,1 , 0, 3,2 ,P Q   1,1, 4R  EXAMPLE 22
  • 47.  0 0 0 Hint : What we need? ? Point , , ?x y z N  N PQ PR Since, a11 point , and are points in the p1ane, so just pick one of them !! P Q R
  • 48.   0 0 0 0 0 0 Hint : Equation for 1ine; , , , so, obvious1y, you just have to find the va1ue of , and . and , , x x At y y Bt z z Ct A B C x y z       EXAMPLE 23 Find an equation of the line that passes through the point Q(2,-1,3) and is orthogonal to the plane 3x-7y+5z+55=0 N = Ai + Bj + Ck (2, -1, 3)
  • 49.
  • 50.     1. Find an equation for the p1ane that contains the point 2,1, 1 and is orthogona1 to the 1ine 3 1 . 3 5 2 2. Find a p1ane that passes through the point 1,2, 1 and is para11e1 to the p1ane 2 3 1. 3. Sh x y z x y z          1 1 2 ow that the 1ine 2 3 4 is para11e1 to the p1ane 2 6. x y z x y z        
  • 51.  Find the equation of a 1ine passing through 1,2,3 that is para11e1 to the 1ine of intersection of the p1anes 3 2 4 and 2 3 5.x y z x y z        Equation of a Line Parallel toThe Intersection ofTwo Given Planes EXAMPLE 24
  • 52. Find the standard-form equation of the p1ane determined by the intersecting 1ines. 2 5 1 1 16 and 3 2 4 2 1 5 x y z x y z           Equation of a Plane Containing Two Intersecting Lines EXAMPLE 25
  • 53. Find the point at which the 1ine with parametric equations 2 3 , 4 , 5 intersects the p1ane 4 5 2 18 x t y t z t x y z          Point where a Line intersects with a Plane. EXAMPLE 26
  • 54. INTERSECTING PLANE The acute angle between the planes : 21 21 cos nn nn   EXAMPLE 27 Find the acute angle of intersection between the planes 4326and6442  zyxzyx
  • 55. DISTANCE PROBLEMS INVOLVING PLANES The distance D between a point and the plane is  0000 ,, zyxP 0 dczbyax 222 000 cba dczbyax D   
  • 56.
  • 57. EXAMPLE 28 Find the distance D between the point (1,-4,-3) and the plane 1632  zyx
  • 58. A polar coordinate system consists of : A fix point O, called the pole or origin Polar coordinates where r : distance from P to the origin : angle from the polar axis to the ray OP ),( r  ),( rP Polar axisOrigin O 
  • 60. rahimahj@ump.edu.my 607 For any angle θ; A CT S θ θ +ve -ve          tantan coscos sinsin    THETRIGONOMETRIC RATIOS
  • 61. rahimahj@ump.edu.my 618 Trigonometrical ratios of some special angles; A 1 2 O B 30° 60° 3 B A 1 1 O 45° 45° 2 THETRIGONOMETRIC RATIOS 1/ 2 1/ 2 3 / 2 3 / 2 1/ 2 1/ 2 1/ 3 3 θ 30° 45° 60° sin θ 0 1 cos θ 1 0 tan θ 0 1 undefined 0 90
  • 62. rahimahj@ump.edu.my EXAMPLE 4 Plot the points with the following polar coordinates  0 225,3)(a        3 ,2)(  b Solution: (a) 0 225  0 225,3P x O )(b        3 ,2  P xO 3  
  • 63. Relationship between Polar and Rectangular Coordinates O x y P sinry  cosrx  r  cosrx  sinry  x y yxr  tan222
  • 64. Change the polar coordinates to Cartesian coordinates.      3 ,2   3,1isscoordinateCartesianThe 3 3 sin2sin 1 3 cos2cos then, 3 and2Since           ry rx r EXAMPLE 5 Solution:
  • 65. rahimahj@ump.edu.my EXAMPLE 6 Find the rectangular coordinates of the point P whose polar coordinates are          3 2 ,4,  r Solution: 2 2 1 4 3 2 cos4        x 32 2 3 4 3 2 sin4          y Thus, the rectangular coordinates of P are    32,2, yx
  • 66. Change the coordinates Cartesian to polar coordinates. 1,1                      4 7 ,2and 4 ,2arescoordinatepolarpossibleThe 4 7 or 4 ,1tan 211 thenpositive,betochooseweIf 2222     x y yxr r  1,1  x y EXAMPLE 7
  • 67. rahimahj@ump.edu.my EXAMPLE 8 Find polar coordinates of the point P whose rectangular coordinates are  4,3  Solution:     5or5Thus 2543 22222   rr yxr Then, 3 4 3 4 tan     x y  Therefore, 000 13.23313.53180 
  • 68. Symmetry Tests SYMMETRIC CONDITIONS about the x axis about the y axis about the origin  ,r  ,r    ,r  ,r   ,r   ,r  ,r   ,r  ,r
  • 69.   ),( r ),( r    ),( r ),(  r  0 Symmetry with respect to x axis Symmetry with respect to y axis
  • 70.  ),( r ),( r Symmetry with respect to the origin
  • 71. (c)Given that . Determine the symmetry of the polar equation and then sketch the graph. sin33r (d) Test and sketch the curve for symmetry.2sinr (a) What curve represented by the polar equation 5r (b) Given that . Determine the symmetry of the polar equation and then sketch the graph. cos2r EXAMPLE 9
  • 73. (a) Find the area enclosed by one loop of four petals 2cosr (b) Find the area of the region that lies inside the circle and outside the cardioid sin3r sin1r drA b a  2 2 1 : 8 Answer  :Answer  EXAMPLE 10
  • 74. rahimahj@ump.edu.my TANGENT LINES TO PARAMETRIC CURVES dt dx dt dy dx dy  If 0and0  dt dx dt dy Horizontal If 0and0  dt dx dt dy Infinite slope Vertical If 0and0  dt dx dt dy Singular points
  • 75. rahimahj@ump.edu.my EXAMPLE 14 (a) Find the slope of the tangent line to the unit circle at the point where (b) In a disastrous first flight, an experimental paper airplane follows the trajectory of the particle as but crashes into a wall at time t = 10. i) At what times was the airplane flying horizontally? ii) At what times was it flying vertically? tx cos ty sin 3  t ttx sin3 ty cos34
  • 76. ARC LENGTH OF PARAMETRIC CURVES               b a dt dt dy dt dx L 22 EXAMPLE 15 Find the exact arc length of the curve over the stated interval 2 tx  3 3 1 ty  10  t)(a tx 3cos ty 3sin  t0)(b
  • 77. rahimahj@ump.edu.my Consider the parametric equations, a) Sketch the graph. b) By eliminating t, find the Cartesian equation. 12 3 9, for 3 2x t y t t      2 R EXAMPLE 16
  • 78. 12 3 9, for 3 2x t y t t      2 9 9x y  )(a Solution: )(b
  • 79.     Sketch the graph of 2 4 , 1 5 ,3 So1ution: In this form we can see that 2 4 , 1 5 , 3 Notice that this is nothing more than a 1ine, with a point 2, 1,3 and a vector para11e1 is 4,5,1 . F t t t t x t y t z t               v Graph in 3 REXAMPLE 17
  • 80.
  • 81. rahimahj@ump.edu.my sCoordinatePolarlCylindrica cosrx  sinry  22 yxr          x y1 tan  r0  20   z
  • 88. rahimahj@ump.edu.my EXAMPLE 30 (a) Convert from rectangular to cylindrical coordinates (i) (-5,5,6) (ii) (0,2,0) (b) Convert from cylindrical to rectangular coordinates (c) Convert from spherical to rectangular coordinates (d) Convert from spherical to cylindrical coordinates  9,7(ii)3, 6 ,4)( ,πi                    4 , 6 5(ii) 2 ,0,7)(  π ,i         3 2 , 4 5(ii)0,0,3)( π ,i
  • 89. rahimahj@ump.edu.my Success = 90% Perspiration + 10% Inspiration