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Page 1: Math 54 Reviewer 3rd LE

7/17/2019 Math 54 Reviewer 3rd LE

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

Cartesian coordinate system in three dimensionsPoint – represented by an ordered triple of realnumbers (x, y, z)

Coordinate axes – three perpendicular linespassing thru the Origin, O (x, y, and z axes)Octant – division by the three coordinate planes(xz, yz and xy plane) into eight parts (rst octant

shon belo)!idpoint formula – to nd the midpointconnecting the line segment beteen to points

 M =( x1+ x2

2,

 y1+ y2

2, z1+ z2

2)

"phere – set of all points hose distance frompoint C is r

r2=( x−h )2+ ( y−k )2+( z−l )2

#here r is the radius  (h, $, l) is the coordinates of the centre C

%ectors – &uantity that has both magnitude anddirection, geometrically represented by an arro,and denoted by a letter ith an arro above it' he vector from the origin to the point (a, b,

c) is represented by the vector   ⟨ a ,b ,c ⟩

' *n v=⟨a , b , c ⟩ a, b and c are the

components of v

' + vector v   can be represented by any

directed   line segment that has same

magnitude and direction as v

'   v=⟨a , b , c ⟩∧u= ⟨d , e, f  ⟩   +re e&ual if and

only if a=d ,b=e ,∧c=f 

et v  be the vector from point P (a, b, c) to

- (d, e, f) hen v=⟨d−a , e−b , f −c ⟩   and

ritten as vector  PQ .

' *f third component of vector is zero, e thin$of it as a vector in the xy'plane

.orm of a vector – the magnitude of a vecto

v=⟨a , b , c ⟩

‖v‖=√ a2+b2+c

2

/nit vector – a vector of length or norm 0 1

2ero vector – v=0= ⟨0,0,0 ⟩

3irectional angles – +ngles that a vecto

v=⟨a , b , c ⟩ ma$es ith the positive x, y, and z

axes (α , β ,γ  )

cosα =  a

‖v‖

cosβ=  b

‖v‖

cosγ =  c

‖v‖

Page 2: Math 54 Reviewer 3rd LE

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

%ector operations

 A ± B= ⟨ a1

± b1

, a2

±b2

, a3

± b3 ⟩

c A=⟨ c a1

, c a2

, c a3 ⟩

‖c  A‖=|c|‖ A‖

' *f c>0 , then c A  is in the same direction

as  A

' *f c<0 , then c A   is in the opposite

direction as  A

' *f c=0 , then c A=0

"tandard basic vectors

i=⟨1,0,0 ⟩

 j= ⟨0,1,0 ⟩

k =⟨0,0,1 ⟩

.ormalizing a vector – turning a vector into a unitvector

uni vecor ~v=  v

‖v‖

3ot product – et A= ⟨ a

1, a

2, a

3 ⟩   and

B=⟨ b1

, b2

, b3 ⟩ ,

 A ! B=a1

b1+a

2b2+a

3b3

 A !

B=‖

 A‖‖

B‖cos"

#here "  is the angle beteen  A  and B ,

" ϵ  [0,# ]

' o vectors  A∧B   are orthogonal or

perpendicular if and only if  A ! B=0

Properties4

1)   A ! B=B ! A

5)   A ! c B=c ( A ! B)

6)   A (B ± $ )=( A ! B ) ± ( A ! $ )

7)

  A !0=0

8)   A ! A=‖ A‖2

Pro9ections ( %ro jab)   ' perpendicular

pro9ection of the position representation of

b

onto the line containing the position

representation of a

 %ro j a b=a !b

‖a‖2 a

Cross product ' et A= ⟨ a

1, a

2, a

3 ⟩   and

B=⟨ b1

, b2

, b3 ⟩ ,

 A & B=de ( i    j   k 

a1

  a2

  a3

b1

  b2

  b3

)¿de (a

2  a

3

b2

  b3)i−de (a

1  a

3

b1

  b3) j+de (a

1  a

2

b1

  b2)k 

Page 3: Math 54 Reviewer 3rd LE

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

 A & B=‖ A‖‖B‖sin"

' *f  A & B=0 , then the vectors are parallel

' :or A=

(

a1

  a2

b1   b2

), de A=a

1b

2−a

2b1

' he vector  A & B   is orthogonal to both

 A  and B

Properties4

1)   A & B=−B& A

5)c B=c ( A & B)

 A &¿

6)   A & ( B± $ )=( A & B ) ± ( A &$ )

7)   A &0=0

+rea of a parallelogram

 A=‖a & b‖=‖a‖‖b‖sin"

"calar triple product

 A ! ( B &$ )=( A & B ) !$ 

%olume of a parallelepiped ' =|a ! (b &c )|

Plane (# )   – determined by a point and a

normal vector

a ( x− xo )+b ( y− yo )+c ( z− zo )=0

#here ⟨ a ,b ,c ⟩  is a normal vector to the plane

  ( xo , yo , zo)  is a point on the plane

ine (l) – determined by a point and a parallevector

a) %ector form

r= ⟨ xo , y o , zo ⟩+⟨ a,b ,c ⟩ 

b) Parametric form

{

 x= x0+a 

 y= yo+b 

 z= zo+c 

c) "ymmetric form

 x− xo

a  =

 y− yo

b  =

 z− zo

c

#here ⟨ a ,b ,c ⟩  is a parallel vector to the line

  ( xo , yo , zo)  is a point on the line

*ntersections

1) Of to lines – point; e&uate each componenof parametric e&uations of the to lines, thensolve for the to variables t  and s, substitutethen nd the point

5) Of to planes – line; solve for the system o

e&uations of the to planes, let  x=    and

Page 4: Math 54 Reviewer 3rd LE

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

solve for y and z in terms of it, express line indesired form

6) Of a line and a plane – point; substitute eachcomponent of the parametric e&uation of theline into the e&uation of the plane, solve for t ,then nd the point

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

3istances

1) <eteen points4 P1(a1 , b1 , c1)  and

 P2(a2 , b2 , c2)

d ( P1, P

2 )=√ ( a1−a

2)2+( b

1+b

2 )2+( c

1+c

2 )2

5) <eteen point  P( xo , yo , zo)   and plane

# :ax+by+cd+ z=0

d ( P , # )=|a xo+b y o+c zo+d|

√ a2+b2+c

2

6) <eteen point P( xo , yo , zo)  and line

l :{ x= x

1+a 

 y= y1+b 

 z= z1+c 

d ( P , l )=‖⟨ xo− x

1, y o− y

1, zo− z

1 ⟩ & ⟨ a,b ,c ⟩‖√ a2+b

2+c2

7) <eteen to lines – use a point on one of the

lines, then use formula d ( P , l)

8) <eteen to planes – use a point on one of 

the planes, then use formula d ( P , # )

=) <eteen intersecting lines or planes – zero

Cylinders – surface that consists of all lines thatare parallel to a given line and pass through a

given curve, the graph of an e&uation of tovariables in 6'3

>o to dra41) ?raph the e&uation in to dimensions5) ?enerate the cylinder by moving it along the

direction of the axis of the missing variable

 races ' plane curves that are determined bypro9ecting the surface along a given coordinateplane (x'y, x'z or y'z)

>o to dra41) "ubstitute the e&uation of the plane

( xn=k ) to the e&uation of the surface

5) ?raph the resulting e&uation in todimensions

"urface of revolution – surface generated hen aplane curve is revolved about a xed line

?enerating

curve

+xis of  

revolution

"urface

e&uation

 y=f ( x)

 z=f ( x)

x'axis  y2+ z

2=[ f  ( x ) ]2

 x=f ( y)

 z=f ( y)

y'axis  x2+ z

2= [ f  ( y ) ]2

 x=f ( z)

 y=f ( z)

z'axis   ( z )f  ¿¿

 x2+ y

2=¿

-uadric sections

1) @llipsoidA"pheroid

(2

( +

(2

( +

(2

( =1

5) >yperboloid of one sheet

(2

( +

(2

( −

 (2

( =1

6) >yperboloid of to sheets

(2

( −

(2

( −

(2

( =1

7) @lliptical cone

(2

( +

(2

( −

 (2

( =0

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University of the Philippines Chemical Engineering Society, Inc. (UP KEM)Math 5 ! "r# $ong E%am &evie'er

8) @lliptic paraboloid

(2

( +

(2

(  ±

 (2

( =0

=) >yperbolic paraboloid

(2

( −

(2

(  ±

 (2

( =0

>o to dra41) ?raph the traces of the &uadric in the x'y, y'z

and x'z planes by letting the unanted

variable be e&ual to zero5) *f the trace produces a point, let the unanted

variable e&ual a value that ill produce a

conic section

6) Combine traces on the three'dimensiona

Cartesian coordinate system


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