Vector Calculus: Are you ready?
Vectors in 2D and 3D Space: Review
ƒ Purpose:
Make certain that you can define, and use in context, vector terms,
concepts and formulas listed below:
Section 7.1-7.2
¾find the vector defined by two points and determine the norm
of the vector.
¾add two vectors
¾multiply a non-zero vector by a non-zero scalar.
¾represent a non-zero vector in the xy-plane in terms of its
magnitude and the angle it makes with the positive x-axis. 1
Vectors in 2D space
ƒ Vector algebra and vector calculus have resulted from practical
engineering applications: Mechanics, Fluid flows, Wireless
Communications
ƒ Scalar: is described by a single quantity such as work, energy,
potential, speed, temperature, blood pressure ..
ƒ Vector: is described by a magnitude and direction such as
velocity, electric force, position of a robot …
ƒ There are many quantities that are vector functions:
Some Daily Use of Vectors
¾A wind of 80 km/h from the Southeast.
¾A car going 80 km/h East.
¾A vertical velocity of 20 m/s.
¾A plane traveling 1000 km/h on a 180 heading.
These issues are described by a magnitude and a direction.
2
Some Applications of Vectors
ƒ Mechanics: Force, Torque, position, speed,
acceleration, …
ƒ Electromagnetism: Electric and magnetic fields,
current density, pointing vector,…
ƒ Example Walking and Different Forces
ƒ Example Mechanical System in Equilibrium
Other Examples of vector quantities
ƒ Notation
r
v, u
⎯
⎯→
AB
Acknowledgment: Most
figures included in class
notes are copied from
the textbook by Zill and
3
Cullen.
Notation and Terminology
⎯
⎯→
ƒ A vector with starting point A and ⎯
end
point B is written as AB
⎯→
⎯
⎯→
ƒ Magnitude of AB is written as: || AB ||
ƒ Example: In 2D Cartesian Coord.:
r
a = a1iˆ + a2 ˆj = < a1 , a2 > = [a1 , a2 ]
r
2
2
Magnitude : a = a1 + a2
ƒ Two vectors with the
same magnitude and
direction are equal
ƒ Parallel vectors: nonzero
scalar multiples of each other
4
A note about notation
ƒ The textbook uses boldface to represent vectors,
ƒ I may place an arrow above general vectors and a hat
over unit vectors.
ƒ I would like you all to clearly identify vectors in your
work.
F
r
F
u
û
i = iˆ
5
Addition of Vectors
⎯
⎯→
⎯
⎯→
ƒ Consider two vectors AB and AC
with common initial point A
ƒ The sum of two vectors is the main
diagonal of the parallelogram with the
vectors as sides
⎯
⎯→
⎯
⎯→
⎯
⎯→
AD = AB + AC
Example:
( 4i + 4 j ) + (6i − 6 j ) = 10i − 2 j
4i + 6 j + 6i − 6 j = 10i − 0 j = 10i
6
Subtraction
⎯
⎯→
ƒ Subtraction: The difference of AB and
⎯
⎯→
⎯
⎯→
⎯
⎯→
⎯
⎯→
AC
is defined by
⎯
⎯→
AB − AC = AB + (− AC )
ƒ
⎯
⎯→
⎯
⎯→
AB − AC is the main diagonal of the
⎯
⎯→
⎯
⎯→
parallelogram with sides AB and − AC
⎯
⎯→
⎯
⎯→
⎯
⎯→
ƒ Or CB = AB − AC is a vector from the end
of the second vector toward the end of
the first vector
7
Review Exercise (page 346): Prob. 48
Sphere
weight=50 lb
Find the magnitude of F1 and F2.
At equilibrium: F1 + F2 + w = 0
w = -50 j lb
2 supporting planes
∴ F1 = 25.9 lb, F2 = 36.6 lb
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Properties of Vectors
ƒ Magnitude, length, or norm of a vector a: ||a||
ƒ If a =< a , a > then: || a ||=
1
2
a +a
2
1
2
2
ƒ A vector that has magnitude 1 is called unit vector.
A unit vector in the direction of a is:
⎞⎟ a with
uˆ = ⎛⎜ 1
⎝ || a || ⎠
u =1
ƒ The i, j unit vectors: i=<1,0>, j=<0,1>
j
u
i
< a1 , a2 > = < a1 ,0 > + < 0, a2 > = a1i + a2 j
ƒ Example: Given a=<3,-4>, form a unit vector
• in the same direction as a.
Answer: <0.6,-0.8>
• In the opposite direction of a.
Answer: <-0.6,0.8>
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7.2 Vectors in 3-Space
ƒ Rectangular or Cartesian Coordinate
2D-Space: Two orthogonal axes
3D-Space: Three mutually orthogonal axes
The three axes follow the Right Hand Rule
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3D-Space
ƒ Coordinate Plane: Each pair of coordinate axes determines a
coordinate plane (xy,xz and yz).
ƒ Octant:The coordinate planes divide the 3-space into 8 parts
known as Octants.
ƒ First octant: x, y, z>0
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ƒ Position Vector:
For a point P, the position vector is
r
r
r1 = OP =< x1 , y1 , z1 >
ƒ Distance Formula between two points:
Given 2 points:
P1 ( x1, y1, z1 ) & P2 ( x2 , y2 , z2 )
d (P1 , P2 ) = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z 2 − z1 ) 2
P2
P1
ƒ Vector between two points:
r
r
r
P1P2 = OP2 − OP1
=< x2 − x1 , y2 − y1 , z2 − z1 >
Examples: P1 = (1,2,3) & P2 = (1,-1,-1)
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Component Definitions in 3D-Space
Let a =< a1 , a2 , a3 > and b =< b1 , b2 , b3 > be vectors in R 3
(i) Addition:
a + b =< a1 + b1 , a2 + b2 , a3 + b3 >
ka =< ka1 , ka2 , ka 3 >
a=b if an only if a1 = b1 , a2 = b2 , a3 = b3
(iii) Equality:
− b =< −b1 ,−b2 ,−b3 >
(iv) Negative of a vector:
(ii) Scalar Multiplication:
(v) Subtraction:
a − b = a + (−b) =< a1 − b1 , a2 − b2 , a3 − b3 >
(vi) Zero vector:
0 = <0,0,0>
(vii) Magnitude:
|| a ||= a + a + a
2
1
2
2
2
3
a =< a1 , a2 , a3 >= a1i + a2 j + a3k
13
ƒ Unit Vectors in 3D space:
i = <1,0,0>,
j = <0,1,0>,
k = <0,0,1>
a = < a1 , a2 , a3 > = a1i + a2 j + a3k
Example: If a = 3i - 4j + 8k and b = I - 4k,
b = i- 0j - 4k
Æ
find 5a - 2b.
2b = 2i + 0j - 8k
5a = 15i - 20j + 40k
5a - 2b = 13i - 20j + 48k
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7.3 Dot (scalar or inner) Product
2. Section 7.3
¾define the dot (inner) product (a . b) and interpret it
geometrically.
¾use the dot product to determine:
– work done by a force,
– the angle between two vectors,
– whether two vectors are perpendicular to one
another,
– projections and components of vectors,
– direction angles and direction cosines
15
Dot (scalar or inner) Product
ƒApplications: Mechanics and Electromagnetism
Definition:
The dot product of two vectors a and b
is the scalar
a ⋅ b =|| a || ⋅ || b || cosθ & 0 ≤ θ ≤ π
θ is the angle between a and b
Example Dot product
i . i=1,
j . j=1,
k . k=1 since ||i||=||j||=||k||=1 and θ = 0
i . j=0,
j . k=0,
k . i=0 since θ = 90o
Example Given:
a=10i+2j-6k, b=-0.5i+4j-3k Æ a . b=(10)(-0.5)+(2)(4)+(-6)(-3)=21
16
Physical Interpretation of the Dot Product
ƒ A constant force of magnitude F moves an object a distance d in
the direction force, the work done by the force (W):
r r
W = F ⋅ d =|| F || || d ||
ƒ When a constant force F applied to a body acts at an angle θ to
the direction of motion, the work done by the force (W):
r r
W = F ⋅d
= (|| F || cosθ ) || d ||
=|| F || || d || cosθ
Note: if F and d are orthogonal, W=0.
Examples:
17
P7.3-47: Given
|| F ||= 30 N
r
d =< 4,3 > m
weight
Work done by w (gravity force) = w . d = 0 (w
¦
d)
Work done by F (applied force) = F . d = |F|.|d| cos θ (d // F)
= 150 N.m
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Properties of Dot Product
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
a . b = 0 if a=0 or b=0
a.b=b.a
(commutative law)
a . (b+c) = a . b+a . c
(distributive law)
a . (kb) = (ka) . b = k(a . b) k a scalar
a.a≥0
a . a = ||a||2
For nonzero vectors a and b
(i) a . b > 0 if and only if θ is acute
(ii) a . b < 0 if and only if θ is obtuse, and
(iii) a . b = 0 if and only of if cos θ =0 (Orthogonal vectors)
Theorem 7.1 Criterion for Orthogonal Vectors
Two nonzero vectors a and b are orthogonal if and only if a . b=0
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Angle Between Two Vectors:
r r
a•b
a1b1 + a2b2 + a3b3
cosθ =
=
|| a || || b ||
|| a || || b ||
ƒ Example
Find the angle between
a = 2i+3j+k
&
b = -i+5j+k.
Solution:
a . b=14,
|| a ||= 14 , || b ||= 27
14
cosθ =
2 27
⎛ 42 ⎞
⎟ ≈ 0.77 ≈ 44.9o
∴θ = cos ⎜⎜
⎟
9
⎝
⎠
−1
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Direction Angles
For a nonzero vector in 3D-Space the angles α, β and γ with i, j,
and k are called direction angles of a.
Direction cosines for a = a1i + a2 j + a3k
r ˆ
a •i
a1
=
cos α =
|| a || || iˆ || || a ||
a3
cos γ =
|| a ||
a2
cos β =
|| a ||
cos α + cos β + cos γ = 1
2
2
2
ƒ Example Find the direction cosines and direction
angles of the vector a = 2i+5j+4k. Æ ||a||=6.71
⇒ α ≈ 72.7 , β ≈ 41.8 , γ ≈ 53.4
o
o
o
21
Component of a on b:
a⋅b
comp ba =|| a || cosθ =
= a ⋅ bˆ
|| b ||
= scalar
Projection of a in the direction of b:
( )
⎛ b ⎞ r ˆ ˆ
projba = (comp ba )⎜
⎟⎟ = a ⋅ b b
⎝ || b || ⎠
= vector
Example: a = < -1,-2,7 > & b = < 6,-3,-2 >
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7.4 Cross (Vector) Product
3. Section 7.4
¾ define the vector (cross) product (a x b) and interpret it
geometrically.
¾ determine the cross product of vectors and combinations
of vectors, use to determine torque
¾ find unit vectors that are perpendicular to two given
vectors.
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Cross (Vector) Product
ƒ The vector product of 2 vectors A and B is given by
ˆi ˆj
kˆ
A × B = A x A y A z = iˆ( Ay Bz − Az B y ) − ˆj ( Ax Bz − Az Bx )
AxB
ˆ
Bx B y Bz
+ k( A B − A B )
x
y
y
x
n
= A ∗ B ∗ sin(θ ) nˆ
|| A × B ||= area of paralleogram
B
θ
A
ƒ where n is a unit vector perpendicular to A and B, pointing in the
direction given by the right hand screw rule (i.e. the direction in
which a screw would advance if it were turned from A to B.
Example: a = < -1,-2,7 >
& b = < 6,-3,-2 >
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Typical Applications
a) AREA OF A PARALLELOGRAM with edges a and b:
Area = ⎪⎢a × b ⎪⎢= ⎪⎢a⎪⎢⎪⎢b⎪⎢ sin θ
b) AREA OF A Triangle with edges a and b:
Area = 1/2⎪⎢a × b ⎪⎢= 1/2 ⎪⎢a⎪⎢⎪⎢b⎪⎢ sin θ
Example
(p7.4, # 48): Find area of the triangle through:
p1 = (0,0,0), p2 = (0,1,2), P3 = (2,2,0)
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Typical Applications
ƒ Volume of a parallelepiped (with edges: a, b & c)
Volume = (area of base) . (height)
= ⎪⎢b × c ⎪⎢ .⎮comp b × c a⎮
= ⎪⎢b × c ⎪⎢. ⎮a • (b × c) ⎮ / ⎪⎢b × c ⎪⎢
∴Volume =⎮a • (b × c)⎮
Example: a = < 3,1,1 >,
b = < 1,4,1 >
& c = < 1,1,5 >
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Typical applications
ƒ
MOMENT OF A FORCE
In mechanics the moment m of a force F about a point Q is
defined as the product
m =⎪⎢F ⎪⎢ d
where d is the (perpendicular) distance
between Q and the line of action L on F.
If r is the vector from Q to any point A
on L, then
d = ⎪⎢r ⎢⎢ sin θ
r r
r
∴ m = F ∗ r sin θ dˆ = F × r
Q
d=r si
nθ
m is called the moment vector or vector moment of F about Q
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More Applications
• Torque
T=rxF
• Force on a moving charge due
to a magnetic field due to
F=qvxB
• Velocity of a rotating body
F
B
q
v
w
v= wxr
v
ω
ω= angular speed
r
|w| = ω and directed along axis of rotation
o
28
• Circular Mnemonic:
iˆ × ˆj = kˆ
ˆj × iˆ = − kˆ
ˆj × kˆ = iˆ
kˆ × ˆj = −iˆ
kˆ × iˆ = ˆj
iˆ × kˆ = − ˆj
• More Cross Product Properties:
r r
¾ Two non-zero vectors A and B are parallel if & if : A × B = 0
r r r r
¾ Cross product is not commutative: A × B ≠ B × A
r r
r r
A × B = −B × A
r r r
r r r
¾ Cross product is not associative: A × ( B × C) ≠ ( A × B) × C
Example
(p7.4, # 13): A = <2,7,-4>, B = <1,1,-1>
Find a vector that is perpendicular to A and B
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7.5 Lines and Planes
ƒ Purpose:
Make certain that you can define, and use in context, vector
terms, concepts and formulas listed below:
4. Section 7.5
¾ express a line as a: vector parameterization, and scalar
parameterization,
¾ use vectors to determine whether two lines intersect, and
if so, the point of intersection.
¾ use vectors to find the distance from a point to a line.
¾ express a plane as a scalar equation and as a vector
equation.
¾ find whether two planes intersect, and if so, the angle of
intersection and a vector parameterization of the line
formed by the intersection.
¾ unit normal for a plane.
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Equation of a “straight” Line
Given two points in 3D:
r
P1 : r1 =< x1 , y1 , z1 > &
r
P2 : r2 =< x2 , y2 , z2 >
Two forms for the line through P1 & P2:
1. Vector equation of the line through r1 & r2:
r r
r r
r = r2 + t ( r2 − r1 ),
t = scalar parameter
r r r
r
r = r2 + ta , the line is in the direction of a
If a is a unit vector, then its components are direction cosines of the line.
2. Parametric & symmetric equations of the line:
x − x2
y − y2
z − z2
=
=
x2 − x1 y2 − y1 z2 − z1
Examples to
follow:
31
Examples: 7.5, # 3
Find the vector equation of a line through:
(1/2, -1/2, 1) & (-3/2, 5/2, -1/2).
r r
r r
r = r1 + t ( r1 − r2 ),
t = scalar parameter
Examples: 7.5, # 27
Show that the two lines:
r = t <1,1,1> and r = <6,6,6> + t <-3,-3,-3>
are the same.
32
7.5 Equation of a Plane
Two forms:
1. The equation of a plane perpendicular to a normal vector
is given by:
r
n = a iˆ + b ˆj + c kˆ
r r r
(r - r1 ) • n = 0
a x + b y + cz + d = 0
2. The equation of a plane contains 3 points:
P1(r1), P2(r2), P3(r3) is given by:
r r
r r
r r
[( r2 − r1 ) × ( r3 − r1 )] • ( r − r1 ) = 0
a vector form.
Examples to follow:
33
Examples: 7.5, # 39
Find the equation of a plane contains: (5,1,3) &
perpendicular to <2,-3,4>
Two methods:
Or:
a x + b y + cz + d = 0
< r-r1 > . n = 0
Answer:
Examples: 7.5, # 51
Find the equation of a plane contains: (2,3,-5) &
parallel to x + y - 4z = 1
34
Intersection of Two Planes
ƒ Let
a1 x + b1 y + c1 z = d1 &
a2 x + b2 y + c2 z = d2
be two non parallel planes.
We get a system of two
equations and three unknowns.
ƒ Choose one variable arbitrary, say x = t, and solve the new
system of two equations and two unknowns y and z.
Æ parametric equations for the line of intersection
35
Example Find the parametric equation for the line of
intersection of
2x – 3y + 4z = 1 and
x–y–z=5
Solution
Let choose z = t, sub in the 2 equatins
and solve for x and y from
2x – 3y = 1 – 4t
and
x–y=5+t
Then,
x = 14 + 7t,
y = 9 + 6t,
z=t
END of selected materials from Chapter 7.
36