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Vector Calculus: Are you ready?

Purpose: Make certain that you can define, and use in context, vector terms,

concepts and formulas listed below:

Section 7.1-7.2

find the vector defined by two points and determine the norm of the vector.add two vectors multiply a non-zero vector by a non-zero scalar.represent a non-zero vector in the xy-plane in terms of its magnitude and the angle it makes with the positive x-axis.

Vectors in 2D and 3D Space: Review

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Vectors in 2D space

There are many quantities that are vector functions: Some Daily Use of Vectors

A wind of 80 km/h from the Southeast.A car going 80 km/h East.A vertical velocity of 20 m/s.A plane traveling 1000 km/h on a 180 heading.

These issues are described by a magnitude and a direction.

Vector algebra and vector calculus have resulted from practical engineering applications: Mechanics, Fluid flows, Wireless CommunicationsScalar: is described by a single quantity such as work, energy, potential, speed, temperature, blood pressure ..Vector: is described by a magnitude and direction such as velocity, electric force, position of a robot …

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Some Applications of VectorsMechanics: Force, Torque, position, speed, acceleration, …Electromagnetism: Electric and magnetic fields, current density, pointing vector,…Example Walking and Different Forces

Example Mechanical System in Equilibrium

Other Examples of vector quantities

Notation ur,v⎯→⎯

AB

Acknowledgment: Most figures included in class notes are copied from the textbook by Zill and Cullen.

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Notation and TerminologyA vector with starting point A and end point B is written asMagnitude of is written as:

⎯→⎯

AB

⎯→⎯

AB||||

⎯→⎯

AB

Two vectors with the same magnitude and direction are equal

Parallel vectors: nonzero scalar multiples of each other

Example: In 2D Cartesian Coord.:

22

21

212121

:Magnitude

],[ , ˆˆ

aaa

aaaajaiaa

+=

=><=+=r

r

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A note about notation

The textbook uses boldface to represent vectors,I may place an arrow above general vectors and a hat over unit vectors. I would like you all to clearly identify vectors in your work.

Fr

F u u ii ˆ=

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Addition of Vectors

The sum of two vectors is the main diagonal of the parallelogram with the vectors as sides

⎯→⎯⎯→⎯⎯→⎯

+= ACABAD

Consider two vectors and with common initial point A

⎯→⎯

AB⎯→⎯

AC

Example:

ijijijijijiji

100106664210)66()44(

=−=−++

−=−++

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Subtraction

Subtraction: The difference of and is defined by

)(⎯→⎯⎯→⎯⎯→⎯⎯→⎯

−+=− ACABACAB

⎯→⎯

AB⎯→⎯

AC

⎯→⎯

AB⎯→⎯

− AC

⎯→⎯⎯→⎯

− ACAB is the main diagonal of the parallelogram with sides and

⎯→⎯⎯→⎯⎯→⎯

−= ACABCBOr is a vector from the end of the second vector toward the end of the first vector

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At equilibrium: F1 + F2 + w = 0

Sphere weight=50 lb

2 supporting planes

w = -50 j lb

∴ F1 = 25.9 lb, F2 = 36.6 lb

Review Exercise (page 346): Prob. 48Find the magnitude of F1 and F2.

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Properties of VectorsMagnitude, length, or norm of a vector a: ||a||

If then:>=< 21 aa ,a22

21 aa +=|||| a

1||||1ˆ =⎟

⎠⎞⎜

⎝⎛= u with aau

ji 212121 ,00, , aaaaaa +=><+><=><The i, j unit vectors: i=<1,0>, j=<0,1>

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>

A vector that has magnitude 1 is called unit vector.A unit vector in the direction of a is:

ij u

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7.2 Vectors in 3-SpaceRectangular or Cartesian Coordinate2D-Space: Two orthogonal axes

The three axes follow the Right Hand Rule

: Three mutually orthogonal axesSpace-3D

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Coordinate Plane: Each pair of coordinate axes determines a coordinate plane (xy,xz and yz).Octant:The coordinate planes divide the 3-space into 8 parts known as Octants.First octant: x, y, z>0

3D-Space

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P 1P 2

Given 2 points:

>=<= 1111 ,, zyxPOrrr

Position Vector:For a point P, the position vector is

>−−−=<−=

121212

1221

,, zzyyxxPOPOPPrrr

Vector between two points:

),,(P & ),,(P 22221111 zyxzyx

212

212

212 zzyyxxd )()()()P,P( 21 −+−+−=

Distance Formula between two points:

Examples: P1 = (1,2,3) & P2 = (1,-1,-1)

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Component Definitions in 3D-SpaceLet and be vectors in

(i) Addition:

(ii) Scalar Multiplication:

(iii) Equality: if an only if

(iv) Negative of a vector:

(v) Subtraction:

(vi) Zero vector: 0 = <0,0,0>

(vii) Magnitude:

>=< 321 aaa ,,a >=< 321 bbb ,,b 3R>+++=<+ 332211 bababa ,,ba

>=< 321 kakakak ,,aba= 332211 bababa === ,,

>−−−=<− 321 bbb ,,b>−−−=<−+=− 332211 bababa ,,)( baba

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22

21 aaa ++=|||| a

kjia 321321 aaaaaa ++>==< ,,

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Unit Vectors in 3D space:

i = <1,0,0>,

j = <0,1,0>,

k = <0,0,1>

kji a 321321 ,, aaaaaa ++=><=

Example: If a = 3i - 4j + 8k and b = I - 4k, find 5a - 2b.

b = i- 0j - 4k 2b = 2i + 0j - 8k

5a = 15i - 20j + 40k

5a - 2b = 13i - 20j + 48k

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2. Section 7.3define the dot (inner) product (a . b) and interpret it geometrically.use 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

7.3 Dot (scalar or inner) Product

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Applications: Mechanics and Electromagnetism

πθθ ≤≤⋅=⋅ 0 & cos|||||||| baba

Definition:

The dot product of two vectors a and b

is the scalar

θ is the angle between a and b

Example Dot producti . i=1, j . j=1, k . k=1 since ||i||=||j||=||k||=1 and θ = 0i . j=0, j . k=0, k . i=0 since θ = 90o

Example Given:a=10i+2j-6k, b=-0.5i+4j-3k a . b=(10)(-0.5)+(2)(4)+(-6)(-3)=21

Dot (scalar or inner) Product

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Physical Interpretation of the Dot ProductA constant force of magnitude F moves an object a distance d in the direction force, the work done by the force (W):

θθcos||||||||

||||)cos||(||d F

dF ==

⋅= dFWrr

|||||||| d F=⋅= dFWrr

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):

Note: if F and d are orthogonal, W=0.

Examples:

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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

md

NF

>=<

=

3,4

30||||r

P7.3-47: Given

weight

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Properties of Dot Producta . 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 scalara . a ≥ 0a . 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 VectorsTwo nonzero vectors a and b are orthogonal if and only if a . b=0

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Angle Between Two Vectors:

||||||||||||||||cos 332211

b ab ababababa ++

=•

=rr

θ

27214cos =θ o9.4477.0

942cos 1 ≈≈⎟⎟

⎠

⎞⎜⎜⎝

⎛=∴ −θ

ExampleFind the angle between a = 2i+3j+k & b = -i+5j+k.

27||||,14|||| == baa . b=14,

Solution:

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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

ooo 4.53,8.41,7.72 ≈≈≈⇒ γβα

kjia 321 aaa ++=

||||||ˆ||||||ˆ

cos 1

a aa

iia

=•

=r

α

||||cos 2

aa

=β||||

γcosa

3a=

1coscoscos 222 =++ γβα

Example Find the direction cosines and direction angles of the vector a = 2i+5j+4k. ||a||=6.71

Direction Angles

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Component of a on b:

( ) vector

ˆ ˆ||||

)comp(proj

=

⋅=⎟⎟⎠

⎞⎜⎝

⎛= bba

bbaa bb

r

scalar

ˆ||||

cos||||comp

=

⋅=⋅

== bab

baaab θ

Projection of a in the direction of b:

Example: a = < -1,-2,7 > & b = < 6,-3,-2 >

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3. Section 7.4

define the vector (cross) product (a x b) and interpret it geometrically.determine the cross product of vectors and combinations of vectors, use to determine torquefind unit vectors that are perpendicular to two given vectors.

7.4 Cross (Vector) Product

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Cross (Vector) ProductThe vector product of 2 vectors A and B is given by

amparalleogr of area ||BA|| BA

B B BA A Ak j i

BA

zyx

zyx

=×

∗∗=

−−−==×

n

BABAjBABAi xzzxyzzy

ˆ)sin(

)(ˆ)(ˆˆˆˆ

θ

)(ˆ xyyx BABAk −+

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.

A

Bθ

AxB

n

Example: a = < -1,-2,7 >

& b = < 6,-3,-2 >

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Typical Applications

b) AREA OF A Triangle with edges a and b:

Area = 1/2⎪⎢a × b ⎪⎢= 1/2 ⎪⎢a⎪⎢⎪⎢b⎪⎢ sin θ

a) AREA OF A PARALLELOGRAM with edges a and b:

Area = ⎪⎢a × b ⎪⎢= ⎪⎢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

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 applicationsMOMENT OF A FORCE

In mechanics the moment m of a force F about a point Q is defined as the product

m =⎪⎢F ⎪⎢ d

d = ⎪⎢r ⎢⎢ sin θ

rFdrFm rrr×=∗=∴ ˆ sin θ

m is called the moment vector or vector moment of F about Q

where d is the (perpendicular) distance between Q and the line of action L on F.

Q d=r sin θ

If r is the vector from Q to any point Aon L, then

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• Torque T = r x F

• Force on a moving charge due to a magnetic field due to

FB

vqF = q v x B

• Velocity of a rotating body

v = w x rω= angular speed

|w| = ω and directed along axis of rotation

w

r

v

ω

o

More Applications

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Two non-zero vectors A and B are parallel if & if : 0BA =×rr

jkiijkkij

jikikjkjiˆˆˆ ˆˆˆ ˆˆˆ

ˆˆˆ ˆˆˆ ˆˆˆ

−=×−=×−=×

=×=×=×

• Circular Mnemonic:

• More Cross Product Properties:

Cross product is not commutative:

ABBA

ABBArrrr

rrrr

×−=×

×≠×

Cross product is not associative: C)BA()CB(Arrrrrr

××≠××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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Purpose: Make certain that you can define, and use in context, vector

terms, concepts and formulas listed below:

4. Section 7.5express a line as a: vector parameterization, and scalar parameterization,use vectors to determine whether two lines intersect, and if so, the point of intersection.use vectors to find the distance from a point to a line.express a plane as a scalar equation and as a vector equation.find whether two planes intersect, and if so, the angle of intersection and a vector parameterization of the line formed by the intersection.unit normal for a plane.

7.5 Lines and Planes

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Equation of a “straight” Line

>=<>=<

22222

11111

,,:P& ,,:P

zyxrzyxr

r

r

aatrrtrrtrr

rrrr

rrrr

of direction thein is line the,parameterscalar ),(

2

122

+==−+=

12

2

12

2

12

2

zzzz

yyyy

xxxx

−−

=−−

=−−

1. Vector equation of the line through r1 & r2:

2. Parametric & symmetric equations of the line:

Given two points in 3D:

Two forms for the line through P1 & P2:

If a is a unit vector, then its components are direction cosines of the line.

Examples to follow:

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Examples: 7.5, # 3

parameter scalar =−+= trrtrr ),( 211rrrr

Find the vector equation of a line through: (1/2, -1/2, 1) & (-3/2, 5/2, -1/2).

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.

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7.5 Equation of a Plane

kcjbian ˆ ˆ ˆ ++=r

0 z y x =+++ dcba

1. The equation of a plane perpendicular to a normal vector

is given by:

2. The equation of a plane contains 3 points: P1(r1), P2(r2), P3(r3) is given by:

[ ] 0)()()( 11312 =−•−×− rrrrrr rrrrrr

Examples to follow:

Two forms:

a vector form.

0( =• n)r-r 1rrr

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Examples: 7.5, # 39

0 z y x =+++ dcba

Examples: 7.5, # 51

Find the equation of a plane contains: (5,1,3) & perpendicular to <2,-3,4>

< r-r1 > . n = 0

Two methods:

Find the equation of a plane contains: (2,3,-5) & parallel to x + y - 4z = 1

Or:

Answer:

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Intersection of Two PlanesLeta1 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.

Choose one variable arbitrary, say x = t, and solve the new system of two equations and two unknowns y and z.

parametric equations for the line of intersection

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Example Find the parametric equation for the line of intersection of

2x – 3y + 4z = 1 and x – y – z = 5

SolutionLet choose z = t, sub in the 2 equatinsand 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.