lecture notes for applied mathematics

Coordinates and vectors

Joachim Vogt

Course 210221, Fall 2015

Mathematical and Computational Concepts in ESS

Joachim Vogt (Jacobs University Bremen) Coordinates and vectors Course 210221, Fall 2015 1 / 36

Coordinates and vectors — Contents

1 Vector operationsAddition, dot product, cross productParallel and perpendicular components

2 Coordinate systemsCartesian coordinatesPolar and cylindrical coordinatesSpherical coordinatesDistances and angles

3 Complex numbersAddition and complex multiplicationComplex exponential function

4 Spreadsheet exercise on coordinate transformation (homework)

5 AppendixSolutions of the sample problems, references


Coordinates and vectors — Section 1

Vector operations


What are vectors ?

Vectors are objects in a linear space: one can add (and subtract) vectorsin a meaningful way, and also multiply them with scalar values.

Important example: n-tuples (ordered lists) of real numbers.

Notation: V , ~V , V .

In three dimensions (n = 3):

V =

V1V2V3

= (V1, V2, V3)

T , W =

W1

W2

W3

= (W1,W2,W3)

T .

v + w

v

w

Addition of two vectors:

V1V2V3

+

W1

W2

W3

=

V1 +W1

V2 +W2

V3 +W3

.


Scalar multiplication, dot product, length, angle

Multiplication with a scalar value a ∈ R: a

V1V2V3

=

aV1aV2aV3

.

Dot (scalar, inner) product of two vectors: V ·W = V TW

V1V2V3

·

W1

W2

W3

= (V1, V2, V3)

W1

W2

W3

= V1W1 +V2W2 +V3W3 .

Length of a vector : ‖V ‖ =√V · V =

√(V1)2 + (V2)2 + (V3)2 .

Also: magnitude, (euclidean) norm, |V |, V .

v

w

α

Angle between two vectors:

cosα = cos∠(V ,W ) =V ·WVW


Examples

Addition of two vectors:

42−4

+

−213

=

23−1

.

Multiplication with a scalar: 2

−213

=

−426

.

Dot (scalar) product between two vectors:

201

·

30

− 1

= 6 + 0− 1 = 5 .

Lengths:

∥∥∥∥∥∥ 2

01

∥∥∥∥∥∥ =√

4 + 0 + 1 =√

5 ,

∥∥∥∥∥∥ 3

0− 1

∥∥∥∥∥∥ =√

9 + 0 + 1 =√

10 .

For the angle α between the two vectors we obtain

cosα =5√

5√

10=

5

5√

2=

1√2,

thus α = 45.


Cross product

Vector sum, multiplication with a scalar, and dot product can be formed inlinear spaces with arbitrary dimensions.

Cross product and triple product: three-dimensional space only.

Cross (vector, outer) product of two vectors:

V ×W =

V1V2V3

×

W1

W2

W3

=

V2W3 − V3W2

V3W1 − V1W3

V1W2 − V2W1

.

The vector V ×W is directed perpendicular to both vectors V and W .

Length: ‖V ×W ‖ = V W sinα = V W sin∠(V ,W ) .

Cross product example:

020

×

304

=

80−6

.


Triple product and further remarks

The triple product U · (V ×W ) gives the volume of the parallelepipedformed by the three vectors U , V , and W . The triple product vanisheswhen the three vectors are co-planar.

Note that the

dot product produces a scalar,

cross product produces a vector,

dot product vanishes if the two vectors are perpendicular,

cross product vanishes if the two vectors are aligned.

α 0 30 45 60 90

sinα√

0/2 = 0√

1/2 = 1/2√

2/2 = 1/√

2√

3/2√

4/2 = 1

cosα√

4/2 = 1√

3/2√

2/2 = 1/√

2√

1/2 = 1/2√

0/2 = 0


Parallel and perpendicular components of a vector

One often needs to find the components of vectors parallel andperpendicular to a given direction D (unit vector: ‖D‖ = 1).

For a vector V , its component V‖ parallel to D is given by

V‖ = (V · D) D .

For a vector V , its component V⊥ perpendicular to D is given by

V⊥ = V − V‖ .

For three-dimensional vectors, we can write

V⊥ = D × (V × D) .

The latter identity can be shown by means of the so-called bac-cab rule

A× (B ×C) = B(A ·C)−C(A ·B)

that allows to convert double cross products.


Sample problems

Dot products and cross products

Compute the angles α = ∠(V ,W ) between the vectors V and W .

(a) V = (2, 0,−2)T, W = (2, 2, 0)T.

(b) V = (1, 2, 2)T, W = (2, 1,−2)T.

(c) V = (1, 1,−2)T, W = (−2,−2, 4)T.

Compute also the cross products V ×W .

Vector identitiesUse the bac-cab rule A× (B ×C) = B(A ·C)−C(A ·B) to compute the result of

a× (b× c) + b× (c× a) + c× (a× b) . (1)

Furthermore, show that

(a× b) × (c× d) = c [(a× b) · d]− d [(a× b) · c] , (2)

(a× b) · (c× d) = (a · c)(b · d)− (a · d)(b · c) . (3)



Coordinate systems


What are coordinates ?

Coordinates are sets of numbers to specify positions. Examples:

Local planar (2D) cartesian coordinates (e.g.: east, north).

Local (planar, 2D) polar coordinates (distance, angle).

Local three-dim. (3D) cartesian coordinates (e.g.: east, north, up).

Geographic coordinates: latitude, longitude, elevation.

Celestial coordinates: declination, right ascension, distance.

[Wikipedia Commons / Djexplo (1)] [Wikipedia Commons / Tfr000 (2)]


Cartesian coordinates

Cartesian coordinates are rectilinear(rectangular) coordinates denotingthe distances from coordinate lines(2D) or coordinate planes (3D).

In 2D: x, y.

In 3D: x, y, z.

Position vector: r =

xyz

.

y

x

xy

=

76

xy

=

72

Freitag, 2. August 13

To compute the distance between two points,

form position vectors r1 = (x1, y1, z1)T and r2 = (x2, y2, z2)

T,

compute the difference vector r2 − r1, and then

its norm ‖r2 − r1‖ =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.


Polar and cylindrical coordinates

Freitag, 2. August 13

Polar coordinates r, φ give theposition in a plane using

the radial distance r from areference point (origin), and

the angle φ (azimuth) betweenthe radial vector and a referencedirection (polar axis).

Cylindrical coordinates in 3D: r, φ, zwhere z is the cartesian coordinateperpendicular to the polar plane.

Transformation formulas:x = r cosφy = r sinφ

↔

r =

√x2 + y2

φ = arctan(y/x)


Spherical coordinates

[Wikipedia Commons / Andeggs (3)]

Spherical coordinates r, θ, φ give the 3Dposition as follows:

radial distance r from a referencepoint (origin),

angle θ (polar angle, co-latitude) ofthe radial vector with the z-axis,

angle φ (azimuth, longitude) of theprojected radial vector with the x-axis.

The latitude β = 90 − θ may replace θ.

Transformation formulas:

x = r sin θ cosφ = r cosβ cosφ ,y = r sin θ sinφ = r cosβ sinφ ,z = r cos θ = r sinβ .

In spherical coordinates, r =√x2 + y2 + z2.


Angles

The most popular units to express angles are degrees and radians.

One degree () is 1/360 of a full circle.One radian is the angle that corresponds to an arc of unit length on acircle of unit radius. The full circle gives an angle of 2π.

One radian is 180/π degrees (' 57.2958).

[Wikipedia Commons / Tfr000 (4)]

Celestial coordinates

Declination (DE, δ, celestial latitude) isusually given in degrees (), arcminutes (′)and arcseconds (′′). Conversion to decimaldegree representation: 1 = 60′, 1′ = 60′′.

Right ascension (RA, α, celestiallongitude) is usually given in hours (h),minutes (m), and seconds (s). Here24 hours correspond to 360 degrees of anarc: 1h = 15, 1m = 15′, 1s = 15′′.


Distances

On a spherical surface, the shortest connection between two points (r1 andr2) is through a great circle, and its length is called great-circle distance.

If α = ∠(r1, r2) denotes the angle between the two points expressed inradian, then their great-circle distance is simply Rα where R is the radiusof the sphere:

Rα = R arccos [cosβ1 cosβ2 cos(φ1 − φ2) + sinβ1 sinβ2] .

To compute the (usual) straight-line distance between two points given inspherical coordinates, convert first to cartesian coordinates, then proceedas described above. E.g., for the distance between two stars,

‖r2 − r1‖ =

∥∥∥∥∥∥

r2 cos δ2 cosα2

r2 cos δ2 sinα2

r2 sin δ2

−

r1 cos δ1 cosα1

r1 cos δ1 sinα1

r1 sin δ1

∥∥∥∥∥∥

= · · ·


Sample problems

Great-circle distanceOn a sphere of radius R, the shortest connection between two points is through a greatcircle. Let α denote the angle between the two points.

(a) Derive the formula for the great-circle distance:

Rα = R arccos [cosβ1 cosβ2 cos(φ1 − φ2) + sinβ1 sinβ2] .

(b) The earthquake observatory Bensberg is located at β1 = 50.96 and φ1 = −7.18.On 27 February 2008 the station registered an earthquake of Richter magnitude5.1. The epicenter was located in Lincolnshire (GB) at β2 = 53.166 andφ2 = −0.287. Give the epicentral distance in degrees and also in km. Note thatthe radius of the Earth is RE = 6371 km.

(c) According to the International Geomagnetic Reference Field (IGRF) model, in theyear 2005 the geomagnetic pole on the Northern hemisphere was located at aboutβ1 = 79 44′ and φ1 = −71 47′. Compute the geomagnetic latitude γ (i.e.,angular distance to the geomagnetic equator) of Jacobs University Bremen usingthe geographical coordinates β2 = 53 10′ and φ2 = 8 39′.



Complex numbers


What are complex numbers ?

Complex numbers (C) are pairs of real numbers (R2) that can bemultiplied in the usual sense (as in R).

Notation:

R2 : r =

(xy

)= x

(10

)+ y

(01

)

C : z = x + iy = x · 1 + y · iwhere

z is a complex variable (do not confuse with cartesian coordinate),x is the real part of z,y is the imaginary part of z,i =√−1 is the imaginary unit (i2 = −1).

Addition:

R2 : r1 + r2 = (x1 + x2)

(10

)+ (y1 + y2)

(01

)

C : z1 + z2 = (x1 + x2) · 1 + (y1 + y2) · iJoachim Vogt (Jacobs University Bremen) Coordinates and vectors Course 210221, Fall 2015 20 / 36

Complex exponential function

Define the complex exponential function

exp(Ω) = exp(ψ + iφ) = eψ(cosφ+ i sinφ)

where eψ denotes the real-valued exponential function. Here ψ ∈ R andφ ∈ R denote the real part and the imaginary part of the complex variableΩ = ψ + iφ, respectively. The right-hand side is thus entirely determinedthrough real-valued functions.

Using eψ1+ψ2 = eψ1eψ2 and addition theorems for trigonometric (cos, sin)functions one can show that

exp(Ω1 + Ω2) = exp(Ω1) · exp(Ω2)

also for complex variables Ω1,Ω2 ∈ C.

Furthermore, differentiating the complex exp(Ω) yields the same function,analogous to the real-valued case.


Multiplication of complex variables

Polar representation:

R2 : r =

(r cosφr sinφ

)= r

(cosφsinφ

)

C : z = r cosφ + i r sinφ = r (cosφ+ i sinφ) = r eiφ

Multiplication

In R2, there is no proper multiplication of the type

R2 × R2 → R2

(r1, r2) 7→ r1 ⊗ r2 =

(· · ·· · ·

)

In C, we define

z1 z2 = (x1 + iy1) (x2 + iy2) = x1x2 + ix1y2 + iy1x2 + i2y1y2

= (x1x2 − y1y2) + i (x1y2 + y1x2)

z1 z2 = r1eiφ1 r2e

iφ2 = r1r2 eiφ1+iφ2 = r1r2 ei(φ1+φ2)


Complex conjugation and inversion

The product z1 z2 in polar form gives r1r2 ei(φ1+φ2) which shows thatmultiplication in the complex plane is a combination of stretching (r1r2)and rotation (ei(φ1+φ2)).

Complex conjugate: z = x+ iy = x− iy = r e−iφ.

Modulus: |z| = r =√zz =

√x2 + y2.

Argument (phase): arg(z) = φ.

Reciprocal (inverse): z−1 =1

z=

z

zz=

z

|z|2 =z

r2.

Useful identities:

cos(α+ β) + i sin(α+ β) = (cosα + i sinα) (cosβ + i sinβ) ,

(cosnα + i sinnα) = (cosα + i sinα)n .


Sample problems

Representations of complex numbersHere you are supposed to convert between the two representations of complex numbers,namely, the cartesian representation z = x+ iy and the polar representation z = r eiφ.

Complete the following table.

Real part x Imag. part y Modulus r Phase φ

z1 3/2√

3/2

z11/z1

z2 3 2π/3

z21/z2

z1 + z2z1z2



Spreadsheet exercise on coordinatetransformation (homework)


Spreadsheet programs: relative addresses

Only basic knowledge of spreadsheet programs (such as Excel,OpenOffice-Calc, Numbers, . . . ) is required to solve the exercise.

Before you start working on the spreadsheet assignments, familiarize withthe mathematical and statistical functions that are available.

You are also advised to review the two different types of cell referencingthat are implemented in spreadsheet programs: addresses can be absoluteor relative.

Example of a relative address

Click on cell A1 and enter the number 10.Click on cell B1, enter the formula “=A1+5, and check if you get thenumber 15.Copy the formula in cell B1, paste it to B2, and you’ll get the value”5”. Check the formula in B2, it should “=A2+5”: Excel interpretsthe formula as ‘add 5 to the cell on the left” which is A2 and empty(the value is zero).


Spreadsheet programs: absolute addresses and graphing

An absolute address is characterized by $-symbol(s).

Formula “=$A1+5”: the column is fixed when the formula is copied.

Formula “=A$1+5”: the row is fixed when the formula is copied.

Formula “=$A$1+5”: both the row and the column remain fixed.

Finally, familiarize with the graphing options in your spreadsheet program.

This spreadsheet exercise deals with the transformation from spherical tocartesian coordinates. The application should then be used to find thedistances between stars in our neighborhood.

‖r2 − r1‖ =

∥∥∥∥∥∥

r2 cos δ2 cosα2

r2 cos δ2 sinα2

r2 sin δ2

−

r1 cos δ1 cosα1

r1 cos δ1 sinα1

r1 sin δ1

∥∥∥∥∥∥

= · · ·


Homework assignment

Stars in the cosmic neighborhood

The following table gives celestial coordinates (right ascension RA, declination DE, anddistance from Earth) of nine bright stars in our cosmic neighborhood.

Star Right ascension Declination Distance [pc]

Alpha Centauri 14h 39m 36.5s −60 50′ 2.3′′ 1.34Sirius 6h 45m 8.92s −16 42′ 58′′ 2.64Procyon 7h 39m 18.12s 5 13′ 30′′ 3.51Altair 19h 50m 47s 8 52′ 6′′ 5.13Vega 18h 36m 56.34s 38 47′ 1.3′′ 7.68Formalhaut 22h 57m 39.05s −29 37′ 20.1′′ 7.70Pollux 7h 45m 18.95s 28 1′ 34.3′′ 10.36Arcturus 14h 15m 39.67s 19 10′ 56.7′′ 11.24Capella 5h 16m 41.34s 45 59′ 52.8′′ 12.9

For each star, give right ascension and declination in radians, compute the cartesiancoordinates x, y, z, and then the distance to a reference star, namely Vega, in parallaxseconcs (parsec, pc). Plot the star positions in the (RA,DE) plane.



Appendix


Solutions of the sample problems

Dot products and cross products

The angle between the vectors V and W is given by

α = ∠(V ,W ) = arccos

(V ·WVW

)= arccos

(V ·W|V | |W |

)hence we have to compute the norms of the vectors as well as the dot product.

For the vectors in (a), V ·W = 2 · 2 + 0 · 2 + (−2) · 0 = 4, and

|V | =√

22 + 02 + (−2)2 =√

8 = 2√

2 ,

|W | =√

22 + 22 + 02 =√

8 = 2√

2 .

This gives cosα =4

8=

1

2, and thus α =

π

3= 60 .

For the vectors in (b), V ·W = 1 · 2 + 2 · 1 + 2 · (−2) = 4− 4 = 0, and

|V | =√

12 + 22 + 22 =√

9 = 3 ,

|W | =√

22 + 12 + (−2)2 =√

9 = 3 .

This gives cosα =0

9= 0, and thus α =

π

2= 90 .


Solutions of the sample problems (continued)

Dot products and cross products (continued)

For the vectors in (c), V ·W = 1 · (−2) + 1 · (−2) + (−2) · 4 = −12, and

|V | =√

12 + 12 + (−2)2 =√

6 ,

|W | =√

(−2)2 + (−2)2 + 42 =√

24 = 2√

6 .

This gives cosα = −12

12= −1, and thus α = π = 180 .

The cross products are as follows.

(a) V ×W =

20− 2

×

220

=

4− 4

4

.

(b) V ×W =

122

×

21− 2

=

−66

− 3

.

(c) V ×W =

11− 2

×

−2− 2

4

=

000

.



Vector identities

For the three terms in (1), the bac-cab rule yields

a× (b× c) = b(a · c)− c(a · b) ,

b× (c× a) = c(b · a)− a(b · c) ,

c× (a× b) = a(c · b)− b(c · a) .

When we compute the sum of the three contributions, we find that pairs of termscancel, and the final result is zero.

The identity (2) follows for A = a× b, B = c, and C = d.

To derive identity (3), we first rearrange the triple product of the vectors a× b,c, and d to yield

(a× b) · (c× d) = [(a× b) × c] · d = −[c× (a× b)] · d .

The term in square brackets can be rearranged by means of the bac-cab rule toobtain

c× (a× b) = a(c · b)− b(c · a)

which finally leads to identity (3).



Great-circle distance

If a point on a sphere of radius R has the spherical coordinates φ (longitude, azimuthalangle) and β (latitude, angular distance to the equator), its cartesian coordinates aregiven by x = R cosβ cosφ, y = R cosβ sinφ, and z = R sinβ.

(a) The angular distance between two position vectors r1 and r2 is given by

α = ∠(r1, r2) = arccos

(r1 · r2

|r1| |r2|

).

Since for two points on the sphere, the lengths of the vectors are |r1| = |r2| = R,we are left with the dot product

r1 · r2 =

R cosβ1 cosφ1

R cosβ1 sinφ1

R sinβ1

·

R cosβ2 cosφ2

R cosβ2 sinφ2

R sinβ2

= R2 [cosβ1 cosβ2 (cosφ1 cosφ2 + sinφ1 sinφ2) + sinβ1 sinβ2] .

Using the trigonometric identity cosφ1 cosφ2 + sinφ1 sinφ2 = cos(φ1 − φ2), wefinally arrive at

cosα = cos∠(r1, r2) = cosβ1 cosβ2 cos(φ1 − φ2) + sinβ1 sinβ2 .



Great-circle distance (continued)

(b) Inserting β1 = 50.96, φ1 = −7.18 and β2 = 53.166, φ2 = −0.287 into theformula derived in part (a) yields

cosα = 0.99653 ⇒ α = 4.775 = 0.0833 (radian) .

The epicentral distance in km is thus

RE · 0.0833 = 531 km .

(c) With β1 = 79 44′ = 79.73, φ1 = −71 47′ = −71.78 andβ2 = 53 10′ = 53.17, φ2 = 8 39′ = 8.65 the formula from part (a) yields

cosα = 0.80536 ⇒ α = 36.35 = 0.6345 (radian) .

(c) Here α denotes the geomagnetic colatitude (i.e., the angular distance of a locationto the geomagnetic pole). The geomagnetic latitude (i.e., angular distance to thegeomagnetic equator) is γ = 90 − α, hence γ = 53.65.



Representations of complex numbers

The completed table is as follows.

Real part x Imag. part y Modulus r Phase φ

z1 3/2√

3/2√

3 π/6

z1 3/2 −√

3/2√

3 −π/61/z1 1/2 −1/

√12 1/

√3 −π/6

z2 −3/2√

27/2 3 2π/3

z2 −3/2 −√

27/2 3 −2π/3

1/z2 −1/6 −√

3/6 1/3 −2π/3

z1 + z2 0√

12√

12 π/2

z1z2 −9/2√

27/2√

27 5π/6


Figure references

(1) Image credit: Wikipedia Commons, author Djexplo, date 30 May 2011, fileLatitude and Longitude of the Earth.svg. ”Illustration of geographic latitudeand longitude of the earth. Download from http://en.wikipedia.org on31 July 2013.

(2) Image credit: Wikipedia Commons, author Tfr000, date 14 March 2012, fileEarths orbit and ecliptic.png. ”Diagram of Earth’s orbit, showing eclipticplane and celestial equator on the celestial sphere. Download fromhttp://en.wikipedia.org on 31 July 2013.

(3) Image credit: Wikipedia Commons, author Andeggs, file 3D Spherical.svg.”3D spherical coordinates. Download from http://en.wikipedia.org on2 August 2013.

(4) Image credit: Wikipedia Commons, author Tfr000, date 15 June 2012, fileRa and dec on celestial sphere.png. ”Diagram of a star’s right ascensionand declination as seen from outside the celestial sphere. Depicted are thestar, the Earth, lines of RA and dec, the vernal equinox, the ecliptic, thecelestial equator, and the celestial poles. Download fromhttp://en.wikipedia.org on 31 July 2013.


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