If the vector calculation is used outside of physics

Extension of vector calculation - theoretical physics

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Mathematics preliminary course

F. Krause

Chapter 6

extensiontheVector calculation

The content of this chapter:

• 6.1

• 6.2

• 6.3

The scalar product

The vector product

Complex numbers

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Copyright F. Krause

1


Chapter 6.1: The scalar product

Overview of contents chap. 6.1

• 6.1.1 The component form of the scalar product

- 6.1.1a: The amount of a vector

- 6.1.1b unit vectors

- 6.1.1c The scalar product of perpendicular vectors

- 6.1.1d The calculation rules of the scalar product

• 6.1.2 The geometric form of the scalar product

• 6.1.3 Angle and projection

• 6.1.4 Application examples

- 6.1.4a Description of equations for planes in space

- 6.1.4b The angle between two straight lines in space

- 6.1.4c Other applications

- 6.1.4d Generalization of the scalar product

- 6.1.4e Term construction and formulas

- 6.1.4f Some triangular formulas

- 6.1.4g ​​Determination of intersections with the help of coordinate equations

With help the terms introduced so far theVector calculation are we in the Able to a number

to quantitatively describe and determine geometric quantities and objects in the configuration space:

Points, straight lines and parabolas, planes and intersections of such figures, meeting points

etc. Some more sizes that we use in the usual geometric handling are

however, vectorially inaccessible to us. This includes in particularthee the Concept of angle. How determined

about the angle between two geometric arrowsthe form? how can we

decide whether a straight line is perpendicular to a plane? Another deficit is that we do

not in the Are able to systematically determine the length of vectors. Othe equivalent to,

the distance between two points. The calculation rules introduced for vectors do not give any

Indication of how one could answer these questions.

A little more specifically: It is not yet possible for us to vectorially bisect the angle of a triangle

to describe as we did for the bisector.

We will see that these questions can be answered with the help of the scalar product to be introduced below

can answer. This is a construction that turns two vectors into a number.

The scalar product comes to the rules theVector calculation in addition, does not follow from them.

We choose the following route to the scalar product: We discuss an obvious question

and consideration of the linear equations. In doing so, you discover the construction you are looking for.

A little open play around as well as the awareness of the mentioned problems leads one

then to the sought structure.

2


6.1.1 The component form of the scalar product

(6.1.1) We begin with an obvious consideration of the linear systems of equations. In fact

let us consider a homogeneous linear 1 × 3 system of equations:

a 11 x 1 + a 12 x 2 + a 13 x 3 = 0.

The system of equations is given in the Have rule m == 1: A condition for three indeterminates. So it follows

k = 2. That is, one level is typically to be expected as the solution set.

¤ Why typically? When is the statement k = 2 wrong?

(6.1.2) So far we have combined the indefinite into a tuple and only the solutions of the

Systems interpreted geometrically as coordinate vectors. But it is not obvious, neither do the three

To combine coefficients into a triplet and to interpret them as a coordinate vector?

That turns out to be a fruitful idea. We use likethe Row vectors and set a = (a 11, a 12, a 13).

As long as we only use a fixed coordinate system, we leave out the index K. We provide the

following question:

• Suppose we interpret a as the coordinate vector of a geometric arrow. Then through

a, on the one hand, defines the system of equations and its solution set. Can we then

Solution leveltheon the other hand also determine geometrically without first solving the system?

Is there any geometric connection between a and the Solution level?

If you ask yourself what could be considered as a geometrical solution level at all,

Actually only the following possibility remains: That the solution plane is the plane perpendicular to a

by the origin is. Because the vertical plane and only this is already complete through a alone

certainly. All other levels require additional information to be fully defined.

One could also think of a (one, not the) plane that contains a. But there are

many of which liketheby no specialthes is excellent.

(6.1.3) Simple examples support this assumption immediately. Let a = e 3 = (0, 0, 1). The associated

The system of equations is x 3 = 0with the Solution x L (x, y) = xe 1 + ye 2. This is actually the one perpendicular to e 3

Level.

¤ Think about the case a = (1, 1, 0) for yourself.

(6.1.4) This leads one to consider the following construction, which is just the formation the left

page the The equation from the two triples describes:

• For a = (a 1, a 2, a 3) and b = (b 1, b 2, b 3) the following number (= scalar) should be formed:

a 1 b 1 + a 2 b 2 + a 3 b 3.

For example, if a = (1, 2, 3) and b = (4, −1, 0), the number 1 4 + 2 (−1) + 3 0 = 2 results.

(6.1.5) Abstracted as assignment:

We order each (ordered) pair of vectors a = (a 1, a 2, a 3) and b = (b 1, b 2, b 3) of vectors

assign a number from R 3 or R 3 K, which we denote by a · b and which we call the scalar product of a and b

want.

The calculation takes place via the assignment

(a, b) 7 → a b = a 1 b 1 + a 2 b 2 + a 3 b 3.

(6.1.6) With the help of this scalar product, our output equation is written a · x = 0. And the preliminary considerations

suggest the following assumption (hypothesis), which will actually be confirmed:

a · b = 0 is equivalent to the fact that a and b are relatedthe stand vertically.

(6.1.7) Please keep apartthe:

3


To load that later

• An inner link between two vectors turns them into a vector again.

talking vector product is an example.

• The scalar product turns two vectors into a number, a scalar.

• The multiplication of a vector by a number makes a scalar and a vector again one

Vector.

⇓ Next, let's just try what you can do with the can start a new assignment. physicshe call

like to mess around with something like that.

6.1.1a: The amount of a vector

(6.1.8) What can you do with the Start assignment? The second geometrically interesting case (besides a

perpendicular b) is a = b. Then both vectors have the same direction. For this our assignment gives

(a, a) 7 → a · a = a 2 1 + a 2 2 + a 2 3.

Two right-angled

Triangles.

(x, x) 7 → xx = x 2 + y 2 + z 2

x

| x |

x 2 + y 2

z

y

| x | 2 = (x 2 + y 2) + z 2

The sketch shows (through the double application of Pythagoras) that this - a · a - equals the square

the Length of the geometric arrow is a. Let us denote the length of the geometric arrow a with | a |

(= Amount of a), then we get:

The length othethe The amount of a is given by:

| a | = √ a a = p a 2 1 + a2 2 + a2 3

(6.1.9) Instead of a · a one likes to write a 2. This leads to the useful equation a 2 = | a | 2. Left

stands the scalar product of a with itself, on the right the square of a real number. So never do math

With the nonsensical equation √ a 2 = a.

Useful usage rule: a 2 = | a | 2 othe | a | = √ a · a.

(6.1.10) Expressions like a 3 o are not explained and meaningfulthe a 4, even if you put them in sloppy

Representations always likethe encountered. a is a vector, but a 2 is a number. Meaningful (and intended) terms are

then a 2 a and (a 2) 2.

(6.1.11) In physical texts one writes instead of | a | often simply a (thesame letter).

In particularthee for | r | Othe | x | if you write r. R stands for radius. (But not x for | x |)

(6.1.12) Let P and Q be two points of E 3 and x K P and xK Q ∈ R3 K corresponding coordinate vectors.

The distance between P and Q is understood to be the quantity ¯¯xK P - x K ¯

Q, the length of the difference vector.

This difference vector x K P −xK Q should be called a distance vector and not confuse the two. The shortest

The distance of a straight line from the origin is therefore a number, not a vector. (Compare question and figure from (3.3.11)).

4


6.1.1b unit vectors

(6.1.13) With the help of the amount one can choose any ³

a´ vectors (6 = 0) according to a formula according to direction and length

1

disassemble. To do this, just write a = | a |

| a |

. (Targeted factoring out, Section 1.2.3.)

| A vector of magnitude 1 is called a unit vector. For example, e 1 is one, as well as

e 2 and e 3. And generally 1

| a |

a is a unit vector in the direction of a.

For every a 6 = 0, 1 is a

| a |

a =

| a |

Unit vector in the direction of a.

¤ Determine the unit vectors for (1,1,0) and (1,1,1) and (1,2,3).

¤ What is the interpretation of a− b

| a− b |

(6.1.14) In physical formulas one likes to separate direction and magnitude in this way in vector terms

from one anotherthe. A typical example is the formula for the Coulomb force, which describes the force that causes a

electrical point charge (at the origin) exerts on a second with position vector r). The following applies to this force:

F =

e1e2

4πε 0 r

r = 3

e1e2 r

4πε 0 r 2 r.

¤ How great is the force if the force-generating charge is not in the origin, sonthen in one

Point with position vector R?

¤ Now we can also find the bisector between two given vectors. Consider it

even that theen direction through a

| a | + b

| b |

is determined. What is the difference to a + b

| a + b | ? Which one

additional factor has to be added in order to get the vector the Bisecting the triangle?

6.1.1c The scalar product of perpendicular vectors

m (6.1.15) After we have seen by playing around that the introduced assignment scalar product

useful, let's turn the Conjecture (6.1.6). To do this we have to do this in some way

determine how vertical standing is to be understood geometrically at all. We do this without further justification

and justification about Pythagoras, which we now use vectorially with the help of the concept of distance

be able to formulate:

Let a, b 6 = 0 geometric arrows.

Then a is perpendicular to b if and only if for the triangle generated by a and b the Pythagoras

applies. In formulas:

| a | 2 + | b | 2 ¯

= ¯a - ¯

b¯2

a

a

c = b-a

c 2 = a 2 + b 2

b b = | b |

(6.1.16) According to (6.1.9) we can rewrite this equation as follows:

a 2 + b 2 = (a - b) * (a - b).

The right-hand side of this equation first invites you to calculate it distributively. Is that permissible? We

have to ask us whether the distributive laws apply to the scalar product. We'll show that in a moment

the the Case is. More generally, the following calculation is permissible with the a b = b a is also used:

a 2 + b 2 = (a - b) * (a - b) = a 2 - a * b - b * a + b 2 = a 2 + b 2 - 2a * b

5


Taken together, this chain of equations implies 0 = a · b. Now we argue like this: If that

The triangle generated by a and b fulfills Pythagoras (1st equals sign!), the above calculation applies. And the

is called a · b = 0. The scalar product the both vectors is zero.

Conversely, if a b = 0, then it follows:

a 2 + b 2 = a 2 + b 2 - 0 = a 2 + b 2 - 2a · b = (a - b) · (a - b).

And that's right now the Pythagoras.

| (6.1.17) Is at least one the both vectors a othe b the Zero vector, then a · b = 0 also applies.

In this case the triangle degenerates. To avoid unpleasant case distinctions, one says

(as agreed) that the Zero vector is perpendicular to each vector.

⇑ (6.1.18) With this we can formulate the following summarizing result:

Let a, b from R 3 and R 3 K, respectively.

Then a · b = 0 applies precisely when a and b meetthe stand vertically.

! ⇑ Our original assumption, which provided the introduction to the considerations, is thus confirmed: Is

a 6 = 0, then there is the solution level the linear equation a x = 0 the Plane perpendicular

to a.

¤ Let a 6 = 0 with given a K = (a 1, a 2, a 3). You are looking for two independent vectors s 1 and s 2, which lead to a

stand vertically. (Two, not all!). You can write down a possible solution immediately - without an invoice,

s K 1 = (.,.,.) and sK 2 = (.,.,.). Namely?

6.1.1d The calculation rules of the scalar product


Does the scalar product actually fulfill the rules just used? That question remains to be examined.

(6.1.19) First the commutative law. It is obviously fulfilled:

a b = a 1 b 1 + a 2 b 2 + a 3 b 3 = b 1 a 1 + b 2 a 2 + b 3 a 3 = b a.

The equality in the Middle is crucial for justification. There a 1 b 1 = b 1 a 1 is used. This is

the commutative law for number multiplication.

(6.1.20) What about the distributive laws? We would like to prove a · (b + c) = a · b + a · c. We

calculate as follows, where we obviously denote the 1-component of a + b with (a + b) 1:

a (b + c) = a 1 (b + c) 1 + a 2 (b + c) 2 + a 3 (b + c) 3

= a 1 (b 1 + c 1) + a 2 (b 2 + c 2) + a 3 (b 3 + c 3)

!

= a 1 b 1 + a 1 c 1 + a 2 b 2 + a 2 c 2 + a 3 b 3 + a 3 c 3

= (a 1 b 1 + a 2 b 2 + a 3 b 3) + (a 1 c 1 + a 2 c 2 + a 3 c 3) = a b + a c.

At the beginning and the end the Bill will be likethe only definitions used. In the middle line (! =)

becomes - decisive for the validity the Calculation - the distributive law used for real numbers.

How do you finally come to the penultimate reshaping?

6


(6.1.21) In addition the The following insert on the tunnel method.

The tunnel method

Experience has shown that beginners have to check - othethe Verification - of hypotheses like

“For the scalar product, the distributive laws” initially have great difficulties. Though

can they understand the individual calculation steps, but cannot see how to arrive at them,

how to find this step yourself. Up in the Invoice is the penultimate step an example.

As an aid in thelike cases should be remembered.

The tunnel through the darkness may, yes, should be started from both sides!

The equation to be tested is known - as a hypothesis. Both sides are calculated - in ours

Case a · (b + c) and a · b + a · c - separately as far as possible and finally try the results

to match. In our case, you can easily get up to from both sides

the critical transformation and immediately sees the equality.

. In the In the final formulation of the proof, one then only has to transform the second part into

the written down in reverse order, which makes these deformations often difficult

appear. (In the extreme case it is composed the Readers to the head and asks: How is it the Author only on

did this ingenious transformation come about? I can never do that. What is wrong. You only have to have one

part the Invoice in the reverse direction.) For the validity equality is this

reversal the Order of course permitted.

We will come across a number of examples of how this method was used in the course of the course.

(Compare for example (6.2.14))

(6.1.22) The first distributive law is thus proven. The second follows via commutativity.

¤ Prove the following rule that we have already come up with from the Know matrix linearity (5.1.27):

(αa) b = α (a b) = a (α b) for α ∈ R.

(6.1.23) The rule turns out to be very useful. In general, it is advisable to consider common factors

possibly also to exclude main denominators from tuples. With an additional scalar product formation,

then thelike factors are simply pre-contracted. Notice what this rule does: Is

c = (1, 2, 2) with | c | = 5, then | 3c | = 3 x 5 = 15.

¤ To do this, calculate the scalar product the the following two vectors in two ways: x = (1 2, 1 3, 1 5) =

1

30

(15, 10, 6) and y = (14, 21, 35) = 7 (2, 3, 5).

One meets leithe always likethe an astonishing urge to do cumbersome arithmetic, probably because you

thereby without application the abstract formula (αa) · (β b) = αβ (a · b). Let's say ¡1

¡ 2 , 2 3 , 2¢ 1 ·

1

5 , 1 7 , − ¢ 3

7 =

1

10 + 2 21 - 3 14 = .... instead of ... 1 6 1

21+20−45

35

(3, 4, 3)(7, 5, 15) =

210

= − 2

105

(6.1.24) The distributive laws and the last proved rule are for computing with scalar products

extremely important. Their shape is analogous to thethe Linearity rules from the matrix area, only that

they apply here to both factors. It is therefore characterized as bilinearity. In summary received

we have the following set of calculation rules for the scalar product:

Commutativity a b = b a for all a, b ∈ R 3

a * (x + y) = a * x + a * y

Bilinearity (a + b) x = a x + for all a,

b x

b, x, y ∈ R 3

(αa) b = α (a b) = a (α and α ∈ R.

b)


This proves all the rules that we apply to execution the Needed Pythagorean calculus. This in

(6.1.18) is proven.

(6.1.25) Another pending problem: How do you usually calculate with the scalar product, what

are the usage rules to be remembered? Which transformations are to be avoided and which are faulty? For this

if one compares again with the usual arithmetic rules for real numbers, where one finds that more for

they valid rules are not transferrable. Take the associative law. So the two terms (a · b) c

and a (b * c). The first term is a vector in the direction of c, the second one in the direction of a. The expressions

cannot be generally the same. The associative law and the consequences derived from it apply

Not.

7


In summary, we get the following set of usage rules for dealing with the scalar product:

¨

¨

¨

With the help of the scalar product, only products from two vectors can be formed

become. The scalar product is carefully of the product of two numbers and of the product

of a vector with a number to distinguish.

One must never divide by a vector. That is, (a x) = b has many solutions.

All term transformations that only use the calculation rules from (6.1.24) are permitted.

In particularthee the rule Je applies to the product of two sumsthe with everyone being pure

Number factors can still be preferred. With it, many bills can be completely

perform analogously to real calculations.

(6.1.26) One finds violations of the first rule not infrequently. So you always encounter howthe Transformations

like ”(a · x) x = ax 2. Othe (ax) 2 - a 2 x 2 = 0. Something like that then leads to adventurous geometric

and physical consequences that are overlooked. For example: “Every triangle has the area

0.”

¤ Why are the following two terms permissible, no violation of the second rule?

(a x) 2 (a x) x

a 2 and

x 2.

The following term conversions, on the other hand, are cruel rule violations:

(a x) 2

a 2 = x 2 (a x) x

x 2 = a.

What about x2

| x | = | x | ?

¤ Calculate (a +2 b - 3c) (3a - 2 b + c). (Think about the sensible arrangement!)

¤ How can the scalar product be included in the method the Insert history diagrams? Cape. 3.1.1 and special

(3.2.6). Use it to analyze some terms and mathematical laws.

| So far we have always indicated the scalar product by a point, i.e. written a · b. This one

Point is usually left out (safely). You shouldn't skimp on brackets. It is better for (a · b) c

just write (a b) c, but not a · bc.

¤ From 4 number factors you can form 5 permissible brackets, from 5 you can form 14. (Question in (3.3.13).)

If the associative law applies, all these brackets give the same result. How many different

Assignments (arithmetic expressions) are obtained if one takes 4 or 5 vector factors and and occurs

Product symbol as a scalar product, as a product of a vector with a number, etc.the as

Interpreted product of two real numbers? (Fixed order the Factors. Use different ones

Names for the various products. Work with history charts.)

6.1.2 The geometric form of the scalar product

(6.1.27) We now come to an extremely important result, which gives us a geometrical interpretation

the Scalar product formation will deliver. Let a K and b K be two non-zero vectors from R 3 K. We educate

the configuration the Sketch. I.e. we split b K into one parallel to a K and one perpendicular to a K

Component. So b K = p K + s K, where p K should be parallel to a K and s K perpendicular to a K and thus p K

stands. The angle between a K and b Kthe we denote by ϑ. It just depends on the two

Arrows down, not from the choice the Coordinate axes.

Decomposition of a K into a to

b K parallel and perpendicular

Component: a K = p K + s K

(p K · s K) = 0 and p K || b K.

ϑ angle between a and b.

8


Then it follows from the Sketch immediately ¯¯p K¯¯ ¯

= ¯b K¯¯¯ ´

cos ϑ and thus p K =

³¯¯¯ b K¯

¯ cos ϑ

(6.1.18) that a K s K = 0. Now we calculate under utilization the Bilinearity as follows:

µ ´

a K b K = a K (p K + s K) = a K p K + a K s K = a K p K = a K

³¯¯¯ b


a

K

¯ cos ϑ

| a K |

=

´ ³¯¯¯ b


¡A K ¢ 2

¯ cos ϑ

| a K |

= ¯¯a K¯¯ ¯¯¯ b

K

¯ cos ϑ.

a K

. In addition, after

| a K |

It is a K coordinate vector of the geometric arrow a. Then ¯¯a K¯¯ = | a | . The length of a is the same

the number calculated using the scalar product ¯¯a K¯¯ = √ a K · a K.

(6.1.28) So it follows in total:

a K · b K ¯

= | a | ¯b ¯ cos ϑ

!! That is the promised geometric interpretation of the scalar product. The left side becomes coordinate-dependent

determined as before. The right side, on the other hand, has a purely geometric meaning,

there only the lengths the two arrows and the Angles between them.

(6.1.29) So far we only have the scalar product for tuples from R 3 or for coordinate vectors from R 3 K

Are defined. Now we can extend the definition to geometric arrows from V 3 0 and V 3. We sit

easy:

The geometric shape of the scalar product

For a, b from V 3 0 or V 3 ¯

be

a · b = | a | ¯b ¯ cos ϑ.

This allows the scalar product to be calculated for a configuration of two geometric arrows. In many

Examples - espthes thephysics - are the amounts and the Known angle and then you can do this

Apply formula.

(6.1.30) If one introduces a full coordinate system K in addition to the origin, then one can do that

Scalar product eitherthe geometric othe calculate over the components. And both ways result

same number:

a b = a K b K.

This means that the calculation rules formulated above also apply to the geometric form of the scalar product

be valid.

(6.1.31) One introduces atheit is Cartesian coordinate system L with the same origin, then

änthen the components the involved vectors, but not the geometric configuration. I.e. it

applies

a K · b K = a L · b L.

¤ Specify this fact using a simple example in the Level!

The latter result is of great value to the applications. Because this ensures that you are with help

of the scalar product can describe geometric and physical phenomena independently of the observer.

(6.1.32) Work is a typical quantity that is described by a scalar product of two vectors.

Imagine dragging a stone up a slope. The work is a measure of that

Effort that you have to make. Observer dependency would mean the following:

At the top of the hanging wall it says the Observer, the defines the coordinate axes. He likes you personally, turns

he set the axes so that the Labor value becomes small, you can go up without effort. He doesn't like you

he turns the axes so that you can barely move forward.

If one disregards descriptive variables such as the coordinate vector, experience has shown that physical variables are

not observer dependent, shouldn't be. At least one should ask the question the eventual

Carefully analyze observer dependency in each case. We saw in (4.2.7) that the by

the observer-dependent center of gravity vector specified point from E 3 was observer-independent. Othe:

"Free vectors" do not depend on the Choosing the origin of coordinates, bound ones do it in whole

certain way.

¤ What about the Speed?

9


¤ It is not uncommon for the scalar product to be formed by analogy as follows:

(a 1, a 2, a 3) · (b 1, b 2, b 3) = (a 1 b 1, a 2 b 2, a 3 b 3).

Use simple concretizations to show that the resulting vector (= geometric arrow) at fixed

The origin of the observer depends on what disqualifies this formation as an inherently obvious vector product.

(Imagine the observer here as a wind god!)

(6.1.33) The argumentation from (6.1.27) delivers another result. We have the vector b there

decomposed into a component parallel to a and a perpendicular component. b = p + s. For p and s one can find formulas

indicate which are often useful: ³

´

It was p =

³¯¯¯ b¯¯¯ cos ϑ´

a ¯

a

| a |

= | a | ¯b ¯ cos ϑ = (a · b)

| a | 2 a

a. So if a and b are known, one can use the

2

Determine projection p by forming scalar products. s is then obtained by s = b - p.

The decomposition into parallel and perpendicular components.

⇒ Let a and b be vectors with b 6 = 0.

! Then one can clearly divide a into a component p parallel to b and one perpendicular to b

Disassemble component s.

This means that a = p + s with p parallel to b and p · s = 0.

p and s are given by the following formulas:

!!! p = (a b)

b 2 b and s = a - p.

¤ (6.1.34) Let K be a Cartesian coordinate system and x ∈ V 3 or V 3 0. Prove and discuss

the following formulas:

x = (x e 1) e 1 + (x e 2) e 2 + (x e 3) e 3

So x 1 = (x e 1), x 2 = (x e 2) and x 3 = (x e 3),

6.1.3 Angle and projection

(6.1.35) At the beginning it was said that we are not yet in the Are able to vectorially determine angles. The

is on nowtheThe geometric form of the scalar product solves the problem immediately, one has to use the formula

only dissolve for cosϑ:

⇒ Let a and b be two vectors 6 = 0.

⇒ ϑ let the Angle between the two vectors with 0 ≤ ϑ ≤ π.

!!! Then cos ϑ = (a b)

| a || b | .

The scalar products required to determine the angle can be specified on a case-by-case basis using the component form or similarthe

the geometric shape can be determined.

Example: a K = (1, 1, 1) and b K = (2, −3, 2). It follows (without any additional written invoice, immediately!)

1

cosϑ = √ √

3 17

. We strongly advise against invoices and representations the of the following type,

which the application the To make a formula for a writing retreat that looks something like this:

(a b)

cosϑ = ¯

=

| a | ¯b ¯

a 1 b 1 + a 2 b 2 + a 3 b

p 3

a

2

1 + a 2 2 + p a2 3 b

1

1 + b 2 2 + =

b2 3

1 · 2 − 1 · 3+1· 2


12 +1 2 +1 2p 2 2 + (- 3) 2 +2 2 = .....

There is no reasonable justification for this kind of pointless job creation, the one leithe too often

encountered.

(6.1.36) We now consider a fixed plane with coordinate system K. Space the associated coordinate vectors

is R 2 K. We put a circle with radius 1 around the origin. Everthe Point on the circle

10


then determines a unit vector e (ϕ), where ϕ the Angle between the 1 direction and the vector

should. Out the From the sketch, one immediately reads the specified representation of the unit vector. (See (4.5.20))

y

e (ϕ)

1 e (ϕ) = cos (ϕ) e 1 + sin (ϕ) e 2

sinϕ

ϕ

e K (ϕ) = (cos (ϕ), sin (ϕ))

cosϕ x position and coordinate vectors

The point of the unit circle

Formation of scalar products gives cos (ϕ) = e 1 · e (ϕ). Now you can see immediately how the Value of the scalar product

(with fixed vector lengths) with the angle änthet: You start with the value 1 at the angle zero.

At acute angles you get a positive value under 1. Vertical belongs to the value zero, a blunt one

Angle gives a negative value and anti-parallelism finally gives the value -1.

1

0.5

θ

0

-0.5

0.5 1 1.5 2 2.5 3

-1

(6.1.37) Summary: The scalar product is a formation that consists of two vectors of the same type

a number that makes a scalar. The evaluation can eitherthe via the component form othe about the

geometric shape. The equality of the result is ensured by (6.1.31).

The calculation rules for dealing with the scalar product are summarized in (6.1.23) and (6.1.25).

You should use them to transform terms, but watch out for differences to number arithmetic.

With the help of the scalar product, the geometric terms length, distance and angle can be converted into the vectorial

Insert description scheme. In particularthee are two vectors on top of each other if and only thenthe perpendicular,

if its scalar product is zero.

A useful construction is the decomposition of a vector into a vector perpendicular to and to a second vector

parallel component. The formulas are given in (6.1.33).

In the following we give some application examples that show how wide the spectrum of problems is

which is covered by a formalism like that of the scalar product. The first application is in

the Rounding off our entry-level idea into the topic. The second is of the routine type, an angle determination.

In the Third application, an angle is determined via an additional idea, i.e. not routinely. And

at the fourth is a generalization of the formalism.

6.1.4 Application examples

6.1.4a Description of equations for planes in space

(6.1.38) Let us return to our entry problem, the geometric interpretation the 1 × 3 equation

back. The homogeneous equation is now written (a · x) = 0 with a 6 = 0. The solution set is the geometrical one

uniquely defined plane of all vectors perpendicular to a. Transition to the inhomogeneous equation (a x) = b

means according to the general results in (5.3.24) parallel displacement the Solution set. How

far is to move in parallel? Let X be some solution to our equation (solution role, so a · X = b).

We decompose X into a component parallel and perpendicular to a and find X = (X a)

a

a + s = b

2

a

a + s. that

2

obviously results for every s the a solution to a perpendicular plane. We calculate with bilinearity

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the square the Length of X and because of a · s = 0 we immediately find: X 2 = b2

a

+ s 2. This means, the shortest

2

Distance belongs to s = 0 and the Shortest distance vector is simply b

a

a. To this vector is

2

move the solution level in parallel.

(6.1.39) The result can still be startedtheinterpret s. In the We had level in Chap. 1.6 straight lines

initially not vectorial, butthen described by associated equations. The solutions of y-mxb = 0

gave the coordinate vectors the Points the Straight lines. How does the generalization look to three

Dimensions? Now the solutions of a · x = b result in the coordinate vectors of a plane perpendicular

to a. A scalar linear equation describes a plane in three dimensions, not a straight line, like

one may naively generalize.

(6.1.40) The plane equation, like the straight line equation, can be broken down into a number of different

Bring shapes from which one can read off certain geometric configuration variables. We

start with the general equation a · x = b and go through equation transformation to three special ones

Forms over. We use the bilinearity (6.1.24) of the scalar product.

Transformation Equation Geom. Interpretation

Division by | a | with n = a

α = b

| a |

n x = α

| a |

shortest distance

n unit vector

the Plane from the origin

Division by b with A A

= a = (

b

A x = 1

1 a, 1 b, 1 c).

a, b, c intercepts

Division by a 3 with N = 1 a 3

a N x =

b z = b

a 3

- a 1

a 3

x - a 2

a 3

y

a 3

Height function

Note: The penultimate equation is written as x a + y b + z c

= 1. The cut with the z-axis is obtained

by setting x and y to zero. It remains zS c

= 1othe z s = c.D.h. As stated, c gives the associated

Intercept. And the last one allows the transition to a parametric representation the level

(x, y) 7 → (x, y, z (x, y)) = (x, y,

b

a 3

- a 1

a 3

x - a 2

a 3

y).

6.1.4b The angle between two straight lines in space

(6.1.41) an onthehe problem circle: Determine the angle between two (possibly crooked)

Straight lines. Is that a sensible problem? Yes, you start with two intersecting and not parallel

Straight lines. Moves a straight line in the direction perpendicular to both straight lines, remains the "Angle"

obviously meaningful and unchangedthet. To do this, imagine that you are heading towards the Direction of displacement

look at the configuration! You can move until the two straight lines meet.

¤ Exception: The two straight lines are parallel. What's wrong Does that matter?

(6.1.42) The actual execution the Determining the angle is then unproblematic. Let g and h be those

Straight lines and x g (a) = x 0 + a d as well as x h (b) = y 0 + b f associated parameterizations. The angle between

the straight line is equal to the angle between the direction vectors d and f. So cos (γ) = (d · f)

| d || f | .

What if instead of f one had taken the opposite vector - f? Then you would have that

negative cos value and thus the supplementary angle π - γ.

Beware of instead the Direction vectors the starting point vectors othe even the position vectors in

insert the angle formula. That makes a lot of nonsense. (Origin-dependent angle)

6.1.4c Other applications

(6.1.43) How does one get the straight line that is perpendicular to two given independent vectors

stands? (Was required in (6.1.41).) Given the results of this chapter, the answer is not difficult. Be

a and b the two given vectors. a x = 0 determines all vectors perpendicular to a. You also want

perpendicular to b, one must still b x = 0forthen. Dh. the straight line we are looking for follows as the solution of

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following 2 × 3 system a x = 0, b x = 0. If = 2, then k = 3− 2 = 1. The solution set is

a straight line with the desired properties.

(6.1.44) How big is the Tetraetheangle? The tetraethe is a regular symmetrical body

with four corner points. We put the origin in the center of the body. The position vectors the

four corner points should be denoted by x 1, ..., x 4. For reasons of symmetry, they all have the same 4

Length a = | x 1 | = | x 2 | = | x 3 | = | x 4 | . The angle between two different of these position vectors is the

Tetraetheangle τ. For reasons of symmetry, it also always has the same value. The center agrees

the center of gravity (with the same masses). Then x 1 + x 2 + x 3 + x 4 = 0 applies to our choice of origin.

Othe x 1 = −x 2 - x 3 - x 4. From both sides of this equation we form the square (in the sense of the

Scalar product). Note that as a result the Bilinearity according to such products of two the usual rule