Lösungshinweise zum Kapitel 4

Winkelbestimmungen

 

40-1 Bodendreieck ABC gleichseitig, also ist $S=(0/(\sqrt{3}/3)/h)$, finde h aus SA'*SC = 0 = [-0.5 $\sqrt{3}/6$ -h]' * [ 0.5 $\sqrt{3}/6$ -h], h = sqrt(1/6) = 0.4082, Betrag(SA) = sqrt(1/2) = 0.7071

40-2 A,B.C.D = ($\pm$ 6 ($\pm$ 6 0 ), S=(0 0 10), n= SAxSB, Neigung = acos([0 0 1]*n/Betrag(n)) = 59.03 $^{\mathrm{o}}$. m= SBxSC, Winkel ABS-BCS = acos(m'*n/(Betrag(n)*Betrag(m)) = 74.65 $^{\mathrm{o}}$.

40-3 Im Würfel ABCD-EFGH bilden die Diagonalen AC, AF, AH mit der K-Diag. AG den Winkel 35.26 $^{\mathrm{o}}$. BE, DE, BD stehen senkrecht auf AG.

Vektorgeometrie in der Ebene

 

40-4 OA = [4 0]'; OB = [0 3]'; AB = [-4 3]' ; N = [3 4]'; en = [3/5 4/5]';
en'*OP - en'*[4 0] = en'*OP - 12/5 = 0; en'*[0 3]' - en'*[4 0] = 0;

40-5 siehe Beispiel-M-Files ``showhessenf.m''

Vektorgeometrie in 3D

 

40-6 A=(0 0 0) ; B= (1 0 0) ; C= (0.5 sqrt(3)/2 0) ; S= (0.5 sqrt(3)/6 sqrt(2/3) );
enABC = [0 0 -0.866]' ; enABS = [0 -0.8165 0.2887]'; enACS = [ -0.707 0.4082 0.2887]'; enBCS = [ 0.707 0.4082 0.2887]';

40-7 A=[4 0 0]', B=[0 4 0]', C=[-4 0 0]', D=[0 -4 0]', E=[0 0 4]' F=[0 0 -4]';
Linienzug: ABCDAFAEABFDEB; T= [0 /2 /2]' ; S=[4/3 4/3 4/3]; OP = OT + w*r = [0 /2 /2]'+ w*[4 4 4]'; V = [-2 -2 0] bei w=-0.5

40-8 4-seitige Pyramide aus gleichseitigen Dreiecken = oberer Teil eines Oktaeders h = s*sqrt(2)/2 = 7.07 m. ds = 7.07/tan(15*pi/180) 27.6157 m Schattengrenzen-Winkel = 2*15 Grad = 30 Grad.

40-9 Ebene durch ABC, Hesse'sche NF: n = cross(C-A,B-A) = [8 16 16]'; en = [1/3 2/3 2/3]' ; d = 2.666 = 8/3; x*1/3 = 8/3; x=8; sABC = [-4/3 7/3 7/3] sACD = [8/3 4/3 4/3] aABC = 24;whos aACD = 48 stot = (sABC*24+sACD*48)/(24+48)= [4/3 5/3 5/3]

40-10 Ebene in HesseNF en=[0.64 0.48 0.60]' ; en'*OP - 1.92 = 0.

Ausflug in höhere Dimensionen

 

40-11 function wbk = muldiwi(v,w) wbk = acos( v'*w/norm(v)/norm(w))*180/pi;
alle Winkel in Grad w(r,v)=30 ; w(r,f)=45; w(r,k1)=w(r,k3)=60;
w(v,f)=35.2644; w(v,k1)= w(v,k3)=54.7356; w(f,k1)=45; w(f,k3)=90;

40-12 w(k,f)= 45 ; w(k,r)= 90 ; w(k,v)= 63.435 ;
w(f,r)= 90 ; w(f,v)= 50.7685; w(r,v)= 39.2315;

40-13 Quadrat-Diagonalen: [1 1]' und [1 -1]'
Würfel-Kantenvektoren: [1 0 0]', [0 1 0]', [0 0 1]', je 4-fach
Würfel-Flächendiagonalen z.B. [1 1 0]' (2 Einsen, 1 Null), Raumdiagonalen [1 1 1]' drei Einsen.
Bei 16 Ecken gibt es 15*16/2 Verbindungsstrecken = 120. Dabei sind 4 Typen von Kanten, je 8 fach (3 konstante Bits in allen Kombinationen), 12 Arten von Flächendiagonalen (2 bits aus 4 auswählen = 6, dann je ++ oder +-), je 4fach, 16 3D-Körperdiagonalen (3 Bit aus 4, dann +++, ++-, +-+, -++) je 2 fach (Startpunkt bei 4. Dimension 0 oder 1) und 8 Arten 4D Raumdiagonalen, je einfach. (an jedem der 4 Plätze + oder -, geteilt durch 2 wegen vorwärts- rückwärts Identität). Also, 8+32+48+32=120.
Winkel Flächendiagonalen $\pm45^{\mathrm{o}}$ $\pm135^{\mathrm{o}}$ $\pm90^{\mathrm{o}}$
Winkel 3D-Raumdiagonalen $\pm54.73^{\mathrm{o}}$, $\pm125.26^{\mathrm{o}}$ $\pm90^{\mathrm{o}}$
Winkel 4D-Raumdiagonalen $\pm 60^{\mathrm{o}}$, $\pm 120^{\mathrm{o}}$

Kongruente Abbildungen der Ebene

 

40-14 My = [-1 0 ; 0 1]; Lmy = [ -5 -5 -6; 2 0 0];
Mp = [-1 0 ; 0 -1]; Lmp = [ -5 -5 -6; -2 0 0];
R45 = [0.707 -0.707; 0.707 0.707 ]; Lr45 = [ 2.12 3.53 4.24; 4.45 3.53 4.24];
R90 = [0 -1; 1 0]; Lr90 = [-2 0 0; 5 5 6];

Homogene Koordinatentransformationen in der Ebene

 

40-15 Ta1 = [1 0 -5 ; 0 1 5; 0 0 1]; Ta2 = [1 0 -5 ; 0 1 0; 0 0 1];
Ta3 = [1 0 -10 ; 0 1 0; 0 0 1]; Tb = [0 -1 0; 1 0 0; 0 0 1];
Tc1 = [0.707 -0.707 1.465; 0.707 0.707 -3.54; 0 0 1]; Lc1 = [3.58 5 5.707; 1.414 0 0.707; 1 1 1];
Tc2 = [1 -1 5; 1 0 -5; 0 0 1]; Lc2 = [ 3 5 5; 0 0 1; 1 1 1]

40-16

Ra1 = [0.707 -0.707 0; 0.707 0.707 0; 0 0 1]; Ra2 = Ra1^2; Ra3 = Ra1^3;
La1 = [ 2.12 3.54 4.24; 4.95 3.54 4.24; 1 1 1]; 
Tb1 = [1 0 -3.54; 0 1 -3.54; 0 0 1];
Tb1b= [1 0  3.54; 0 1  3.54; 0 0 1];
La2 = [-2 0 0; 5 5 6; 1 1 1]
Tb2 = [1 0 0; 0 1 -5; 0 0 1]
Tb2b= [1 0 0; 0 1  5; 0 0 1]
La3 = [ -4.95 -3.54 -4.24; 2.12 3.54 4.24;1 1 1]
Tb3 = [1 0  3.54; 0 1 -3.54; 0 0 1];
Tb3b= [1 0 -3.54; 0 1  3.54; 0 0 1];
Rb1 = [0.707 0.707 0; -0.707 0.707 0; 0 0 1]; Rb2 = Rb1^2; Rb3 = Rb1^3;
G1 = Tb1b*Rb1*Tb1*Ra1  % = [ 1 0 -1.47; 0 1 3.54; 0 0 1] 
G2 = Tb2b*Rb2*Tb2*Ra2  % = [ 1 0  -5; 0 1 5; 0 0 1] 
G3 = Tb3b*Rb3*Tb3*Ra3  % = [ 1 0 -8.54; 0 1 3.54; 0 0 1] 

40-17 Interessante Lösungsvariante: zuerst Gegenrotation um Ecke des ``L''
Tc = [1 0 -5; 0 1 0 ; 0 0 1] ; R = [cos(w) sin(w) 0; -sin(w) cos(w) 0 ; 0 0 1]; Tb = [1 0 5; 0 1 0 ; 0 0 1] ;
Gesamt Gegendrehung G= Tb*R*Tc
Dann Drehung um (0/0) um w R = [cos(w) -sin(w) 0; sin(w) cos(w) 0 ; 0 0 1]

40-18 a) Tges = [1 0 5*(cos(w)-1); 0 1 5*sin(w) ; 0 0 0]
b)Tr2= [cos(w) sin(w) 5*(cos(2*w)-1); -sin(w) cos(w) 5*sin(2*w) ; 0 0 1]

40-19

 
Dr = [0 10 5; 0 0 5*sqrt(3); 1 1 1] ; T = [1 0 -5; 0 1 -5*sqrt(3)/3; 0 0 1]
w = 2*pi/3 ; R = [cos(w) -sin(w) 0; sin(w) cos(w) 0; 0 0 1]
Tb = [1 0 5; 0 1 5*sqrt(3)/3; 0 0 1]; TT = Tb*R*T; TT^3
Dr2 = TT*Dr ;  Dr3 = TT^2*Dr;
[Dr Dr2 Dr3]

40-20

 
ABCD = [0 10 10 0 0; 0 0 2 2 0; 1 1 1 1 1]; T1a = [1 0 0; 0 1 -2; 0 0 1];
T1b = [0 -1 0; 1 0 2; 0 0 1]; ABCD1 = T1b*T1a*ABCD 
 T2a = [1 0 -2; 0 1 -12; 0 0 1];  T2b = [0 -1 2; 1 0 12; 0 0 1]; 
ABCD2 = T2b*T2a*ABCD1  
T3a = [1 0 -12; 0 1 -10; 0 0 1];  T3b = [0 -1 12; 1 0 10; 0 0 1]; 
ABCD3 = T3b*T3a*ABCD2 
T4a = [1 0 -10; 0 1 0; 0 0 1];  T4b = [0 -1 10; 1 0 0; 0 0 1]; 
T4b*T4a* T3b*T3a* T2b*T2a*T1b*T1a
ABCD4 = T4b*T4a*ABCD3
plot(ABCD(1,:),ABCD(2,:),'k'); hold on; plot(ABCD1(1,:),ABCD1(2,:),'r')
plot(ABCD2(1,:),ABCD2(2,:),'g'); plot(ABCD3(1,:),ABCD3(2,:),'b'); axis equal

40-21

 
F = [0 8 8 0 0; 0 0 4 4 0; 1 1 1 1 1]; Tr4 = [1 0 0; 0 1 -4; 0 0 1]
Mi = [1 0 0; 0 -1 4; 0 0 1]; F1 = Mi*Tr4*F
Tr44 = [1 0 -4; 0 1 -4; 0 0 1]; R = [-1 0 4; 0 -1 4; 0 0 1];
F2 = R*Tr44*F

Korrespondenz: A1=B2, B1=A2, C1=D2, D1=C2.

Homogene Koordinatentransformationen im Raum

 

40-22

 
A=[0 0 0 1]'; B=[10 0 0 1]'; C=[0 10 0 1]'; D=[0 0 10 1]'; 
Db = [B A C B D A C D];
Mp = [[-eye(3) [0 0 0]']; 0 0 0 1]; Dbt = Mp*Db;
plot3(Db(1,:), Db(2,:), Db(3,:)); hold on; 
plot3(Dbt(1,:), Dbt(2,:), Dbt(3,:),'r');   axis equal 
set(gca,'CameraPosition',[-150 45 70]); hold off

40-23

 
Q=[50 50 50 50 150 150 150 150 150 150  50 50  50 150 150 50 50; ...
   50  0  0  0   0   0   0  50  50  50  50 50   0   0  50 50  0 ; ...
    0  0 18  0   0  18   0   0  18   0   0 18  18  18  18 18 18 ; ...
    1  1  1  1   1   1   1   1   1   1   1  1   1   1   1  1  1  ];
plot3(Q(1,:),Q(2,:),Q(3,:),'k');  hold on
w = 15*pi/180;
Trl = [cos(w) -sin(w) 0 0; sin(w) cos(w) 0 0; 0 0 1 18; 0 0 0 1];
Q1 = Trl*Q;  plot3(Q1(1,:),Q1(2,:),Q1(3,:),'k');
Q2 = Trl*Q1; plot3(Q2(1,:),Q2(2,:),Q2(3,:),'k');
Q3 = Trl*Q2; plot3(Q3(1,:),Q3(2,:),Q3(3,:),'k');
Q4 = Trl*Q3; plot3(Q4(1,:),Q4(2,:),Q4(3,:),'k');
Q5 = Trl*Q4; plot3(Q5(1,:),Q5(2,:),Q5(3,:),'k');
Q6 = Trl*Q5; plot3(Q6(1,:),Q6(2,:),Q6(3,:),'k');
axis equal; hold off

40-24

 
W = [-2 -2 -2 -2  2  2  2  2  2  2 -2 -2 -2  2  2 -2 -2 ;... 
      2 -2 -2 -2 -2 -2 -2  2  2  2  2  2 -2 -2  2  2 -2 ;...
     -2 -2  2 -2 -2  2 -2 -2  2 -2 -2  2  2  2  2  2  2 ;...
      1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1 ]
plot3(W(1,:),W(2,:),W(3,:),'k');  hold on
for ix = 0:1 ; for iy = 0:1; for iz = 0:1
t = [2-4*ix 2-4*iy 2-4*iz]'; Ta=[eye(3) t; 0 0 0 1];
Tb = [-eye(3) -t; 0 0 0 1]; Wt = Tb*Ta*W;
plot3(Wt(1,:),Wt(2,:),Wt(3,:),'r'); 
end; end; end; axis equal; hold off

40-25 siehe Beispiel-M-Files

40-26

 
A=[-1 -1 -1 1]'; B=[1 -1 -1 1]'; C=[1 1 -1 1]';D=[-1 1 -1 1]';
E=[-1 -1  1 1]'; F=[1 -1  1 1]'; G=[1 1  1 1]';H=[-1 1  1 1]';
T=[0 0 6 1]'; U=[0 0 -6 1]'; N=[-6 0 0 1]'; S=[6 0 0 1]';
O=[0 6 0 1]'; W=[0 -6 0 1]';
liz = [E T G F T H E N D H N A E F S C G S B F ...
       W A B W E A U C D U  B C O H G O D C D A]
plot3(liz(1,:),liz(2,:),liz(3,:),'k') ; hold on
cols = ['r','b';'g','c']
sh = -15;
for k=1:2
  Tr= eye(4);   Tr(1,4)= sh;   li = Tr*liz;
  plot3(li(1,:),li(2,:),li(3,:),cols(1,k))
  sh = 15;
end
sh = -15;
for k=1:2
  Tr= eye(4);   Tr(2,4)= sh;   li = Tr*liz;
  plot3(li(1,:),li(2,:),li(3,:),cols(2,k))
  sh = 15;
end ; axis equal ; hold off

Lösungen zu den Selbsttests

 

T411 -sqrt(v'*v)   - Die unterste Zeie ist [0 0 1]   - P=[-1 0 0; 0 -1 0; 0 0 1]   - v*v'/(v'*v)

T412 [-4 -2 0]', [-2 0 2]', [0 2 4]'

T413 en = [0.48 0.64 0.6]'
E: en'*OP -1.92 = 0    F: en'*OP -3.84 = 0

T414 en1 = [0 -0.971 0.2425]' [-0.971 0 0.2425]'
Neigung, beide 75.96 $^{\mathrm{o}}$, Zwischenwinkel 86.63 $^{\mathrm{o}}$.

T415 V=[1 0 -4; 0 1 -4; 0 0 1]
R=[0 -1 4; 1 0 4; 0 0 1], T=R*V= [0 -1 8; 1 0 0; 0 0 1] R^4=I

T421 - $\vert v\vert \cdot \vert w\vert \sin(\alpha)$   - Der Abbildungsteil kongruenter Abbildungsmatrizen ist orthogonal   - R=[0 -1 0; 1 0 0; 0 0 1]}  - Die Ebenen liegen auf verschiedenen Seiten des Koordinatenursprungs.

T422 R=[0.5 0.5 0]' und S=[0.5 0.5 1]'

T423 Distanz = 9.6, OF=[5.76 6.144 4.608]'

T424 Neigung Fläche 70.5288 $^{\mathrm{o}}$ , Neigung Kante 54.7356 $^{\mathrm{o}}$,

T425 P = [0 8 2 0; 0 0 2 0];plot(P(1,:),P(2,:))
s=sqrt(2)/2 ; hold on; M=[s -s; s s]; plot(P(1,:),P(2,:))
for k=1:7; Q=M^k*P; plot(Q(1,:),Q(2,:))