Lösungen / Solutions

1

v={5,4};
vPerp[v_]:={-v[[2]],v[[1]]};
vPerp[v]

2

Remove[a,b]

a={5,4}; b[x_,y_]:={x,y};
a.b[x,y]

vLen[v_]:=Sqrt[v.v]; vLen[a]

gl1 = ( a.b[x,y] == vLen[a]  5 Cos[30 Degree])

gl2 = ( a.vPerp[b[x,y]] == vLen[a]  5 Cos[60 Degree] )

Solve[{gl1,gl2},{x,y}]

%//N

3

Remove[a,b]

vLen[v_]:=Sqrt[v.v];vPerp[v_]:={-v[[2]],v[[1]]};

a={8,3}; b[x_,y_]:={x,y};

gl1 = ( a.b[x,y] ==  vLen[a]  3 )

gl2 = ( vPerp[a].b[x,y] ==  vLen[a]  Tan[36 Degree] )

Solve[{gl1,gl2},{x,y}]

%//N

4

Remove[a,b]

a={7,-3}; b={2,12};

A = Abs[a.vPerp[b]]

5

Remove[a,b]

a[y_]:={5,y}; b={10,6}; c={1,8};

Solve[a[y].b==b.c,{y}]

6

Remove[a,b]

a ={10,4}; b={5,-2}; c=a-b;

Solve[b.c==vLen[c] proj, {proj}]

%//N

7

Remove[a,b,g]

a ={1,6,5}; b={5,-2,1}; g[t_]:= t a;

solv7 = Solve[(g[t]-b).a == 0, {t}] // Flatten

H = g[t] /.solv7

%//N

8

Remove[a,b,g]

a ={1,6,5}; b={5,-2,1}; g[t_]:= t a;

solv8 = Solve[(g[t]-b).a == vLen[g[t]-b] vLen[a] Cos[30 Degree], {t}] // Flatten

H = g[t] /.solv8

%//N

9

Remove[a,b,g]

a ={1,6}; b={5,-2}; c={3,5}; g[t_]:= t a;

solv81 = Solve[(g[t]-b).a == 0, {t}] // Flatten

H1 = g[t] /.solv81

%//N

solv82 = Solve[(g[t]-c).a == 0, {t}] // Flatten

H2 = g[t] /.solv82

%//N

vLen[H2-H1]

%//N

10

Remove[a,b,d,g]

<<Graphics`Arrow`

o={0,0};a={5,4};
d[α_] := {{Cos[α Degree],-Sin[α Degree]},{Sin[α Degree],Cos[α Degree]}};
d[20]//MatrixForm

pS = d[20].a;    d[20].a//MatrixForm

d[20].a//N//MatrixForm

Show[Graphics[{Arrow[o,a],
    Hue[0], Arrow[o,pS]}],AspectRatio->Automatic];

pSS=pS+a; pSS//N//MatrixForm

pSSS=d[-20].pSS; pSSS //N//MatrixForm

pSSSS = pSSS-a; pSSSS //N//MatrixForm

Show[Graphics[{Arrow[o,a],
    Hue[0], Arrow[o,pS], Hue[0.3], Arrow[pS,pSS], Hue[0.5], Arrow[o,pSS],Arrow[o,pSSS],Hue[0.8], Arrow[pSSS,pSSSS],Arrow[o,pSSSS],Line[{a,pSS}],Line[{a,pSSS}]}],AspectRatio->Automatic];

11

<<Graphics`Polyhedra`;
Show[ Polyhedron[ Tetrahedron ]];
Show[ Polyhedron[ Cube ]];
Show[ Polyhedron[ Hexahedron ]];
Show[ Polyhedron[ Octahedron ]];
Show[ Polyhedron[ Dodecahedron ]];
Show[ Polyhedron[ Icosahedron ]];

a

vLen[v_]:=Sqrt[v.v];

?Append

p1={1,0,0}; p2=Append[d[120].{1,0},0]

p3=Append[d[240].{1,0},0]

s = 2 Sqrt[1^2-(1/2)^2]

p4={0,0,Sqrt[s^2-1^2]}

v1=p4;v2=(p2+p3+p4)/3 - p1;
w=ArcCos[v1.v2/(vLen[v1] vLen[v2])] / Degree //N

b

p1={0,0,0}; p2={1,0,0}; p3={1,1,0}; p4={0,1,0};
p5={0,0,1}; p6={1,0,1}; p7={1,1,1}; p8={0,1,1};

v1=p7-p1; v2=p8-p2; v3=p5-p3; v4=p6-p4;

w1=ArcCos[v1.v2/(vLen[v1] vLen[v2])] / Degree //N

w1=ArcCos[v1.v3/(vLen[v1] vLen[v3])] / Degree //N

w1=ArcCos[v1.v4/(vLen[v1] vLen[v4])] / Degree //N

c

Raumdiagonalen senkrecht.
Flächendiagonalen = Raumdiagonalen des Würfels

12

Die Lösung ist im Script "Script Math Architektur III, Auswahl von Materialien", Seite 210ff, nachzulesen.