Lösungen

1

1a

pA={x,0};
pB={0,1};
pC={0,0};
pE= 1/2 {-1,1};
pD= 1/2 {x,-x};
pF1=pA+{1,x};
pF=pB+ 1/2 (pF1-pB) //Simplify

pE-pD

(pE-pD).(pF-pC)//Expand

Rechtwinklig

1b

Norm[pE-pD]//Simplify

Norm[pF-pC]

(Norm[pE-pD]//Simplify) == (Norm[pF-pC]//Simplify)

2

2 a

A={{1,2,3},{3,2,1},{1,3,2}}; A//MatrixForm

Det[A]

2 b

B={{0,0,1,2,3},{1,0,u,v,w},{0,0,3,2,1},{0,2,g,h,j},{0,0,1,3,2}}; B//MatrixForm

Det[B]

2 c

Remove[a,b,c]

v={r,s,t}; w={a,b,c};
M={w,w+v,w+2v}; Det[M]

Zeilen linear abhängig, da Differenz von Zeilen linear abhängig, nach den Elementarsubstitutionen!

3

pA={12,10,0};
pB={9,7,12};
pC={-2,2,8};

3 a

(pB-pA).(pB-pC)

==> rechtwinklig

Norm[(pB-pA)]

Norm[(pB-pA)]//N

Norm[(pB-pA)]==Norm[(pB-pC)]

3 b

pD = pA+(pC-pB)

3 c

pM = pA+1/2 (pC-pA)

g[t_]:= pM+t Cross[pB-pA,pB-pC]; g[t]

solv1=Solve[Det[{pB-pA,pB-pC,g[t]-pM}]/3==1944,{t}]//Flatten

%//N

pS1=g[t]/.solv1

solv2=Solve[Det[{pB-pA,pB-pC,g[t]-pM}]/3==-1944,{t}]//Flatten

%//N

pS1=g[t]/.solv2

4

pA = {-1,9,8}; pB = {1,10,10}; pC = {-5,5,8};

k[x_,y_,z_]:= x^2+y^2+z^2-2z-8;
k[{x_,y_,z_}]:= k[x,y,z];
ϕ[λ_,μ_]:= pA+ λ (pB-pA) + μ (pC-pA);
pM = {0,0,1};
r=3;

k[x,y,z] == (x-0)^2 + (y-0)^2 + (z-1)^2 -r^2//ExpandAll

gϕ[t_]:= pM + t Cross[(pB-pA),(pC-pA)]

4 a

solv1 = Solve[ϕ[λ,μ]==gϕ[t],{λ,μ,t}]//Flatten

pS0 = gϕ[t]/.solv1

%//N

k[gϕ[t]]==0

solv2=Solve[k[gϕ[t]]==0,{t}]//Flatten

%//N

pS1 = gϕ[t]/.solv2[[1]]

pS2 = gϕ[t]/.solv2[[2]]

%//N

4 b

Norm[pS0-pS1]

%//N

Norm[pS0-pS2]

%//N

4 c

gMAB[t_]:= pM + t r (pB-pA)/Norm[(pB-pA)]

gMAB[1]

gMAB[-1]

gAB[t_]:= pA + t (pB-pA)

Simplify[(gAB[t]-gMAB[1]).(gMAB[1]-gMAB[-1])]==0

solv1=Solve[(gAB[t]-gMAB[1]).(gMAB[1]-gMAB[-1])==0,{t}]//Flatten

%//N

pP1 = gAB[t]/.solv1

%//N

solv2=Solve[(gAB[t]-gMAB[-1]).(gMAB[1]-gMAB[-1])==0,{t}]//Flatten

%//N

pP2 = gAB[t]/.solv2

%//N

4 d

pM = {0,0,1};
r=3;

4 e

Norm[Cross[pA,pM]]/2

Norm[Cross[pA,pM]]/2 //N