20. zadatak

Collect[x+y+z+λ*(2x-2z+3),{x,y,z}]

nπ={1+2 λ,1,1-2 λ};

nα={1,-4,-8};

Solve[((Abs[nπ.nα])/(Norm[nπ]*Norm[nα]))==Cos[(π/4)], λ]

y+z (1-2 λ)+3 λ+x (1+2 λ)/.{λ->-(1/792)}

TraditionalForm[%==0]

nπ2={2,0,-2};

ArcCos[((Abs[nπ2.nα])/(Norm[nπ2]*Norm[nα]))]

Rešenja su π: ((395 x)/396)+y+((397 z)/396)-(1/264)==0 i π2: 2x-2z+3=0

21. zadatak

tM={1,1,1};

d=(5/(Sqrt[14]));

Collect[5(x-2)-(y-11)==0,{x,y}]

Collect[5(z-2)-(y-11)==0,{x,z}]

Collect[1+5 x-y+λ*(1-y+5 z),{x,y,z}]

f[{x_,y_,z_}]:=1+5 x+y (-1-λ)+λ+5 z λ

n={5,-1-λ,5 λ};

Solve[((Abs[f[tM]])/(Norm[n]))==d,λ]

1+5 x+y (-1-λ)+λ+5 z λ/.{λ->(2/3)}

TraditionalForm[%==0]

1+5 x+y (-1-λ)+λ+5 z λ/.{λ->(3/2)}

TraditionalForm[%==0]

((Abs[1-1+5 *1])/(Sqrt[(1^2)+(5^2)]))

Rešenja su: α1: 5 x-((5 y)/3)+((10 z)/3)+(5/3)==0  , α2: x-((5 y)/2)+((15 z)/2)+(5/2)==0

24. zadatak

nα={1,2,-3};

nβ={2,-3,4};

ugao=ArcCos[((Abs[nα.nβ])/(Norm[nα]*Norm[nβ]))]

ArcCos[8 (Sqrt[(2/203)])]//N

%/ Degree

25. zadatak

nα={1,-4,2};

nβ={1,2,-3};

nγ={1,0,-1};

vp=Cross[nβ,nγ]

ugao=(π/2)-ArcCos[((Abs[nα.vp])/(Norm[nα]*Norm[vp]))]

ugao//N

%/ Degree

30. zadatak

f[{x_,y_,z_}]:=x-y+2z-3

tA={1,-2,0};

tB={-1,2,3};

tC={2,1,3};

AB=tA+t*(tB-tA)

f[AB]

Sve tačke duži AB pripadaju ravni α.

AC=tA-s*(tC-tA)

f[AC]

f[AC] = 0 za s = 0, a za s = 0 dobijamo tačku A.
-> Nijedna tačka duži AC, osim tačke A, ne pripada ravni α.
Nema potrebe da proveravamo za duž BC.
Presek ravni α i trougla ABC je duž AB.

Solve[x-y+2z-3==0,z]

Show[
Graphics3D[{
{Red,Opacity[0.5],EdgeForm[Thick],Triangle[{tA,tB,tC}]},
{ Text[Style["A", Large], tA+{0.2,0.2,-0.2}]} ,
{ Text[Style["B", Large], tB+{0.2,0.2,0.4}]} ,
{ Text[Style["C", Large], tC+{0.2,0.2,0.2}]}
}],
Plot3D[(1/2) (3-x+y),{x,-3,3},{y,-3,3}, Mesh->None],
 Axes->True]