COURSE OF PROJECTIVE GEOMETRY

§ 5: answers

*O17*

i) λ(a_{2}b_{3}-a_{3}b_{2},a_{3}b_{1}-a_{1}b_{3},a_{1}b_{2}-a_{2}b_{1}),
short: λ__a__⊗__b__. Indeed, __a__⊗__b__ is normal of the plane subtended by __a__ and __b__.

ii) __l__⊗__m__. Indeed, __l__⊗__m__ is direction vector of the intersecting line of two planes with normals through __l__ and __m__, respectively.

*O18*

i) det(__a__,__b__,__c__)=0. Indeed, this is the condition that __c__ belongs to the plane subtended by __a__ and __b__.

ii) det(__l__,__m__,__n__)=0. Indeed, three planes through the origin go through one line if their normals lie in one plane.

*O19* If det(__a__⊗__a__',__b__⊗__b__',__c__⊗__c__')=0 then
det((__a__⊗__b__)⊗(__a__'⊗__b__'),(__a__⊗__c__)⊗(__a__'⊗__c__'),(__b__⊗__c__)⊗(__b__'⊗__c__'))=0.

*O20* If det(__a___{1},__a___{2},__a___{3}) = det(__b___{1},__b___{2},__b___{3}) = 0, then
det((__a___{1}⊗__b___{2})⊗(__a___{2}⊗__b___{1}),(__a___{1}⊗__b___{3})⊗(__a___{3}⊗__b___{1}),
(__a___{2}⊗__b___{3})⊗(__a___{3}⊗__b___{2})) = 0.

*O21* Start from λ(a,b,c) on *l*, this is λ(a,-2,1) or λ(1,0,0). The image point is λ(13-3a,16-4a,2a-8). This is the point λ(a,-2,1) if
3a-13 = a(8-2a). We get a =(5__+__√129)/4.