204 4 Geometry and Trigonometry
579.Does there exist a bijectionf of (a) a plane with itself or (b) three-dimensional
space with itself such that for any distinct pointsA,Bthe linesABandf (A)f (B)
are perpendicular?
580.On so( 3 )we define the operation∗such that ifAandBare matrices corresponding
to the vectors−→a =(a 1 ,a 2 ,a 3 )and
−→
b =(b 1 ,b 2 ,b 3 ), then theijentry ofA∗Bis
equal to(− 1 )i+ja 4 −jb 4 −i. Prove the identityCBA−BCA=(A∗C)B−(A∗B)C.581.Prove that there is a bijectionffromR^3 to the set su( 2 )of 2×2 matrices with
complex entries that are skew symmetric and have trace equal to zero such that
f(−→v ×−→w)=[f(−→v),f(−→w)].We present two applications of vector calculus to geometry, one with the dot product,
one with the cross-product.
Example.Given two trianglesABCandA′B′C′ such that the perpendiculars from
A, B, ContoB′C′,C′A′,A′B′intersect, show that the perpendiculars fromA′,B′,C′
ontoBC,CA,ABalso intersect.
Solution.This is theproperty of orthological triangles. Denote byOthe intersection of
the first set of three perpendiculars, and byO′the intersection of perpendiculars fromA′
andB′. Note that if the vector
−→
XYis orthogonal to a vector−−→
ZW, then for any pointPin
the plane,
(−→
PX−
−→
PY)·
−−→
ZW=
−→
XY·
−−→
ZW= 0 ;
hence
−→
PX·
−−→
ZW=
−→
PY·
−−→
ZW. Using this fact we can write
−−→
O′C′·−→
OB=
−−→
O′A′·
−→
OB=
−−→
O′A′·
−→
OC=
−−→
O′B′·
−→
OC=
−−→
O′B′·
−→
OA=
−−→
O′C′·
−→
OA.
Therefore,
−−→
O′C′·(
−→
OB−
−→
OA)=
−−→
O′C′·
−→
AB=0, which shows thatO′C′is perpendicular
toAB. This proves that the second family of perpendiculars are concurrent.
Example.LetABCDbe a convex quadrilateral,M, Non sideABandP,Qon side
CD. Show that ifAM=NBandCP=QD, and if the quadrilateralsAMQDand
BNPChave the same area, thenABis parallel toCD.
Solution.Throughout the solution we refer to Figure 24. We decompose the quadrilaterals
into triangles, and then use the formula for the area in terms of the cross-product.
In general, the triangle determined by−→v 1 and−→v 2 has area equal to half the magnitude
of−→v 1 ×−→v 2. Note also that−→v 1 ×−→v 2 is perpendicular to the plane of the triangle, so for