质:kC+md=0,即 b=0 整理后为 )a+(k-m)b=0. 可,b、共,故可,b、性无关所重 3k+2m=0 解得k 即z,d、性无关 6.、明。个向量k1-k2b,k2b-kE,k7-k回共面 交明:、等否 (k17-k2b)+(k2b-k37)+(k37-k1a)=0 可这3个向量、性关所重共面 7.O/、个定点、明:对、在直上的3个点A,B,C,点M位1面ABC上的充分-性 条件/存在实数k1,k2,k3,质得 OM=k1OA+k2OB+k3OC, E k1+k2+k3=1 交明:已/A、B、C点共,故AB,AC、性无关.任意点M位-′面ABC上当且仅当 AM,AB,AC共面即:AM,AB,AC、性关,当且仅当存在、全为0的实数m,m2m3,质 C=0, 当且仅当对定点O m1(OM-OA)+m2(0B-OA)+m3(OC-OA)=0, 当且仅当 显然m1≠0,、然与BAC、性无关矛盾因此若记 k (m1+m2+m3),k2 OM=k10A+k20B+koC 且k1+k2+k3 8.O/、个定点、明:点M位一△ABC上(包括它的边)的充分-性条件/存在负实数 k1,k2,k3,质得 OM=k10A+koB+k3OC, Ek1+k+k3=1 交明:延长AM,-可交BC-D点因此AM=1AD,其中0≤1≤1、-D在段BC上根据 例2.1,存在实数m1,m2,质得 +m 1 AM: G k￾ m ': k −→c + m −→d = 0,  3k −→a + k −→b + 2m−→a − m −→b = 0, +" (3k + 2m) −→a + (k − m) −→b = 0. N< −→a , −→b U(t, ! −→a , −→b t&,*, #$ ( 3k + 2m = 0 k − m = 0 -P k = m = 0,  −→c , −→d t&,*. 6. ST4f k1 −→a − k2 −→b , k2 −→b − k3 −→c , k3 −→c − k1 −→a (. : NV) (k1 −→a − k2 −→b ) + (k2 −→b − k3 −→c ) + (k3 −→c − k1 −→a ) = 0, >w 3 f t&e*, #$(. 7. O Hf, ST: <UkH.ty 3 f A, B, C,  M /< ABC y0@& 121k2 k1, k2, k3, 'P −−→OM = k1 −→OA + k2 −→OB + k3 −→OC, ? k1 + k2 + k3 = 1. :  A￾ B￾ C 4U(t, ! −→AB, −→AC t&,*. ￾
 M /< ABC yb?cb −−→AM, −→AB, −→AC (, : −−→AM, −→AB, −→AC t&e*, b?cb1kU3" 0 2 m1, m2, m3, ' m1 −−→AM + m2 −→AB + m3 −→AC = 0, b?cb< O G: m1( −−→OM − −→OA) + m2( −→OB − −→OA) + m3( −→OC − −→OA) = 0, b?cb m1 −−→OM = (m1 + m2 + m3) −→OA − m2 −→OB − m3 −→OC.  m1 6= 0, UB −→AB, −→AC t&,*45. !O: k1 = 1 m1 (m1 + m2 + m3), k2 = − m2 m1 , k3 = − m3 m1 , J −−→OM = k1 −→OA + k2 −→OB + k3 −→OC, ? k1 + k2 + k3 = 1. 8. O Hf, ST:  M /< 4ABC y (6788) 0@&121kn92 k1, k2, k3, 'P −−→OM = k1 −→OA + k2 −→OB + k3 −→OC, ? k1 + k2 + k3 = 1. : :; AM, @> BC < D . !O −−→AM = l −→AD, < 0 6 l 6 1. N< D ktx BC y, => j 2.1, 1k2 m1, m2, 'P −→OD = m1 −→OB + m2 −→OC, m1 + m2 = 1, m1, m2 > 0. < −−→OM = −→OA + −−→AM = (1 − l) −→OA + l −→OD = (1 − l) −→OA + lm1 −→OB + lm2 −→OC. · 6 ·