令k1=1-l,k2=lm1,k3=lm2,即得 OM=k,0A+k20B+k30C, k1+k2+k3=1, k,k2, k3>0 反之,不妨设k1≠1,解方程组 1-1=k1 l=1-k1 m1=k2可得{m1=1-k Im2= k3 n. 则有 m1+m2=1,m1,m2≥0,0<l≤1. 令 OD=m,oB +m,O 则D点在线段BC上.由 DM=(1-1)0A+10D 可以得出AM=1AD,因此M在线段AD上,从而在△ABC上 9.证明:任意不同的三点A,B,C共线的充分必要条件是存在不全为零的实数k1,k2,k3,使得 0=k1OA+k20B+k3OC, Ek1+k2+k3=0 证明:A、B、C共线,当且仅当LAB+mAC=0(,m都不为零),当且仅当 OA)+m(OC-O 当且仅当 -(L+ m)OA+lOB +mOC=0. 令k1=-(+m),k2=l,k3=m,显然它们不全为零,且: k,0A+k20B+k30c=0, k1+k2+k3=0 10.证明:任意不同的四点A,B,C,D共面的充分必要条件是存在四个不全为零的实数,使得 0=k0A+k0B+kaoC+kOD. k+k+katk=o 证明:A、B、C、D共面当且仅当AB,ACD线性相关,当且仅当有不全为零的数1,m,n使 LAB+mAC+nAD=0 当且仅当 l(OB-OA)+m(OC-0A)+n(OD-OA)=0 当且仅当 (L+m+n)Oa+loB+mOC +nOD=0 记k1=-(+m+m),k2=l,k3=m,k4=n,显然它们不全为零,使得 610A+k20B +k30C+kOD=0, k1+k2+k3+k=0I k1 = 1 − l, k2 = lm1, k3 = lm2, P −−→OM = k1 −→OA + k2 −→OB + k3 −→OC, k1 + k2 + k3 = 1, k1, k2, k3 > 0. M9, U? k1 6= 1, -@AB    1 − l = k1 lm1 = k2 lm2 = k3 >P    l = 1 − k1, m1 = k2 1 − k1 , m2 = k3 1 − k1 , JG m1 + m2 = 1, m1, m2 > 0, 0 < l 6 1. I −→OD = m1 −→OB + m2 −→OC, J D ktx BC y. N −−→OM = (1 − l) −→OA + l −→OD >$P% −−→AM = l −→AD, !O M ktx AD y, C%k 4ABC y. 9. ST: ￾
UC4 A, B, C (t0@&121kU3"o2 k1, k2, k3, 'P 0 = k1 −→OA + k2 −→OB + k3 −→OC, ? k1 + k2 + k3 = 0. : A￾ B￾ C (t, b?cb l −→AB + m −→AC = 0 (l, m mU"o), b?cb l( −→OB − −→OA) + m( −→OC − −→OA) = 0, b?cb −(l + m) −→OA + l −→OB + m −→OC = 0. I k1 = −(l + m), k2 = l, k3 = m, 8U3"o, ?: k1 −→OA + k2 −→OB + k3 −→OC = 0, k1 + k2 + k3 = 0. 10. ST: ￾
UCl A, B, C, D (0@&121klfU3"o2, 'P 0 = k1 −→OA + k2 −→OB + k3 −→OC + k4 −→OD, ? k1 + k2 + k3 + k4 = 0. : A￾ B￾ C￾ D (b?cb −→AB, −→AC, −→AD t&e*, b?cbGU3"o l, m, n ': l −→AB + m −→AC + n −→AD = 0, b?cb l( −→OB − −→OA) + m( −→OC − −→OA) + n( −→OD − −→OA) = 0, b?cb −(l + m + n) −→OA + l −→OB + m −→OC + n −→OD = 0.  k1 = −(l + m + n), k2 = l, k3 = m, k4 = n, 8U3"o, 'P k1 −→OA + k2 −→OB + k3 −→OC + k4 −→OD = 0, k1 + k2 + k3 + k4 = 0. · 7 ·