试题:
已知A、B、C三点共线,则对空间任一点O,存在三个不为零的实数λ、m、n使λ
OA
+m
OB
+n
OC
=
0
,那么λ+m+n的值等于______.

答案:

我来补答
∵A、B、C三点共线,∴存在实数k,使得
AB
=k
BC

AB
=
OB
-
OA
BC
=
OC
-
OB

OB
-
OA
=k(
OC
-
OB
),化简整理得:
OA
-(k+1)
OB
+k
OC
=
O

∵λ
OA
+m
OB
+n
OC
=
O

∴①当k=-1时,比较系数得:m=0且λ=-n,所以λ+m+n=0
②当k≠-1时,可得
λ
1
=
m
-k-1
=
n
k
,得m=(-k-1)λ,n=kλ
由此可得:λ+m+n=λ+(-k-1)λ+kλ=0
综上所述,λ+m+n的值为0
故答案为:0
 
 
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