问题 解答题
已知斜三棱柱ABC-A′B′C′,设
AB
=
a
AC
=
b
AA′
=
c
,在面对角线AC′和棱BC上分别取点M、N,使
AM
=k
AC′
BN
=k
BC
(0≤k≤1),求证:三向量
MN
a
c
共面.
答案

如图所示:

AN
=
AB
+
BN
=
AB
+k
BC

=

AB
+k(
AC
-
AB

=

a
+k(
b
-
a
)

=(1-k)

a
+k
b

AM
=k
AC′
=k(
AA′
+
AC
)=k
b
+k
c

MN
=
AN
-
AM
=(1-k)
a
-k
c

又∵向量

a
c
不共线,∴
MN
a
c
共面.

名词解释

ADR

单项选择题