问题 解答题
已知
i
j
分别是x轴,y轴方向上的单位向量,
OA1
=
j
OA2
=10
j
,且
An-1An
=3
AnAn+1
(n=2,3,4,…)
,在射线y=x(x≥0)上从下到上依次有点Bi=(i=1,2,3,…),
OB1
=3
i
+3
j
且|
Bn-1Bn
|=2
2
(n=2,3,4…).
(Ⅰ)求
A4A5

(Ⅱ)求
OAn
OBn

(III)求四边形AnAn+1Bn+1Bn面积的最大值.
答案

(Ⅰ)∵

An-1An
=3
AnAn+1
AnAn+1
=
1
3
An-1An

A4A5
=
1
3
A3A4
=(
1
3
)2
A2A3
=(
1
3
)3
A1A2
=
1
27
(
OA2
-
OA1
)=
1
3
J
.(3分)

(II)由(1)知

AnAn+1
=
1
3n-1
A1A2
=
1
3n-3
j

OAn
=
OA1
+
A1A2
+…
An-1An
=
j
+
A1A2
+…+
An-1An

=

j
+9
j
+3
j
+…+
1
3n-3
j
=
j
+
9[1-(
1
3
)
n-1
]
1-
1
3
j
=
29-(
1
3
)
n-4
2
j
.(6分)

∵|

Bn-1Bn
|=2
2
Bn-1Bn均在射线y=x(x≥0)上,

Bn-1Bn
=2
i
+2
j
.∴
OBn
=
OB1
+
B1B2
+
B2B3
+…+
Bn-1Bn
=3i+3
j
+(n-1)(2
i
+2
j
)

(III)∵|

AnAn+1
|=
1
3n-3
,△AnAn+1Bn+1的底面边AnAn+1的高为h1=2n+3.

又|

BnBn+1
|=2
2
An(0,
29-(
1
3
)
n-4
2
)到直线y=x的距离为h2=
29-(
1
3
)
n-4
2
2

∴Sn=

1
2
•(2n+3)•
1
3n-3
+
1
2
•2
2
29-(
1
3
)
n-4
2
2
=
29
2
+
n
3n-3
,(10分)

而Sn-Sn-1=

n
3n-3
-
n-1
3n-4
=
-2n+3
3n-3
<0,

∴S1>S2>…>Sn>…

∴Smax=S1=

29
2
+
1
3-2
=
29
2
+9=
47
2
.
(12分)

单项选择题
单项选择题