Question

已知|

OA

|=1，|

OB

|=k，∠AOB=

2π

3

，點(diǎn)C在∠AOB內(nèi)，

OC

•

OA

=0，若

OC

=2m

OA

+m

OB

，|

OC

|=2

3

，則k=（　�。�

• A．1
• B．2
• C．

  3

• D．4

Answer 1

分析：由已知|

OA

|=1，|

OB

|=k

OC

•

OA

=0，

OC

=2m

OA

+m

OB

，|

OC

|=2

3

兩邊同時(shí)平方可得m，k的關(guān)系式，然后再由

OC

-2m

OA

=m

OB

兩邊平方可得關(guān)于m，k的關(guān)系，從而可求k

解答：解：|

OA

|=1，|

OB

|=k
由

OC

•

OA

=0可得∠AOC=90°
∵∠AOB=

2π

3

，點(diǎn)C在∠AOB內(nèi)
∴

OA

•

OB

=|

OA

||

OB

|cos

2π

3

=-

1

2

k，且∠BOC=30°
∵

OC

=2m

OA

+m

OB

，|

OC

|=2

3

∴

OC

2=4m2

OA

2+4m2

OA

•

OB

+m2

OB

2
∴12=4m²+4m²×(-

1

2

k)+（km）²
∵

OC

=2m

OA

+m

OB

∴

OC

-2m

OA

=m

OB

同上平方可得，12+4m²=m²k²
兩式聯(lián)立可得，k=4
故選D

點(diǎn)評(píng)：本題主要考查了向量的數(shù)量積的運(yùn)算性質(zhì)的應(yīng)用，解題的關(guān)鍵是對(duì)已知式子兩邊同時(shí)平方結(jié)合數(shù)量積的定義進(jìn)行求解．屬于向量知識(shí)的綜合應(yīng)用