圍 1-1 向量 題 ( 每題

範 圍 一、單選題

（ ）

解答 解析

（ ）

解答 解析

二、填充題

1. 設

K

a +2 解答

解析

高 範

圍 1-1 向量 題 ( 每題

1.如圖﹐下

於零向量

3

由圖 PO+

K

∴ CO QR

K K

+

2.下圖為五 向量的內 (1)

K

a

1

內積

K K

a⋅ b 與 e

K

在

K

題 ( 每題

2

K K K

b + c = 0

4 3

∵

K K

a +2 b

2

c = −

K K

∴ c

K

=

高雄市明誠中 的基本應用 題分 )

下面哪一選項

量﹖ (1) AO

K

QO QO +

K K

= +

0 R=

K K

﹒

五個等長的向 內積最大﹖

(2)

K

b (3)

K

b 即 a

K

的長

K

a

方向上的

題 10 分 )

K

0

且

K

a =4

0 b + c =

K K K

2

2 a − b

K K

₌

48=4 3﹒

中學 高二數 用 班級 二年

座號

項中的向量與

K

O

(2) BO

K

OR

K K

=QR_﹐

向量﹐試問向

)

K

c

(4)

K

d 長度 a

K

與 b

K

的 5 個投影量

4 ﹐

K

b =2

﹐

﹐∴ c

K

= −

2

4 a + b

K K

數學平時測 年____班

與另兩個向量

(3) CO

K

(4

向量 a

K

與下

K

(5)

K

e

﹒

K

b 在 a

K

方向

量中﹐以 a

K

若 a

K

﹐ b

K

2 a − b

K K

2

4 a b +

K K

⋅

測驗 日期 姓

名

量 PO

K

﹑ QO

K

4) DO

K

(5) E

下列哪一個

向上的投影量

的投影量最

K

b

之夾角為

cos 60 b ⋅ °

K

=

期：99.09.0

之和等

EO

K

﹒

量之乘積﹐又

最大﹐ a

K K

⋅ a 60° ﹐則 c

K

16 16 4

= + + ⋅

09

又 a

K

﹐ b

K

﹐

K

最大﹒故選

K

_{= _______}

4 2 1 48

⋅ ⋅ ⋅ =2

K

c

﹐ d

K

選(1)﹒

______﹒

8 ﹐

2. 已知向量 a

K

﹐ b

K

滿足條件﹕

K

a =1

﹐

K

b =2

﹐ 2 a

K K K K

− b = a + b

﹐求

(1)設 a

K

﹐ b

K

的夾角θ ﹐則 cosθ = ____________﹔(2) a

K K

+ b =

____________﹒

解答 (1)1

4;(2) 6 解析 (1)

2 2

2 a

K K K K

− b = a + b ^{2 2 2 2}

4 a b 4 a b a b 2 a b

⇒

K K K K K K K K

+ − ⋅ = + + ⋅ 4 4 4 a b 1 4 2 a b

⇒ + −

K K

⋅ = + +

K K

⋅ _⇒ 1 a⋅ b =2

K K

﹐

∴

1 2 1

cos 1 2 4

a b a b

θ = ^⋅ = =

⋅

K K K K

^﹒

(2)

2 2 2

2 1 4 2 1 6

a + b = a + b + a⋅ b = + + ⋅ =2

K K K K K K

﹐ ∴

K K

a + b = 6

﹒

3. 已知

K

a =4

﹐

K

b =3

﹐ a

K

﹑ b

K

的夾角為 60° ﹐求 (1) a

K K

+t b

的最小值為____________﹔(2)此時 t= ____________﹒

解答 (1) 2 3 ;(2) 2

− 3 解析

2

a +t b

K K

=⎛⎜⎝ a +t b ⎞ ⎛⎟ ⎜⎠ ⎝⋅ a +t b ⎞⎟⎠

K K K K

² 2 ²

2

a t a b t b

=

K K K K

+ ⋅ + =4²+ ⋅ ⋅ ⋅2 4 3 cos 60°⋅ + ⋅t 3² t² =9t²+12t+16

2 2

9 12

t 3

⎛ ⎞

= ⎜⎝ + ⎟⎠ +

當 2

t= − 時﹐可得最小值為 2 3 ﹒ 3 4. 設

K

a =3

﹐

K

b =5

﹐

K

c =7

﹐且

K K K K

a + b + c = 0

﹐求﹕ b

K K

⋅ c

之值﹒____________

解答 65

− 2

解析 (1)∵

K K K K

a + b + c = 0

﹐∴ b

K K K

+ c = − a

﹐ |

K K K

b + c |²= −| a |²

﹐

2 2 2

2

b + b ⋅ c + c = a

K K K K K

﹐

25 2+

K K

b ⋅ c +49=9 65 b c 2

⇒

K K

⋅ = −

﹒

5. 平面上有三向量 OA

K

﹐OB

K

﹐OC

K

﹐若OA

K

=1

﹐OB

K

= 3

﹐OC

K

=2

且OA OB OC

K K K K

+ + = 0

﹐則△ ABC

的面積為 解答

解析

6. 設 a

K

= 解答

解析

7. 已知正三 (1) AB AC

K K

⋅ 解答

解析

8.

K

u =3 解答 解析

為_________

3 3 2 ∵ OA

K

+OB

K

OA OB

K K

+ =

2

OA

K K

+OB

△ ABC =

1﹐

K

b =2 2

5 a b

⎛ + ⎞

⎜ ⎟

⎝ ⎠

K K

2

⇒

K

a − 1 t 1

⇒ − + 三角形 ABC C=

K

________

(1)25 2 ;(2) − (1) AB AC

K K

⋅

(2) AB GA

K K

⋅

﹐

K

v =4

﹐ 120°

2

2 u

K K

+ v

____﹒

0 B OC+ =

K K K

OC OA

= −

K K

⇒

2

2 B + OA OB⋅

K K K

3 △ OAB =

﹐且 a b

K K

⋅

a t b

⎛ ⎞

⋅⎜⎝ − ⎟⎠

K K

t⎛⎜⎝ a⋅ b ⎞⎟⎠+

K K

4t 0

− = ﹐∴

每邊長為 5 _____﹔(2) A

K

25

− 2

5 5 cos 6 C= ⋅ ⋅

A= −AB AG⋅

K K K

2 u

K K

+ v =

2

28 4 u

= ⇒

K

⇒36 1+

﹐∴ O 為△

2

A OB+ = O

K K K

2

B= OC

K K

_⇒

3 1 1 3 2

⎛ ⋅ ⋅ ⎞

⎜ ⎟

⎝ ⎠

=1

K

﹐若 a

K

⎞ =0

⎟⎠

a⋅ b −t b

K K K

∴ 2

t= ﹒ 5

﹐ G 為重心 AB GA⋅ =

K K

___

0 25

° = 2 ﹒ 5 5 3 G ⎛ 2

= − ⋅ ⋅⎜⎜⎝

=2 7﹐則

K

2 2

u + v +

K K

(

16+4 3 4 c⋅ ⋅

△ ABC 之重心

2

OC

K

﹐

⇒ 1 3 2OA+ +

K

⋅ 3 3

⎞ = 2

⎟⎠ ﹒

+

K

b

與 a

K

−

2

=0

K

心﹐求下列各 __________

3 2 cos 30 2 3

⋅ ⎞⎟⎟⎠⋅

K

u

﹐ v

K

之夾

4

K K

u ⋅ v =2

)

cosθ =28 ⇒ 心﹐

4

OB O

⋅

K

= ⇒

K

−t b

K

互相垂

各式之值﹕

﹒

0 25

° = − 2 ﹒

夾角為_____

8

cosθ 2

⇒ = −

0 OA OB

K K

⋅ = _⇒

垂直﹐則 t 值為

________﹒

1

2﹐∴θ =12

OA OB

⇒

K K

⊥

﹐

為________

20° ﹒

_____﹒

9. 平行四邊 解答 解析

10.設 a

K

與

解答

解析

11.正六邊形

(1) CD

K

=x 解答 解析

(1 (2

12.已知 a

K

(1)若 a

K

﹐ 解答 解析

邊形 ABCD 中 5

AC BD⋅ = ⎜⎛⎝A

K K K

與 b

K

是平面

91

2

2 a + b

K K

3 a

K K

− b 形 ABCDEF

x a

K K

+y b

﹐ (1)

(

^−2,1

)

^;(

1)

CD

K K

=AD−

∴x= −2

﹐ 2)

EA

K K

= −BD

=3 AB

K

∴s=3

﹐

=3

K

﹐ b

K

=

K

b

的夾角 (1) 6− ;(2)5 (1) a

K K

⋅ b =

(2)2 a

K K

+ ∴ a

K K

⋅ b

中﹐AB=2

AB BC+ ⎞ ⎛⎟ ⎜⋅ B

⎠ ⎝

K K K

向量﹐ a

K

=

2

21 9

= ⇒ +

2

9 9 4

= ⋅ + −

中﹐令 AB

K

= 則

( )

^{x y, = _}

(2)

(

^{3, 2−}

)

2 AC BC

−

K K

= −

(

1 y= ⇒ x D= −⎛⎜⎝AD A−

K K K

2AC 3 a

−

K K

=

(

2 ,

t= − ⇒ s

= ﹐求 4 為120° ﹐則

5

3 4 cos12

= ⋅ ⋅

2

52 b = ⇒

K

0 b =

K

﹐∴

K

﹐AD= ﹐3

BC CD+ ⎞⎟

⎠

K K

₌

= ﹐ b =3

K

4 4 4 a⋅ +

K K

⋅ b

( )

6 1 91

− − = ﹐

=

K

a

﹐ AC

K

= ___________

−AC

K

2⎛ b

= ⎜⎝

K

) ( )

, 2,1

x y = − AB⎞

⎟⎠

K

= −⎜⎛⎝2 B 2

a − b

K K ﹐

) (

^{3, 2}

)

t = −

a ⋅ b =

K K

___

0° = − ﹒6

2

4 a

K K

+ b a + b =

K K K

則 AC BD

K K

⋅

2 2

BC

K K

− AB 2 ﹐且 a

K

+

21 b =

K

﹐∴

﹐∴ 3 a

K K

− b

=

K

b

﹐求

__﹔(2) EA =

K

b − a ⎞⎟⎠− b

K K K

﹒

BC AB− ⎞⎟⎠

K K

₌

﹒

____﹔(2)若

2

4 a b +

K K

⋅

2

a + b =

K K

= _________

2

9 4 5

= − =

2

K

b = 21

1 a⋅ b = −

K K

91 b =

K

﹒

s a t b

=

K K

+

2 a b

= − +

K K K

2 AC AB⎛

− ⎜⎝ −

K K

若 2 a

K K

+ b =

=52⇒36+

2 2

a + b

K K

____﹒

5

﹒

﹐則 3 a

K K

−

﹐

﹐則

( )

^{s t, = _}

K

b

﹐

B⎞ +⎟⎠ AB

K K

=2 13﹐則

K

16 4 a b + +

K K

⋅

2

= ﹒ 5

b =

K

______

__________

a + b =

K K

__

52 b =

K

﹐

_______﹒

___﹒

_______﹒

13.△ABC (1) AB AC

K K

⋅ 解答

解析

14.

設

ABC

(1)

AB+

K

(2)

AC+

K

(3)

⎛AC

⎜⎝

K

解答

解析

15.

如右圖 (1)

PE

(2)求

△ 解答 解析

中﹐AB=2 C=

K

________

(1) 3

− ;(2)2 (1) 2 3 cos⋅ ⋅

(2) AB+

K

CD

為一平行

AD

K

₌

_____

+DA

K

=

_____

C BA+ ⎞⎟⎠+DB

K K

(1) AC

K

;(2)

(1) AB

K

+AD

K

(2) AC D

K K

+ (3) AC B⎛⎜⎝ +

K K

圖

ABCD

為一

:

E AP=

____

△PAD

之面

(1) 3 : 8 ;(2) (1)令 AD=

△ABG

又

△A ⇒AP:

又 AE = ∴PE A: (2)△APD

2 ﹐BC= ﹐4 _____﹔(2)

34 22

sA= ⋅ ⋅2 3

2

2AC

K

= 4

行四邊形﹐

________﹔

________﹔

=

________

AB

K

;(3) AB

K

D=AB BC+

K K K

DA

K K K

=AC CB+ BA⎞ +⎟⎠ DB=

K K

一個平行四 ______﹔

面積

:

.

ABC ) 2 :11

=4t﹐

延伸

G

中﹐

∵

~ PD △GP

: PG=AD F

1 2AG

= ﹐令

3 : 8 AP= ﹒

8 D=11△ AD

3 AC= ﹐

2 AB+ AC =

K K

2 2 2

3 4 2 2 3 + − =

⋅ ⋅

4 4 9 4 2

⎛− + ⋅ + ⎜

⎝

求下列各式

_____﹒

C=AC

K K

﹒ B=AB

K K

﹒ AC CD

⎛ + ⎞+

⎜ ⎟

⎝ ⎠

K K

四邊形﹐

CE

CD

之面積

=

伸

AE

和

BC //

CE AB

﹐

PF

4 : 7 FG= ⇒

22 AG= s﹐

DE 8 1

=11 2⋅ △ 則

= __________

3

− ﹒ 2 3 4 36

− ⎞ = +2 ⎟⎠

式﹕

DB AD +

K K

= +

: 1:1 E ED=

且

=

_________

C

交於

G

﹐

∴¹ 2

CE AB

= =

4 AP=11AG

⇒AP=

△ ACD 4

=11⋅ ___﹒

6 6− = 34

DB

K K

=AB

﹒

且

BF FC: =

__﹒

CG BG

= ⇒CG

G ﹐ 4 22 8 11⋅ s= s

1

2 ABCD

⋅

.

﹒

1: 3

﹐則

=4t

﹐

﹐ 1

AE= ⋅2

面積 2

=11

.

22s=11s﹐

.

ABCD^面積積﹐

16. a

K

﹐ b

K

則 t= ____

解答 解析

17.設 a

K

= 解答 解析

如

c

得

表 18.設四邊形

(1) AC

K

= _

解答

∴△ AP

K

b

為平面上二 ________﹒

− 1 ∵^⎡_⎢⎣

K

^{a +}

(

^t2

∴^⎡_⎢⎣

K

^{a +}

(

^t

2

⇒

K

a +

3 3

t t

⇒ + +

⇒ 為實數t 0 b ≠

K

﹐若 30°

∵ a

K K

= b

如右圖﹐△

1 AO

K K

= 2 a

1 DO

K K

=2 a cd代入 a

K

得 2

K

b ⋅cos sin 45 cos°⋅ θ

∵ a

K K

+ b >

表示 a

K

﹑

K

形 ABCD 中

___________

(1) 37 ;(2

PD 面積：

.

二向量﹐ a

K

)

2+3 b ⎤ ⎡⎥ ⎢⎦ ⎣⊥

K

)

2 3

t + b ⎤ ⎡⎥ ⎢⎦ ⎣⋅

K

(

^{2 3}

)

t t +

K

b 4 0 + = ^⇒

(

^t

數﹐∴ t= − 若 a

K K

+ b −

K

b

﹐∴平行

△

AOD

中﹐

b b

+

K K

= ⋅

b b

−

K K

= ⋅ b a

+ − −

K K K

sθ−2

K

b ⋅sin cos 45 s θ− °⋅

a b

>

K K

− ⇒

K

b

之夾角為

﹐ AB

K

=1

﹐

___﹔(2)四邊

2)5 3 2

.

ABCD^面積

⊥ b

K K

﹐ a

K

a t b

⎡ + ⎤

⎢ ⎥

⎣ ⎦

K K

﹐ a +t b ⎤⎥⎦=

K K

2

= （ a ⊥0

K

) (

²

1 t t 4

+ − +

1 ﹒ a − b =

K K

行四邊形 ABC

cosθ ⇒

K

a sinθ ⇒

K

a −

2

b a

−

K K

= nθ = 2 a

K

sin 1

θ =2 ⇒s 0 a b

⇒

K K

⋅ >

為銳角﹐∴ 4 2 AD

K

=

﹐

邊形 ABCD 之

積 2 :11= ﹒

=2

K

﹐ b

K

=

﹐

0

⊥

K

b

﹐∴ a

K

)

4 = 0

2 a

K

﹐則

K

CD 為菱形﹐

2

b b

+ =

K K K

2

b b

−

K K

=

K

a

﹐ 1 cos

2 θ

⇒ ⋅

( )

sin 45° −θ

﹐

45° − =θ 30°

AB

K

_{與 AD}

K

_之

之面積為___

= ﹐若 a +1

K

0 a ⋅ b =

K K

）⇒

K

a

﹑ b

K

之夾

且 AC⊥BD

cosθ

⋅ "

K _c

sinθ

⋅ "

d

1 sin θ− 2 θ

1

= ﹐ 2

θ 15

⇒ = ° ﹐ 之夾角為 60°

__________

(

^t2+3

) K

^b

與

⇒^{4 t t+}

(

²⁺

夾角為_____

D ﹐

1 θ= ﹐ 2

∴ a

K

﹑ b

K

﹐AC

K K

=3A

﹒

與 a

K K

+t b 互

)

3 1 0⋅ =

________﹒

K

之夾角為 3 2

AB+ AD

K K

_﹐則

互相垂直﹐

30° ﹒

則

解析

(

(

19.已知 a

K

解答

解析

20.四邊形 A (1)△AB (2)若 AD

解答

解析

(1)

2

3 AC

K

=

=9

(2)

ABCD

面

K

= ﹐ b =2

K

29 2

a + b −

K K K

∴ a

K K

⋅ c +

ABDC ﹐若 BC ：△ ABD D 與 BC 交於

(1) 4 : 3 ;(2)

(1) 2 AD

K

=4

設

AO

K

=t

∵B

﹐

O

∴AO

K

=

AO

K

=

△ ABC

(2)

(

^{α β,}

)

⁼

2

3AB

K K

+2AD 9 1 4 4 12⋅ + ⋅ +

面積

= ⋅ ⋅4 3

=6 3

= ﹐3

K

c =4

2 2

0 c =

K K

﹐ b c a +

K K K

⋅ −

2AB

K K

+3AC= D 的面積比為 於一點 O ﹐且

) 2 3 5 5,

⎛ ⎞

⎜ ⎟

⎝ ⎠

3 AB+4AC

K K

﹐ 2

4 t AD

K K

= tAB

O

﹐

C

共線

4

5AD

K

⇒AO

2 3

5AB

K K

+5AC :

C △ ABD = 2 3, 5 5

⎛ ⎞

= ⎜ ⎟

⎝ ⎠﹒

2

9 AB 4

=

K

+

(

2 1 2 cos 60⋅ ⋅

sin 60° 1

− ⋅2 2 3 3 3

− − 2 4 ﹐且 a

K K

+ b

4 9 16+ + + 29 b 2

⋅ =

K K

﹒

=4 AD

K

﹐則 為_________

且 AO

K K

=α AB

﹐ 3

4 B+ tAC

K K ﹐

﹐

∴² 4t + : 4 :1 O OD=

﹐

: BO OC

⇒

5

3 ABO

⎛ ⎞

⎜ ⎟

⎝ △ ⎠

2

4 AD

K K

+12A

)

0° =37

﹐

2 4 sin120⋅ ⋅ ° 3 5 3

= 2

﹒

0

− c =

K K K

﹐

2⎛ a b + ⎜⎝ ⋅ −

K K

____﹔

B+β AC

K K

﹐數

3 1 4

t t

+ = ⇒ =

﹐

3 : 2

=

﹐

: 5

4 ABO

⎛ ⎞

⎜⎝ △ ⎠ AB AD⋅

K K

故

AC

K

=

° 1

2 3 si

− ⋅ ⋅ ⋅2

則 a

K K

⋅ c +

b⋅ c − a ⋅

K K K

數對

(

α β,

)

=

4

=5

﹐

1 1: 4 :

⎞ =3 4=

⎟⎠

37

﹒

in120°

c ⋅ b − b ⋅

K K K

0 c ⎞ =⎟⎠

K

﹐

= _________

3 ﹒

a =

K

______

____﹒

_______﹒

21.一圓之圓 解答 解析

作

22 △. ABC (1) BC= _

解答

解析

23.設平行四 解答 解析

24.設 a

K

﹑

= _______

解答

圓心為 O ﹐

− 8

作

OH⊥AB OA AB

K K K

⋅ = −A

= −

= − 中﹐AB=3 ___________

(1) 37 ;(2

(1) BC

K K

=A

2

BC

K

= ∴ BC

K

= (2) AB AC

K K

⋅

(3) CA CB

K K

⋅

四邊形 ABCD 2 21

AC BD⋅ = ⎜⎛⎝A

K K

2

20= AD

K

− 0 b ≠

K K

﹐ ______﹒

± 7

AB 為一弦﹐

B

﹐

AO AB

K K

⋅

c AO

K K

⋅ AB ⋅

1 AB

K K

⋅2 AB

﹐AC= ﹐4 __﹔(2) AB A

K K

⋅ 2) 6− ;(3) 22

AC AB

K K

−

2

AC AB

K K

− =

= 37﹐即 B

C= AB⋅ AC

K K K

2

2 CA CB B

+

=

K K K

D 中﹐已知

AB BC+ ⎞ ⎛⎟ ⎜⎠ ⎝⋅

K K

2

64 AD

− ⇒

K

a = b

K K

_﹐

若AB=4

cos OAB∠ =

4 2 8

= − ⋅ = −

12

∠BAC= AC

K

_{= ______}

2

AC AB

=

K K

+ 37 BC= ﹒

cos120 C ⋅ ° =

2 2

2 B − AB

K K

=

AB = ﹐ A8

K

AD AB− ⎞⎟⎠

K K

₌

=84

﹐

AD

K

5⎛⎜⎝ a − b

K K

﹐則 OA AB

K K

⋅

AB AO

−

K K

⋅

﹒

20° ﹐求 _______﹔(3

2

2

B − AC

K K

⋅ A

3 4 1 2

⎛ ⎞

⋅ ⋅ −⎜ ⎟=

⎝ ⎠

16 37 9 2 + −

=

20 AC BD

K K

⋅ =

AD AB

⎛ + ⎞⋅

⎜ ⎟

⎝ ⎠

K K

=2 21

﹐即

a b ⎞

− + ⎟⎠

K K

B=

K

________

cos OAB

⋅ ∠

3) CA CB

K K

⋅ = _

cos120 AB

K

⋅ °

= − ﹒ 6

=22﹒

﹐則 BC 之長

AD AB

⎛ − ⎞

⎜ ⎟

⎝ ⎠

K K

即

BC=2 2

⎞ =2 a

⎟⎠

K

﹐若

_____﹒

___________

=16 9+ −

長為_______

2

AD A

=

K

− 21

﹒

若 a

K

與 b

K

的 __﹒

2 4 3 1 2

− ⋅ ⋅ ⋅ −⎛⎜⎝

______﹒

2

AB

K _﹐

的夾角為θ ﹐ 1 37 2⎞ =⎟⎠

則 cosθ

解析 5⎛⎜⎝ a − b − a + b ⎞⎟⎠=2 a ⇒

K K K K K

平方

∴

2 2 2 2 2

25⎛ a 2 a b b a 2 a b b 2 a b a b ⎞ 4 a

− ⋅ + + + ⋅ + − − + =

⎜ ⎟

⎜ ⎟

⎝ ⎠

K K K K K K K K K K K K K

_﹐

又

2 2 2

2

a − b = a − b = a − a⋅ b + b

K K K K K K K K

²

2⎛ a a b ⎞

= ⎜⎜⎝ − ⋅ ⎟⎟⎠

K K K

_﹐

同理

2

2

a b ⎛ a a b ⎞ + = ⎜⎜⎝ + ⋅ ⎟⎟⎠

K K K K K

^{4 2}

2

a b a b a ⎛ a b ⎞

⇒ − + = −⎜⎝ ⋅ ⎟⎠

K K K K K K K

_﹐

∴原式⇒25 4 a ² 4 a ⁴ a b ² 4 a ²

⎛ ⎛ ⎞ ⎞

⎜ − −⎜ ⋅ ⎟ ⎟=

⎜ ⎝ ⎠ ⎟

⎝ ⎠

K K K K K

^{2 4 2}

24 a 25 a ⎛ a b ⎞

⇒ = −⎜⎝ ⋅ ⎟⎠

K K K K

4 4 2

2 2

24 a 25 ⎡ a ⎛ a b ⎞ ⎤

⇒ = ⎢⎢⎣ −⎜⎝ ⋅ ⎟⎠ ⎥⎥⎦

K K K K

^{2 4 2}

2

49 7

25 25

a

a b a b a

⎛ ⎞

⇒⎜ ⋅ ⎟ = ⇒ ⋅ = ±

⎝ ⎠

K K K K K K

﹐

∴

7 2

25 7

cos 25

a b a

a b a a

θ= ^⋅ =^± = ±

K K K

K K K K

^﹒