高數(shù)下冊(cè)八章

學(xué)習(xí)改變命運(yùn)，勤奮贏得未來(lái)，細(xì)節(jié)影響成敗，態(tài)度決定一切.第八章知識(shí)點(diǎn)1、空間直角坐標(biāo)系.2、向量代數(shù)：1--2節(jié).3、空間解析幾何：3--6節(jié).點(diǎn)是幾何中最基本的元素，有了平面直角坐標(biāo)系以后，可以用二元有序數(shù)對(duì)表示平面上每一個(gè)點(diǎn)．這樣就可將幾何問(wèn)題與代數(shù)問(wèn)題相互轉(zhuǎn)化．笛卡爾對(duì)數(shù)學(xué)的最大貢獻(xiàn)：建立了代數(shù)與幾何之間的聯(lián)系．如解析幾何是用代數(shù)方法解決幾何問(wèn)題的一門(mén)學(xué)科；而數(shù)形結(jié)合作為一種數(shù)學(xué)基本思想方法，實(shí)際上是利用幾何圖形(方法)幫助解決代數(shù)問(wèn)題.平面直角坐標(biāo)系空間直角坐標(biāo)系有一個(gè)原點(diǎn),三個(gè)坐標(biāo)軸,三個(gè)坐標(biāo)面,八個(gè)卦限.oz上y

右前

x下左后x前y

右左下0常見(jiàn)畫(huà)法：x,

y,z

軸滿足右手系.上z后八章1、2

節(jié)1、若向量

=(x,y,z),則：a2

2

2(1)a的模：|

a|=r

=

x

+y

+z

(2)a的方向角：

a

=a,i

,

b

=a,

j,g

=a,ka

,b,g均為向量間夾角\a

,b,g?

[0,p](3)a的方向余弦：cosa

=

x

,cosb

=

y

,cosg

=

zr

r

ra,

,a

x

y z

r

r r

e

==

(cosa,cosb,cosg)(1)a

–b

=(a1

–b1,

a2

–b2

,

a3

–b3

)

1

2

3a

(a1,

a2

,

a3

),b=(c

,c

,c

),則：=

(b1,b2

,b3

),c2、若

=

2a b

=|

a

| |

b

|

cosa

b=

a1b2

+

a2b2

+

a3b3a a

=|

a

|(2)a

/

/bb1

b2

b3=

=

a1

a2

a3a b

=

0(5)a

^

bb=a

e.b|

b

|

a

bPrj

a

=

aaa

b|

a

|

(6)

Prjb

=

=

b

e；b

bb

b

a

b

m

=

b

=(a e

)e.(7)a在b方向上的投影向量

i

j

k(8)a

·b

=

c

=

a1

a2

a3b1

b2

b3

a2=

ba1

a2j

+

b

b

k21313a3

a

ai

-

1

3b

b

b2(按第一行展開(kāi))

a1

a2

a3b2

b3c1

c2

c3(10)[abc]

=(a

·b)

c

=

b1[abc]

=

0

(11)a,b,c共面

(9)

|

a·b

|=|

a

| |

b

|

sina

b=

Sba例1(1)已知：A(2,2,0),

B(1,3,

2),求與n

=AB同向的單位向量.解：

=

AB

=

(-1,1,

2)n\

r

=|

n

|=

21

1n2

\

e

=

-

,2

2

2

,

為所求n注：–e為與n平行的單位向量\|

a

·b

|=|

m

|=

2

6

i

j

ka

·b

=

m

=

0

-2-2

2

01

=

-2i

-2

j

+4k例1(2)若a

=(0,-2,1),b

=(-2,2,0),求：

a

b及|

a

·b

|.解：a

=(0,-2,1),b

=(-2,2,0)\

a

b

=0·(-2)+(-2)·2+1·0

=-4a

//b求m,n例2(1)若

=(m,-1,2),b

=(6,n,4)且am

-1

2

m

=

3,

n

=

-2解：

=

=6

n

4

(2)若a

=(k,-1,2),b

=(6,-4,4)且a

^

b,求k解：

b

=

0

6k

+

4

+8

=

0

k

=

-2ab

(3)若a

=(6,-1,2),b

=(7,-4,4),求A

=Prj

a

a

b

解：A

=Pr

jb

a

=|

b

|=

42

+4

+8

=

681

求a在b上的投影向量m.例3(1)若

=(6,-1,2),b

=(7,-4,4),a23

解：b b

=

81,

a b

=

-54a

bb

b

\

m

=

b

=-

(-7,

4,-4)例3(2)求

=(1,-2,2)在b

=(1,1,4)方向a上的投影向量.m

解：a b

=

7,b b

=18a

b

7(1,1,

4)b

b

18

\

m

=

b

=c

oa

a

b

+

b c

+

c

a求為單位向量且

+b

+

=a,

b,

c例4(1)設(shè)解：

+b+

=o

+b

=-a

c

a

c

2

=c

(b+a)

=c

(-c)

=-|

c

|

=-1

同理

b

+

b c

=

c a

+

a b

=

-1a32+c a

=-

\

a b

+b

c

\

b

c

+c

a

=c

b+c

a

兩兩垂直且|

a

|例4(2)設(shè),b,ca|

c

|=3,求s

=a

+b

+c

的模.=1,|

b

|=

2,

2\|

s

|

=14

|

s

|=

14

2|

s

|

=

s s

=(a

+b+c)

(a

+b

+c)解：

a,b,c兩兩垂直a

b=b

c

=c

a

=0

2

2

2=a

a+b

b+c

c+2(a

b+b

c

+c

a)=|

a

|

+|

b

|

+|

c

|

+2(a

b+b

c

+c

a)例5(1)設(shè)

=(3,2,1),b

=(2,-1,3),求

·b.a

a

i

j

k解：

·b

=

=

3

2

1a

c2

-1

3

(按第一行展開(kāi))=7i

-7

j

-7k

即：a

·b

=c

=(7,-7,-7).2,

a b

=

2,

例5(2)已知a

=

2,|

b

|=|

a

·b

|

.求

解：記q

=

a,

b

a b

=

a

|

b

|

cosq\

2

=

2 2

cosq

cosq

=222q?[0,p]

\

sinq

=2\|

a

·b

|=

a

|

b

|

sinq

=

2

\

n

=

a

·b

=

(2,2,0)2

22

2n

\

–e

=–,

,0

為所求的單位向量.例5(3)求垂直于

=(1,-1,2)及b

=(2,-2,2)a

2i

+

2

j

i

j

k解：

·b

=

1

-1

2

=a2

-

22

(按第一行展開(kāi))

[abc]

=-1

2

0-1

0

-1

=1?00

1

1b

=

AC

=

(-1,0,-1),c=

A