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平移. 一、向量的內積. 1. 向量的夾角:. ,我們可以將它們平移. 使其始點重合,. . To be continued 例. 例如:在正三角形 ABC 中,. C. C. 120 . 60 . A. B. A. B. 特別地,. 本段結束. 2. 內積的定義:. 注意:. 而是一個「 實數 」。. (2) 內積的記法中,「 」不能省略,. 也不可以寫成「 」。. 本段結束. 3. 範例: 已知 ABC 是邊長為 6 的正三角形,. C. 解:. = 18 。. 60 . B. - PowerPoint PPT Presentation
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To be continued 1.
CAB120CAB60 ABC
2. (2)
CABCAB120603. ABC 6 = 18 = 18 Lets do an exercise !
CAB456
xOyA(x1 , y1)B(x2 , y2)To be continued (2)(3)4.
To be continued
5. = 45Lets do an exercise !
3. k = 1Lets do an exercise !
To be continued (4) (5) (6)1807. (3)
To be continued (6) &
8. = 36= 5 = 22 32= 922 1223cos60 + 432Lets do an exercise !
= 32 + 234cos120 + 42= 13
9. Lets do an exercise !
k = 2 8k 16 = 0 k 00
To be continued (2)10. 30242
3024260606060Lets do an exercise !10.
(4, 4)x + y = 8x + y = 6x y = 0x y = 2(3, 5)(3, 3)(2, 4) 31 (x , y) = (3 , 5)2x + 5y = 23 + 55= 31= 2x + 5y= 1
ABCDABCD2222226011. = 6Lets do an exercise !
12.
DABCTo be continued 13.
753CABDExx ABEC ( ) (2x)2 + 72 = 2( 32 + 52 )53
ABCD234614. = 18
ACBG1215. ABC
ACBHQP16. A H0= 0
LP(x1 , y1)LQ(x2 , y2)To be continued (2) (3)1. P(x1 , y1)Q(x2 , y2) Lax + by + c = 0 = 0
LA(x0 , y0)P(x , y) A(x0 , y0) L P(x , y) L cLax + by + c = 0
2. L3x 4y + 5 = 0 L L3x 4y + 5 = 0 (1) (3) (4) L Lets do an exercise !
L2x + 3y 5 = 0 L (2) (3) (4) L L2x + 3y 5 = 0
L1180L23. L1 L2 180
4. = 45 L1 L2 45 135 Lets do an exercise ! = 30 30 150
30ML 30 x = 1 (1 , 2) L () L m5. x = 1 L(y 2) = m(x 1) mx y m + 2 = 0(1, 2)
dP(x0 , y0)A(x1, y1)Lax + by + c = 0 A(x1, y1) L ax1 + by1 + c = 01.
P(2, 1) L12x 5y 3 = 0 2. P(3 , 4) L3x 4y + 5 = 0 Lets do an exercise !
P(1 , 2)(x , y)L3. = 2Lets do an exercise !
(1) xy L3x + 4y = 15 x2 + y2 (2) O P C(1, 2) C S (1) x2 + y2 L (x , y) O(0 , 0) x2 + y2 = 3yxOA(1, 2)Q(3, 3)B(2, 1)C(1, 2)H(2) S OAQB C S
ALBP4. L2x + 3y +5 = 0A(1 , 3)B(2 , 1)= 16 = 212
dL1L2P(x1 , y1)5.
6. (2) L3x + 4y + 5 = 0 2 Lets do an exercise !
(2) x y = 1 x y = 5 14 ?
Qx y = 5x y = 17(2) = 517
L1L2PTo be continued 7. L12x y + 1 = 0 L2x 2y + 5 = 0 P(x , y) (1) 2x y + 1 > 0 x 2y + 5 < 0 P(x , y) L12x y + 1 = 0 P(x , y) L2x 2y + 5 = 0 x y + 2 = 0 2x y + 1 = (x 2y + 5)
L1L2Q(2) Q(x , y) L12x y + 1 = 0 Q(x , y) L2x 2y + 5 = 0 2x y + 1 < 0 x 2y + 5 < 0 (2x y + 1) = (x 2y + 5) x + y 4 = 0
P(x , y)AQLTo be continued A(x0 , y0) P(x , y) L 8. x2 + y2 + dx + ey + f = 0 x02 + y02 = dx0 ey0 f A(x0 , y0)
(1) x2 + y2 = 5 A(1 2) (2) (x1)2 + (y+2)2 = 25 A(4 2) (1) A(x0 , y0) = (1 , 2) 1 x + 2 y = 5 x + 2y = 5(2) (x1)2 + (y+2)2 = 25 x2 + y2 2x + 4y 20 = 0 A(x0 , y0) = (4 , 2) 3x + 4y = 20
O1O2ABCTo be continued (2) (3)20209. C1(x10)2 + (y8)2 = 25C2(x+12)2 + (y12)2 = 225(2) (1) (3) ()10= 20
PO1O2ABk2k515(3) P (21 6) P Ly 6 = m ( x 21 )(2) (3) () mx y 21m + 6 = 0 7x 24y 3 = 0 3x + 4y 87 = 0
PPPL1L2P10. () P(x0 , y0) Cx2+y2+dx+ey+f=0 P L1L2 C A(x1 , y1)B(x2 , y2) ( 1)AB(1) P(x0 , y0) (2) P(x0 , y0)
To be continued
9x2 + y2 xy2. xy 3x + y = 24 9x2 + y2 288 Lets do an exercise !
xy 4x 3y = 50 x2 + y2 xy x2 + y2 100
To be continued 8x 9y 258x 9y 25 8x 9y xy3. xy 4x2 + 9y2 = 25
cos sin cos2 + sin2 = 1
4.
OABHOA=90BOABHO=HTo be continued 1.
OABH45Lets do an exercise !
OABH120
2. OABH
3. Lets do an exercise !
ABCHABC A(2 , 1)B(6 , 1) C(1 , 4) C AB
OCBA a b ?4. A(a , 1)B(2 , b) C(3 , 4) O
OABH(1) A = 90BO=H(2) OABH(3) 5.
ABCDEFGH6. ABCDEF (2)
COAD7. Lets do an exercise !
COAD
8. O ABC ACBOQPTo be continued
ACBOQP O ABC = 4
9. H ABC ACBHQPTo be continued
ACBHQP H ABC = 4
ACBPEDTo be continued 10. P ABC P ABC (1) (2) (3) (4) P ABC (4)
ACBPED12 P ABC P ABC 2 = 90 1 = 90 BCED ( 1 = 2 = 90 ) 1 = 2 = 90
To be continued To be continued Lets do an exercise != 1