ÀÛ¼ºÀÏ : 14-10-13 18:55
[½Å±Ô ÀÚ·á] [¼Ö·ç¼Ç] ÀüÀÚ±âÇÐ 7ÆÇ / willian H. hayt / ¸Æ±×·ÎÈú
 ±Û¾´ÀÌ : documentview
Á¶È¸ : 447  
[¼Ö·ç¼Ç] ÀüÀÚ±âÇÐ 7ÆÇ / willian H. hayt / ¸Æ±×·ÎÈú
ÀüÀÚ±âÇÐ_7ÆÇ.zip

ÇØ´ç ÀÚ·á´Â ÇØÇÇ·¹Æ÷Æ®¿¡¼­ À¯·á°áÁ¦ ÈÄ ¿­¶÷ÀÌ °¡´ÉÇÕ´Ï´Ù.
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¼³¸í : [¼Ö·ç¼Ç] ÀüÀÚ±âÇÐ 7ÆÇ / willian H. hayt / ¸Æ±×·ÎÈú
CHAPTER 1 ¹éÅÍÇؼ®
CHAPTER 2 Äð·ÕÀÇ ¹ýÄ¢°ú Àü°è¼¼±â
CHAPTER 3 Àü¼Ó¹Ðµµ, °¡¿ì½ºÀÇ ¹ýÄ¢ ¹× º¤ÅÍ°èÀǹ߻ê
CHAPTER 4 Àü°èÀÇ ¿¡³ÊÁö ¹× ÀüÀ§
CHAPTER 5 Àü·ù ¹× µµÃ¼
CHAPTER 6 À¯Àüü ¹× Á¤Àü¿ë·®
CHAPTER 7 ÇÁ¿Í¼Û ¹× ¶óÇÃ¶ó½º ¹æÁ¤½Ä
CHAPTER 8 Á¤»óÀÚ°è
CHAPTER 9 ÀÚ±â·Â°ú ÀÚ¼ºÃ¼, ±×¸®°í ÀδöÅϽº
CHAPTER 10 ½Ã°£¿¡ µû¶ó º¯ÇÏ´Â Àü°¡°è¿Í ¸Æ½ºÀ£ ¹æÁ¤½Ä
CHAPTER 11 Àü¼Û¼±·Î
CHAPTER 12 ±ÕÀÏ Æò¸éÆÄ
CHAPTER 13 Æò¸éÆÄ ¹Ý»ç¿Í ºÐ»ê
CHAPTER 14 µµÆÄ°ü ¹× ¾ÈÅ׳ª ±âº»
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M+ 2N.
−M+ 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a =
(26, 10, 4)
l(26, 10, 4)l
= (0.92, 0.36, 0.14)
b) the magnitude of 5ax +N− 3M:
(5, 0, 0) + (8, 7,−2) − (−30, 12,−24) = (43,−5, 22), and l(43,−5, 22)l = 48.6.
c) lMll2Nl(M+N):
l(−10, 4,−8)ll(16, 14,−4)l(−2, 11,−10) = (13.4)(21.6)(−2, 11,−10)
= (−580.5, 3193,−2902)
µî 1ÀåºÎÅÍ 14Àå±îÁöÀÇ Ç®ÀÌÁý
[¼Ö·ç¼Ç] ÀüÀÚ±âÇÐ 7ÆÇ / willian H. hayt / ¸Æ±×·ÎÈú
CHAPTER 1 ¹éÅÍÇؼ®
CHAPTER 2 Äð·ÕÀÇ ¹ýÄ¢°ú Àü°è¼¼±â
CHAPTER 3 Àü¼Ó¹Ðµµ, °¡¿ì½ºÀÇ ¹ýÄ¢ ¹× º¤ÅÍ°èÀǹ߻ê
CHAPTER 4 Àü°èÀÇ ¿¡³ÊÁö ¹× ÀüÀ§
CHAPTER 5 Àü·ù ¹× µµÃ¼
CHAPTER 6 À¯Àüü ¹× Á¤Àü¿ë·®
CHAPTER 7 ÇÁ¿Í¼Û ¹× ¶óÇÃ¶ó½º ¹æÁ¤½Ä
CHAPTER 8 Á¤»óÀÚ°è
CHAPTER 9 ÀÚ±â·Â°ú ÀÚ¼ºÃ¼, ±×¸®°í ÀδöÅϽº
CHAPTER 10 ½Ã°£¿¡ µû¶ó º¯ÇÏ´Â Àü°¡°è¿Í ¸Æ½ºÀ£ ¹æÁ¤½Ä
CHAPTER 11 Àü¼Û¼±·Î
CHAPTER 12 ±ÕÀÏ Æò¸éÆÄ
CHAPTER 13 Æò¸éÆÄ ¹Ý»ç¿Í ºÐ»ê
CHAPTER 14 µµÆÄ°ü ¹× ¾ÈÅ׳ª ±âº»
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M+ 2N.
−M+ 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a =
(26, 10, 4)
l(26, 10, 4)l
= (0.92, 0.36, 0.14)
b) the magnitude of 5ax +N− 3M:
(5, 0, 0) + (8, 7,−2) − (−30, 12,−24) = (43,−5, 22), and l(43,−5, 22)l = 48.6.
c) lMll2Nl(M+N):
l(−10, 4,−8)ll(16, 14,−4)l(−2, 11,−10) = (13.4)(21.6)(−2, 11,−10)
= (−580.5, 3193,−2902)
µî 1ÀåºÎÅÍ 14Àå±îÁöÀÇ Ç®ÀÌÁý
Ãâó : ÇØÇÇ·¹Æ÷Æ® ÀÚ·á½Ç

 
 

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