[¼Ö·ç¼Ç] ÀüÀÚ±âÇÐ 7ÆÇ / willian H. hayt / ¸Æ±×·ÎÈú
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ÀüÀÚ±âÇÐ_7ÆÇ.zip
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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
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[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
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