[電磁氣學] 電磁氣學 7th 솔루션입니다.
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작성일 21-03-19 22:20
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Download : 전자기학_7th.zip
CHAPTER 1
l(26, 10, 4)l = (0.92, 0.36, 0.14)
c) lMll2Nl(M+N):
25 + 25 + 1 +
√ 9 + 16 + 64 +
2 (−5ax+2az).
2 (A+B) = 1
7.35
The vector from midpoint to midpoint is now MAB −MBC = 1
설명
전자기학, 솔루션
[電磁氣學] 電磁氣學 7th 솔루션입니다. 참고하시면 됩니다.
lMAB −MBCl =
vector is therefore
Thus
b) the magnitude of 5ax +N− 3M:
1.2. The three vertices of a triangle are located at A(−1, 2, 5), B(−4,−2,−3), and C(1, 3,−2).
The vector from the origin to the midpoint of BC is MBC = 1
순서
−M+ 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
어디껀지 잘몰라서 챕터 1에 나온 내용을 같이 올렸습니다.
= (−580.5, 3193,−2902)
a =
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
a) Find the length of the perimeter of the triangle: Begin with AB = (−3,−4,−8), BC = (5, 5, 1),
unit vector is therefore parallel to AC. First we find AC = 2ax +ay −7az, which we recognize
레포트 > 공학,기술계열
where factors of 1/2 have cancelled.
and CA = (−2,−1, 7). Then the perimeter will be = lABl + lBCl + lCAl =
MAB −MBC
Download : 전자기학_7th.zip( 73 )
2 (−2ax − ay + 7az). The unit
= −0.27ax − 0.14ay + 0.95az
√
(5, 0, 0) + (8, 7,−2) − (−30, 12,−24) = (43,−5, 22), and l(43,−5, 22)l = 48.6.
전자기학 7th 솔루션입니다.
BC: The vector from the origin to the midpoint of AB is MAB = 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
(−2ax − ay + 7az)
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
다.
4 + 1 + 49 = 23.9.
(26, 10, 4)
電磁氣學 7th 솔루션입니다.
2 (B+C) = 1
aMM =
참고하시면 됩니다.
a) a unit vector in the direction of −M+ 2N.
√
l(−10, 4,−8)ll(16, 14,−4)l(−2, 11,−10) = (13.4)(21.6)(−2, 11,−10)
2 (−3ax +ay −5az).
만약 내용이 다를시 해피래포트에 환불요청하시면 환불됩니다. 어디껀지 잘몰라서 챕터 1에 나온 내용을 같이 올렸습니다.
