chapter 1a-hayt

# chapter 1a-hayt - Chapter 1 1.1. Given the vectors M = −...

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Unformatted text preview: Chapter 1 1.1. Given the vectors M = − 10 a x + 4 a y − 8 a z and N = 8 a x + 7 a y − 2 a z , find: a) a unit vector in the direction of − M + 2 N . − M + 2 N = 10 a x − 4 a y + 8 a z + 16 a x + 14 a y − 4 a z = (26 , 10 , 4) Thus a = (26 , 10 , 4) | (26 , 10 , 4) | = (0 . 92 , . 36 , . 14) b) the magnitude of 5 a x + N − 3 M : (5 , , 0) + (8 , 7 , − 2) − ( − 30 , 12 , − 24) = (43 , − 5 , 22), and | (43 , − 5 , 22) | = 48 . 6 . c) | M || 2 N | ( M + N ): | ( − 10 , 4 , − 8) || (16 , 14 , − 4) | ( − 2 , 11 , − 10) = (13 . 4)(21 . 6)( − 2 , 11 , − 10) = ( − 580 . 5 , 3193 , − 2902) 1.2. Given three points, A (4 , 3 , 2), B ( − 2 , , 5), and C (7 , − 2 , 1): a) Specify the vector A extending from the origin to the point A . A = (4 , 3 , 2) = 4 a x + 3 a y + 2 a z b) Give a unit vector extending from the origin to the midpoint of line AB . The vector from the origin to the midpoint is given by M = (1 / 2)( A + B ) = (1 / 2)(4 − 2 , 3 + 0 , 2 + 5) = (1 , 1 . 5 , 3 . 5) The unit vector will be m = (1 , 1 . 5 , 3 . 5) | (1 , 1 . 5 , 3 . 5) | = (0 . 25 , . 38 , . 89) c) Calculate the length of the perimeter of triangle ABC : Begin with AB = ( − 6 , − 3 , 3), BC = (9 , − 2 , − 4), CA = (3 , − 5 , − 1). Then | AB | + | BC | + | CA | = 7 . 35 + 10 . 05 + 5 . 91 = 23 . 32 1.3. The vector from the origin to the point A is given as (6 , − 2 , − 4), and the unit vector directed from the origin toward point B is (2 , − 2 , 1) / 3. If points A and B are ten units apart, find the coordinates of point B . With A = (6 , − 2 , − 4) and B = 1 3 B (2 , − 2 , 1), we use the fact that | B − A | = 10, or | (6 − 2 3 B ) a x − (2 − 2 3 B ) a y − (4 + 1 3 B ) a z | = 10 Expanding, obtain 36 − 8 B + 4 9 B 2 + 4 − 8 3 B + 4 9 B 2 + 16 + 8 3 B + 1 9 B 2 = 100 or B 2 − 8 B − 44 = 0. Thus B = 8 ± √ 64 − 176 2 = 11 . 75 (taking positive option) and so B = 2 3 (11 . 75) a x − 2 3 (11 . 75) a y + 1 3 (11 . 75) a z = 7 . 83 a x − 7 . 83 a y + 3 . 92 a z 1 1.4. given points A (8 , − 5 , 4) and B ( − 2 , 3 , 2), find: a) the distance from A to B . | B − A | = | ( − 10 , 8 , − 2) | = 12 . 96 b) a unit vector directed from A towards B . This is found through a AB = B − A | B − A | = ( − . 77 , . 62 , − . 15) c) a unit vector directed from the origin to the midpoint of the line AB . a M = ( A + B ) / 2 | ( A + B ) / 2 | = (3 , − 1 , 3) √ 19 = (0 . 69 , − . 23 , . 69) d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3. Note that the midpoint, (3 , − 1 , 3), as determined from part c happens to have z coordinate of 3. This is the point we are looking for....
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## This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.

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chapter 1a-hayt - Chapter 1 1.1. Given the vectors M = −...

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