chapter01_7th_solution

chapter01_7th_solution - 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 , fnd: 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. The three vertices oF a triangle are located at A ( 1 , 2 , 5), B ( 4 , 2 , 3), and C (1 , 3 , 2). a) ind the length oF the perimeter oF the triangle: Begin with AB = ( 3 , 4 , 8), BC = (5 , 5 , 1), and CA = ( 2 , 1 , 7). Then the perimeter will be ` = | AB | + | BC | + | CA | = 9 + 16 + 64+ 25 + 25 + 1 + 4 + 1 + 49 = 23 . 9 . b) ind a unit vector that is directed From the midpoint oF the side AB to the midpoint oF side BC : The vector From the origin to the midpoint oF AB is M AB = 1 2 ( A + B ) = 1 2 ( 5 a x +2 a z ). The vector From the origin to the midpoint oF BC is M BC = 1 2 ( B + C ) = 1 2 ( 3 a x + a y 5 a z ). The vector From midpoint to midpoint is now M AB M BC = 1 2 ( 2 a x a y + 7 a z ). The unit vector is thereFore a MM = M AB M BC | M AB M BC | = ( 2 a x a y + 7 a z ) 7 . 35 = . 27 a x . 14 a y + 0 . 95 a z where Factors oF 1 / 2 have cancelled. c) Show that this unit vector multiplied by a scalar is equal to the vector From A to C and that the unit vector is thereFore parallel to AC . irst we fnd AC = 2 a x + a y 7 a z , which we recognize as 7 . 35 a MM . The vectors are thus parallel (but oppositely-directed). 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, fnd 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. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector in rectangular components that lies in the xy plane, is tangent to the circle at ( 3 , 1 , 0), and is in the general direction of increasing values of y : A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and y components are t x = a...
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chapter01_7th_solution - CHAPTER 1 1.1. Given the vectors M...

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