chapter01_7th_solution

# chapter01_7th_solution - CHAPTER 1 1.1. Given the vectors M...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## chapter01_7th_solution - CHAPTER 1 1.1. Given the vectors M...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online