Chp 5 - Section 5.1 Chapter 5 Section 5.1 5.1.1 v = 5.1.2 v...

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Section 5.1 Chapter 5 Section 5.1 5.1. 1 v = 7 2 + 11 2 = 49 + 121 = 170 13 . 04 5.1. 2 v = 2 2 + 3 2 + 4 2 = 4 + 9 + 16 = 29 5 . 39 5.1. 3 v = 2 2 + 3 2 + 4 2 + 5 2 = 4 + 9 + 16 + 25 = 54 7 . 35 5.1. 4 θ = arccos u · v u ∥∥ v = arccos 7+11 2 170 = arccos 18 340 0 . 219 (radians) 5.1. 5 θ = arccos u · v u ∥∥ v = arccos 2+6+12 14 29 0 . 122 (radians) 5.1. 6 θ = arccos u · v u ∥∥ v = arccos 2 3+8 10 10 54 1 . 700 (radians) 5.1. 7 Use the fact that u · v = u ∥∥ v cos θ , so that the angle is acute if u · v > 0, and obtuse if u · v < 0. Since u · v = 10 12 = 2, the angle is obtuse. 5.1. 8 Since u · v = 4 24 + 20 = 0, the two vectors enclose a right angle. 5.1. 9 Since u · v = 3 4 + 5 3 = 1, the angle is acute (see Exercise 7). 5.1. 10 u · v = 2 + 3 k + 4 = 6 + 3 k . The two vectors enclose a right angle if u · v = 6 + 3 k = 0, that is, if k = 2. 5.1. 11 a θ n = arccos u · v u ∥∥ v = arccos 1 n θ 2 = arccos 1 2 = π 4 (= 45 ) θ 3 = arccos 1 3 0 . 955 (radians) θ 4 = arccos 1 2 = π 3 (= 60 ) b Since y = arccos( x ) is a continuous function, lim n →∞ θ n = arccos lim n →∞ 1 n = arccos(0) = π 2 (= 90 ) 5.1. 12 v + w 2 = ( v + w ) · ( v + w ) (by hint) = v 2 + w 2 + 2( v · w ) (by definition of length) ≤ ∥ v 2 + w 2 + 2 v ∥∥ w (by Cauchy-Schwarz) = ( v + w ) 2 , so that v + w 2 ( v + w ) 2 Taking square roots of both sides, we find that v + w ∥ ≤ ∥ v + w , as claimed. 217
Chapter 5 5.1. 13 Figure 5.1 shows that F 2 + F 3 = 2 cos ( θ 2 ) F 2 = 20 cos ( θ 2 ) . It is required that F 2 + F 3 = 16, so that 20 cos ( θ 2 ) = 16, or θ = 2 arccos(0 . 8) 74 . Figure 5.1: for Problem 5.1.13. 5.1. 14 The horizontal components of F 1 and F 2 are −∥ F 1 sin β and F 2 sin α , respectively (the horizontal compo- nent of F 3 is zero). Since the system is at rest, the horizontal components must add up to 0, so that −∥ F 1 sin β + F 2 sin α = 0 or F 1 sin β = F 2 sin α or F 1 F 2 = sin α sin β . To find EA EB , note that EA = ED tan α and EB = ED tan β so that EA EB = tan α tan β = sin α sin β · cos β cos α = F 1 F 2 cos β cos α . Since α and β are two distinct acute angles, it follows that EA EB ̸ = F 1 F 2 , so that Leonardo was mistaken. 5.1. 15 The subspace consists of all vectors x in R 4 such that x · v = x 1 x 2 x 3 x 4 · 1 2 3 4 = x 1 + 2 x 2 + 3 x 3 + 4 x 4 = 0.

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