# Chapter 5 Notes - Chapter 5 ISM Linear Algebra Chapter 5...

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Chapter 4 / Exercise 51
Nature of Mathematics
Smith
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Chapter 5 ISM: Linear Algebra Chapter 5 5.1 1. v = 7 2 + 11 2 = 49 + 121 = 170 13 . 04 2. v = 2 2 + 3 2 + 4 2 = 4 + 9 + 16 = 29 5 . 39 3. v = 2 2 + 3 2 + 4 2 + 5 2 = 4 + 9 + 16 + 25 = 54 7 . 35 4. θ = arccos u · v u v = arccos 7+11 2 170 = arccos 18 340 0 . 219 (radians) 5. θ = arccos u · v u v = arccos 2+6+12 14 29 0 . 122 (radians) 6. θ = arccos u · v u v = arccos 2 - 3+8 - 10 10 54 1 . 700 (radians) 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. 8. Since u · v = 4 - 24 + 20 = 0, the two vectors enclose a right angle. 9. Since u · v = 3 - 4 + 5 - 3 = 1, the angle is acute (see Exercise 7). 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. 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 ) 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) 234
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Chapter 4 / Exercise 51
Nature of Mathematics
Smith
Expert Verified
ISM: Linear Algebra Section 5.1 = ( 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. 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. 14. The horizontal components of F 1 and F 2 are - F 1 sin β and F 2 sin α , respectively (the horizontal component 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. 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. These are vectors of the form - 2 r - 3 s - 4 t r s t = r - 2 1 0 0 + s - 3 0 1 0 + t - 4 0 0 1 . The three vectors to the right form a basis. 235
Chapter 5 ISM: Linear Algebra 16. You may be able to find the solutions by educated guessing. Here is the systematic approach: we first find all vectors x that are orthogonal to v 1 , v 2 , and v 3 , then we identify the unit vectors among them. Finding the vectors x with x · v 1 = x · v 2 = x · v 3 = 0 amounts to solving the system x 1 + x 2 + x 3 + x 4 = 0 x 1 + x 2 - x 3 - x 4 = 0 x 1 - x 2 + x 3 - x 4 = 0 ( we can omit all the coe ffi cients 1 2 ) . The solutions are of the form x = x 1 x 2 x 3 x 4 = t - t - t t . Since x = 2 | t | , we have a unit vector if t = 1 2 or t = - 1 2 . Thus there are two possible choices for v 4 : 1 2 - 1 2 - 1 2 1 2 and - 1 2 1 2 1 2 - 1 2 . 17. The orthogonal complement W of W consists of the vectors x in R 4 such that x 1 x 2 x 3 x 4 · 1 2 3 4 = 0 and x 1 x 2 x 3 x 4 · 5 6 7 8 = 0.