In exercise 60 write each scalar triple product as a

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: at u + v is the vector from the initial point of u to the terminal point of v. The shortest distance between two points is along the line joining these points so u+v ≤ u + v . 54. (a): u + v = (u1 i + u2 j) + (v1 i + v2 j) = (v1 i + v2 j) + u1 i + u2 j = v + u (c): u + 0 = u1 i + u2 j + 0i + 0j = u1 i + u2 j = u (e): k (lu) = k (l(u1 i + u2 j)) = k (lu1 i + lu2 j) = klu1 i + klu2 j = (kl)u 55. (d): u + (−u) = u1 i + u2 j + (−u1 i − u2 j) = (u1 − u1 )i + (u1 − u1 ) j = 0 (g): (k + l)u = (k + l)(u1 i + u2 j) = ku1 i + ku2 j + lu1 i + lu2 j = k u + lu (h): 1u = 1(u1 i + u2 j) = 1u1 i + 1u2 j = u1 i + u2 j = u 56. Draw the triangles with sides formed by the vectors u, v, u + v and k u, k v, k u + k v. By similar triangles, k (u + v) = k u + k v. 57. Let a, b, c be vectors along the sides of the 1 1 midpoints of a and b, then u = a − b = 2 2 parallel to c and half as long. triangle and A,B the 1 1 (a − b) = c so u is 2 2 a A c u b B 58. Let a, b, c, d be vectors along the sides of the quadrilateral and A, B, C, D the corresponding midpoints, then 1 1 1 1 u = b + c and v = d − a but d = a + b + c so 2 2 2 2 1 1 1 1 v = (a + b + c) − a = b + c = u thus ABCD 2 2 2 2 is a parallelogram because sides AD and BC are equal and parallel. B a b u v D EXERCISE SET 13.3 √ √ 1. (a) (1)(6) + (2)(−8) = −10; cos θ = (−10)/[( 5)(10)] = −1/ 5 √ √ (b) (−7)(0) + (−3)(1) = −3; cos θ = (−3)/[( 58)(1)] = −3/ 58 (c) (1)(8) + (−3)(−2) + (7)(−2) = 0; cos θ = 0 √ √ √ (d) (−3)(4) + (1)(2) + (2)(−5) = −20; cos θ = (−20)/[( 14)( 45)] = −20/(3 70) 2. (a) u · v = 1(2) cos(π/6) = √ 3. (a) u · v = −34 < 0, obtuse (c) u · v = −1 < 0, obtuse 3 C c A √ (b) u · v = 2(3) cos 135◦ = −3 2 (b) u · v = 6 > 0, acute (d) u · v = 0, orthogonal d Exercise Set 13.3 458 −→ 4. Let the points be P, Q, R in order, then P Q= 2 − (−1), −2 − 2, 0 − 3 = 3, −4, −3 , −→ −→ QR= 3 − 2, 1 − (−2), −4 − 0 = 1, 3, −4 , RP = −1 − 3, 2 − 1, 3 − (−4) = −4, 1, 7 ; −→ −→ −→ −→ −→ −→ since QP · QR= −3(1) + 4(3) + 3(−4) = −3 < 0, P QR is obtuse; since RP · RQ= −4(−1) + (−3) + 7(4) = 29 > 0, P RQ is acute; since P R · P Q= 4(3) − 1(−4) − 7(−3) = 37 > 0, RP Q is acute 5. Since v1 · vi = cos φi , the answers are, in order, √ √ √ √ 2/2, 0, − 2/2, −1, − 2/2, 0, 2/2 6. Proceed as in Exercise 5; 25/2, −25/2, −25, −25/2, 25/2 −→ −→ −→ −→ −→ −→ 7. (a) AB = 1, 3, −2 , BC = 4, −2, −1 , AB · BC = 0 so AB and BC are orthogonal; it is a right triangle with the right angle at vertex B . (b) Let A, B , and C be the vertices (−1, 0), (2, −1), and (1,4) with corresponding interior angles α, β , and γ , then −→ cos α = −→ −→ −→ AB · AC AB cos β = −→ BA · BC BA = √ −2, −4 · 1, −5 √√ = 9/ 130, γ ≈ 38◦ 20 26 −→ −→ −→ CA · CB CA −→ √ −3, 1 · −1, 5 √√ = 4/ 65, β ≈ 60◦ 10 26 BC −→ cos γ = = −→ −→ √ 3, −1 · 2, 4 √√ = 1/(5 2), α ≈ 82◦ 10 20 AC −→ = CB −→ 8. AB · AP = [2i + j + 2k] · [(k − 1)i + (k + 1)j + (k − 3)k] = 2(k − 1) + (k + 1) + 2(k − 3) = 5k − 7 = 0, k = 7/5. 9. (a) v · v1 = −ab + ba = 0; v · v2 = ab + b(−a) = 0 (b) Let v1 = 2i + 3j, v2 = −2i − 3j; 2 v1 3 = √ i + √ j, u2 = −u1 . take u1 = v1 13 13 y 3 u1 x 3 -3 u2 -3 v 10. By inspection, 3i − 4j is orthogonal to and has the same length as 4i + 3j so u1 = (4i + 3j) + (3i − 4j) = 7i − j and u2 = (4i + 3j) + (−1)(3i − 4j) = i +√j each make an angle 7 √ of 45◦ with 4i + 3j; unit vectors in the directions of u1 and u2 are (7i − j)/ 50 and (i + 7j)/ 50. 11. (a) The dot product of a vector u and a scalar v · w is not defined. (b) The sum of a scalar u · v and a vector w is not defined. (c) u · v is not a vector. (d) The dot product of a scalar k and a vector u + v is not defined. 459 Chapter 13 12. (b): u · (v + w) = (6i − j + 2k) · ((2i + 7j + 4k) + (i + j − 3k)) = (6i − j + 2k) · (3i + 8j + k) = 12; u · v + u · w = (6i − j + 2k) · (2i + 7j + 4k) + (6i − j + 2k) · (i + j − 3k) = 13 − 1 = 12 (c): k (u · v) = −5(13) = −65; (k u) · v = (−30i + 5j − 10k) · (2i + 7j + 4k) = −65; u · (k v) = (6i − j + 2k) · (−10i − 35j − 20k) = −65 13. (a) (b) 1, 2 · ( 28, −14 + 6, 0 ) = 1, 2 · 34, −14 = 6 √ 6w = 6 w = 36 (c) 24 5 √ (d) 24 5 14. false, for example a = 1, 2 , b = −1, 0 , c = 5, −3 15. (a) (b) 16. (a) (b) v= √ √ √ 3 so cos α = cos β = 1/ 3, cos γ = −1/ 3, α = β ≈ 55◦ , γ ≈ 125◦ v = 3 so cos α = 2/3, cos β = −2/3, cos γ = 1/3, α ≈ 48◦ , β ≈ 132◦ , γ ≈ 71◦ v = 7 so cos α = 3/7, cos β = −2/7, cos γ = −6/7, α ≈ 65◦ , β ≈ 107◦ , γ ≈ 149◦ v = 5, cos α = 3/5, cos β = 0, cos γ = −4/5, α ≈ 53◦ , β ≈ 90◦ , γ ≈ 143◦ 17. cos2 α + cos2 β + cos2 γ = 2 v1 v2 v2...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online