Homework 4
Transcription
Homework 4
MATH 32A, BRIDGE 2013 M. Wang Homework 4 Solution 1. (Exercise 1, 3) a. Ans: =1 × 3 − 2 × 4 = −5 b. Ans : =−6 × 1 − 9 × 1 = −15 2. (Exercise 5, 7) a. Ans: 1 × −3 − 2 × 4 + 1 × 3 = −8 b. Ans: 1 × 32 − 2 × 22 + 3 × 4 = 0 3. (Exercise 9, 11) a. Ans: v × w = 1i + 2j − 5k b. Ans: v × w = 6i − 8k 4. (Exercises 13, 15) a. Ans: i × k + j × k = −j + i b. Ans: i × j − 3j × j + 2k × j − i × k + 3j × k − 2k × k = k − 2i + j + 3i = i + j + k 5. (Exercises 17, 21) a. Ans: h−1, −1, 0i b. Ans: u × u + 2u × v − 2v × u − 4v × v = 4u × v = h4, 4, 0i 6. (Exercises 31) Ans: e × (e0 × e) = e × (ke0 kkek sin θ) = e × (1 × 1 sin π2 ) = e 7. (Exercises 34) u 1 1 0 Ans: u · (v × w) = det v = det 3 −2 2 = -2+2+0 = 0 w 4 −1 2 8. (Exercises 38) 2 2 1 u Ans: V = |u · (v × w)| = det v = det 1 0 3 = |24 − 0 − 4| = 20 0 −4 0 w 9. (Exercises 45) 1 Proof. v × w = kvkkwk cos θ1 w × v = kwkkvk cos θ2 Since θ1 = π − θ2 =⇒ cos θ1 = − cos θ2 =⇒ kvkkwk cos θ1 = −kwkkvk cos θ2 =⇒ v × w = −w × v 2