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