MATH 32A, BRIDGE 2013
M. Wang
Homework 1
Solution
−−→
1. Find the components of P Q. (13.1 Exercises 5, 6)
a. P = (3, 2), Q = (2, 7)
−−→
Ans: P Q = h2 − 3, 7 − 2i = h−1, 5i
b. P = (1, -4), Q = (3, 5)
−−→
Ans: P Q = h3 − 1, 5 − (−4)i = h2, 9i
2. Calculate (13.1 Exercises 9, 10, 11)
a. h2, 1i + h3, 4i
Ans: h2, 1i + h3, 4i = h2 + 3, 1 + 4i = h5, 5i
b. h−4, 6i − h3, −2i
Ans: h−4, 6i − h3, −2i = h−4 − 3, 6 − (−2)i = h−7, 8i
c. 5h6, 2i
Ans: 5h6, 2i = h5 × 6, 5 × 2i = h30, 10i
3. Sketch v = h0, 2i, w = h−2, 4i, 3v + w, 2v − 2w. (13.1 Exercises 19)
Ans:
11
10
9
8
7
3v+w
6
5
4
3
2
Y Axis
w
-11 -10
-9
-8
-7
-6
-5
-4
-3
-2
1
v
-1
1
2
3
4
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
X Axis
1
2v-2w
5
6
7
8
9
10
11
4. Let R = (-2, 7). Calculate the following: (13.1 Exercises 33, 34, 35, 36)
−−→
a. The length of OR
p
√
−−→
Ans: kORk = (−2 − 0)2 + (7 − 0)2 = 53
−→
b. The components of u = P R, where P = (1, 2).
−→
Ans: u = P R = h−2 − 1, 7 − 2i = h−3, 5i
−→
c. The point P such that P R has components h−2, 7i.
−→
And: Let P = (a, b), therefore P R = h−2 − a, 7 − bi = h−2, 7i, thus P = (a, b) = (0, 0)
−−→
d. The point Q such that RQ has components h8, −3i.
−−→
And: Let Q = (c, d), therefore RQ = hc − (−2), d − 7i = h8, −3i, thus P = (c, d) = (6, 4)
5. Calculate the linear combination.(13.1 Exercises 49, 50, 51)
a. 3j + (9i + 4j)
Ans: 3j + (9i + 4j) = 9i + 7j
b. − 32 i + 5( 12 j − 12 i)
Ans: − 23 i + 5( 12 j − 12 i) = −4i + 25 j
c. (3i + j) − 6j + 2(j − 4i)
Ans: (3i + j) − 6j + 2(j − 4i) = −5i − 3j
2