MATH 32A, BRIDGE 2013
M. Wang
Homework 2
Solution
−−→
1. Find the components of the vector P Q. (13.2 Exercises 5, 6)
a. P = (1, 0, 1), Q = (2, 1, 0)
−−→
Ans: P Q = (2 − 1, 1 − 0, 0 − 1) = (1, 1, −1)
b. P = (-3, -4, 2), Q = (1, -4, 3)
−−→
Ans: P Q = (1 − (−3), −4 − (−4), 3 − 2) = (4, 0, 1)
−−→
2. Calculate the length of OR, where R = (1, 4, 3).(13.2 Exercises 9)
√
√
−−→
Ans: kORk = 12 + 42 + 32 = 26
3. Calculate the linear combinations. (13.2 Exercises 18, 20)
a. 5h2, 2, −3i + 3h1, 7, 2i
Ans: 5h2, 2, −3i + 3h1, 7, 2i = h13, 31, −9i
b. 6(4j + 2k) − 3(2i + 7k)
Ans: 6(4j + 2k) − 3(2i + 7k) = −6i + 24j − 9k
4. Find the given vector. (13.2 Exercises 24, 26)
a. ev , where v = h1, 1, 2i
Ans: ev =
v
kvk
=
√
h1,1,2i
12 +12 +22
√
=h
√ √
6
6
6
,
6
6 , 3 i
b. Unit vector in the direction of u = h1, 0, 7i
Ans: eu =
u
kuk
=
√
h1,0,7i
12 +02 +72
√
√
= h 102 , 0, 7102 i
5. Find a vector parametrization for the line with the given description. (13.2 Exercises 31, 34)
a. Passes through P = (4, 0, 8), direction vector v = 7i + 4k
−−→
Ans: r(t) = OP + tv = h4, 0, 8i + th7, 0, 4i (vector parametrization form)
or: x = 4 + 7t,
y = 0,
z = 8 + 4t (parametric equations form)
b. Passes through (-2, 0, -2) and (4, 3, 7)
−−→
−−→
Ans: r(t) = (1 − t)OP + tOQ = (1 − t)h−2, 0, 2i + th4, 3, 7i (vector parametrization form)
or: x = −2 + 6t,
y = 3t,
z = −2 + 9t (parametric equations form)
6. Show that r1 (t) and r2 (t) define the same line, where
r1 (t) = h3, −1, 4i + th8, 12, −6i
r2 (t) = h11, 11, −2i + th4, 6, −3i
(13.2 Exercises 47)
1
Proof. Since r1 (1) = h3, −1, 4i + h8, 12, −6i = h11, 11, −2i, we know that r1 (t) passes through
h11, 11, −2i. In addition, the direction vector of the two line h8, 12, −6i and h4, 6, −3i are the same
after normalization.
Thus, r1 and r2 pass the same point and have the same direction so that they define the same line.
7. Show that the lines r1 (t) = h−1, 2, 2i + th4, −2, 1i and r2 (t) = h0, 1, 1i + th2, 0, 1i do not intersect.
(13.2 Exercises 51)
Proof. The two lines intersect iff
h−1, 2, 2i + t1 h4, −2, 1i = h0, 1, 1i + t2 h2, 0, 1i
=⇒
x = −1 + 4t1 = 0 + 2t2 ,
y = 2 − 2t1 = 1 + 0t2 ,
(1)
z = 2 + t1 = 1 + t2
Solve the first two equations for t1 and t2 . We get t1 = 21 , t2 = 21 . However, they do not have the
same z-coordinates:
1
1
2 + 6= 1 +
2
2
Therefore, Equation(1) has no solution and the lines do not intersect.
2