Unformatted text preview: Math 23
B. Dodson
Week 1 Homework: (Continued)
12.2 vectors: unit, standard unit, notations
12.3 dot product: orthogonal, proj, comp
12.4 cross product: formula, properties
Problem 12.2.19a:
Find a and a − 2b
when a =< 6, 2, 3 >, b =< −1, 5, −2 > .
Solution:
The length a = √ 36 + 4 + 9 = 7, and a − 2b =< 6, 2, 3 > −2 < −1, 5, −2 >
=< 8, −8, 7 > .
Problem 12.2.25:
Find a unit vector u that has the same direction
as a = 8i − j + 4k.
[variation/continuation: ﬁnd a vector of length 4
in the opposite direction.]
Solution:
The length 8i − j + 4k  =
so the unit vector is u = √ 64 + 1 + 16 = 1
a=
a 1
9 √ 81 = 9, 1
8i − j + 4k = 8 i − 9 j + 4 k.
9
9 2 Likewise, the vector with length 4 is − 4 8i − j + 4k .
9
Problem 12.3.23bc:
Determine whether the given vectors are othogonal, parallel or neither.
(b) a =< 4, 6 >, b =< −3, 2 > .
(c) a = −i + 2j + 5k, b = 3i + 4j − k.
Solution:
(b) Recall that vectors are parallel if one is
a scalar multiple of the other, a = cb, for a scalar c.
This says < 4, 6 >= c < −3, 2 >=< −3c, 2c >, so 4 = −3c, and 6 = 2c.
The second equation gives c = 3, but that’s not a solution of the
4
ﬁrst equation (with c = − 3 ), so there’s no solution c to the vector equation.
Conclusion: the vectors in (b) are NOT parallel.
On the other hand, the dot product a · b =< 4, 6 > · < −3, 2 >
= 4 · (−3) + 6 · 2 = −12 + 12 = 0,
so the cosine of the angle between these vectors is 0,
and the vectors ARE orthogonal (perpendicular!).
part (c): a = cb gives −1 = 3c, 2 = 4c and 5 = −c.
We see c = −5 is inconsistent with the ﬁrst two equations,
so the vectors are not parallel.
We also check a · b = −i + 2j + 5k · 3i + 4j − k
= −3 + 8 − 5 = 0,
so the vectors are orthogonal again. 3 Week 1 Homework (continued): 12.3 dot product: orthogonal,
proj, comp
12.4 cross product: formula, properties
Problem 12.3.37:
Find the scalar and vector projections
of b on a when a =< 4, 2, 0 >, and b =< 1, 1, 1 > .
Solution:
The length a = √ √
16 + 4 = 2 5, and the dot product a · b = 4 + 2 + 0 = 6, so the scalar projection is
compa (b) = a·b
6
= √ . Note: this is a scalar!
a
25 For the vector projection, proja (b) = compa (b) · ua
= a·b
6
3
·a =
< 4, 2, 0 >=
< 4, 2, 0 > .
a2
20
10 Again, note, this one is a vector. Problem 12.4.3: Find the cross product of
a = 2i + j − k and b = j + k .
Solution: a × b =
2
0 1
1 where 21
2 −1
1 −1
>,
,
,−
−1 =<
01
01
11
1
a
c b
= ad − bc.
d =< 1 + 1, −(2 − 0), 2 − 0 >=< 2, −2, 2 > . We check that this vector
is perpendicular to both a and b by ﬁnding the dot products. 4 Homework (Continued): ﬁnish 12.4
cross product: properties
Start of week 2 hw: 12.5 lines, planes
Problem 12.4.25:
For points P (1, 0, 0), Q(0, 2, 0) and R(0, 0, 3)
(a) Find a vector orthogonal to the plane
through P, Q, R.
(b) Find the area of the triangle P QR.
Solution:
The vector P Q =< 0, 2, 0 > − < 1, 0, 0 >=< −1, 2, 0 >
is in the plane through P, Q, R; and so is
P R =< −1, 0, 3 > . We use
P Q × P R for the normal vector.
P Q × P R = −1
−1
=< 2
0 0
3 20
−1 0
−1 2
,−
,
>
03
−1 3
−1 0 =< 6, 3, 2 > . (b) The area of the triangle is half the
area of the parallelogram, so
Area(P QR) = 1
2 ·  < 6, 3, 2 >  = 7 .
2 ...
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 Spring '06
 YUKICH
 Calculus, Linear Algebra, Vectors, Vector Space, Dot Product, 2J

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