Samantha Rodriguez Brewer MAT 267 ONLINE B Fall 2020
Assignment Section 10.4 due 10/21/2020 at 11:59pm MST
1. (1 point) Find the cross product a×bwhere a=h5,4,5i
and b=h−3,5,−1i.
a×b=h, , i
Solution:
SOLUTION
a×b=
i j k
5 4 5
−3 5 −1
=
4 5
5−1
i−
5 5
−3−1
j+
5 4
−3 5
k
= (−4−25)i−(−5+15)j+ (25 +12)k
=−29i−10j+37k
=h−29,−10,37i
Correct Answers:
•-29
•-10
•37
2. (1 point) Find the cross product a×bwhere a=
h−5,4,−2iand b=h2,−1,4i.
a×b=h, , i
Find the cross product c×dwhere c=3i+4j+0kand d=
−2i−2j+1k.
c×d=i+j+k
Solution:
SOLUTION:
a×b=
i j k
−5 4 −2
2−1 4
=
4−2
−1 4
i−
−5−2
2 4
j+
−5 4
2−1
k
= (16 −2)i−(−20 +4)j+ (5−8)k
=14i+16j−3k
=h14,16,−3i
c×d=
i j k
3 4 0
−2−2 1
=
4 0
−2 1
i−
3 0
−2 1
j+
3 4
−2−2
k
= (4−0)i−(3−0)j+ (−6+8)k
=4i−3j+2k
Correct Answers:
•14
•16
•-3
•4
•-3
•2
3. (1 point)
You are looking down at a map. A vector uwith |u|= 3
points north and a vector vwith |v|= 2 points northeast. The
crossproduct u×vpoints:
A) south
B) northwest
C) up
D) down
Please enter the letter of the correct answer:
The magnitude |u×v|=
Solution:
SOLUTION
By the right hand rule, the cross product points down.
|u×v|=|u||v|sinθ=3(2)sin(45o) = 3√2
Correct Answers:
•D
•(3*2*sqrt(2) )/2
4. (1 point) Think of the letter Xas four vectors starting from
the center and pointing outward. Label the four vectors starting
from the top left and proceeding clockwise as u,v,w,z.
Does u×vpoint in or out of the page? (in/out)
Does u×zpoint in or out of the page? (in/out)
Compute u×w
h, , i
Solution:
SOLUTION:
By the right hand rule, u×vpoints into the page.
By the right hand rule, u×zpoints out of the page.
Since uand ware parallel, their cross product is the zero
vector.
Correct Answers:
•IN
•OUT
•0
•0
•0
5. (1 point) Find two unit vectors orthogonal to a=
h0,−2,−1iand b=h−4,5,5i
Enter your answer so that the first non-zero coordinate of the
first vector is positive.
First Vector: h, , i
Second Vector: h, , i
Solution:
1
SOLUTION:
The cross product of two vectors is orthogonal to both vec-
tors. So we calculate
a×b=
i j k
0−2−1
−4 5 5
=
−2−1
5 5
i−
0−1
−4 5
j+
0−2
−4 5
k
= (−10 +5)i−(0−4)j+ (0−8)k
=−5i+4j−8k
=h−5,4,−8i
The magnitude of the cross product is
|a×b|=p(−5)2+ (4)2+ (−8)2=√105,
so two unit vectors orthogonal to both are
±a×b
|a×b|=±−5
√105 ,4
√105 ,−8
√105
, that is,
−5
√105 ,4
√105 ,−8
√105 and +5
√105 ,−4
√105 ,+8
√105
Correct Answers:
•0.487950036474267
•-0.390360029179413
•0.780720058358827
•-0.487950036474267
•0.390360029179413
•-0.780720058358827
6. (1 point) Find the area of the triangle with vertices:
Q(−2,0,4),R(0,−2,3),S(2,1,3).
Solution:
SOLUTION:
Let a=~
QR =h2,−2,−1iand b=~
QS =h4,1,−1i. The area
of the triangle is equal to half the lenght of the cross product of
these two vectors.
The corss product is given by a×b=h3,−2,10i, and it has
magnitude |a×b|=p(3)2+ (−2)2+ (10)2=√113.
Therefore the area of the triangle is 1
2√113.
Correct Answers:
•5.31507290636732
7. (1 point) Find the area of the parallelogram with vertices:
P(0,0,0),Q(−4,−1,−1),R(−4,1,1),S(−8,0,0).
Solution:
SOLUTION:
Choose any three points from the given four. The area of the
parallelogram is given by the magnitude of the cross product of
two vectors built from those three points.
For instance, choosing the points P,Qand R, we obtain the vec-
tors ~
PQ =h−4,−1,−1iand ~
PR =h−4,1,1i.
The cross product of these two vectors is h0,8,−8iwith magni-
tude p(0)2+ (8)2+ (−8)2=√128.
Therefore the area of the parallelogram is √128.
Correct Answers:
•11.3137084989848
8. (1 point) Find the distance the point P(4,7,−5)is to the
line through the two points
Q(5,5,−5 ), and R(2,8,−6 ).
Solution:
SOLUTION:
The distance between a point and a line is the length of the per-
pendicular from the point to the line, here |~
PS|=d. But re-
ferring to triangle PQS,d=|~
PS|=|~
QP|sinθ=|b|sin θ. But
θis the angle between ~
QP =b=h−1,2,0iand ~
QR =a=
h−3,3,−1i. Thus by definition of the cross product, sin θ=
|a×b|
|a||b|and so
d=|b|sinθ=|b||a×b|
|a||b|=|a×b|
|a|=|h2,1,−3i|
|h−3,3,−1i| =√14
√19
Correct Answers:
•0.858395075278952
9. (1 point) Find the volume of the parallelopiped with adja-
cent edges PQ, PR, PS where
P(−2,4,−4),Q(0,7,−1),R(−3,3,−5),S(4,2,−2).
Solution:
SOLUTION:
Let a=~
PQ =h2,3,3i,b=~
PR =h−1,−1,−1iand c=~
PS =
h6,−2,2i.
a·(b×c) =
233
−1−1−1
6−2 2
=2
−1−1
−2 2
−3
−1−1
6 2
+3
−1−1
6−2
=4,
so the volume of the parallelepiped is |a·(b×c)|=4 cubic
units.
Correct Answers:
•4
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