Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.2 due 08/22/2021 at 11:59pm MST
Problem 1. (1 point)
Let a=h−4,4,−2iand b=h2,−2,−4i.
Compute:
a+b=h, , i
a−b=h, , i
2a=h, , i
3a+4b=h, , i
|a|=
Solution:
SOLUTION:
a+b=h−4+2,4−2,−2−4i=h−2,2,−6i
a−b=h−4−2,4+2,−2+4i=h−6,6,2i
2a=h2(−4),2(4),2(−2)i=h−8,8,−4i
3a+4b=h3(−4) + 4(2),3(4) + 4(−2),3(−2) + 4(−4)i=
h−4,4,−22i
|a|=p(−4)2+ (4)2+ (−2)2=√36
Answer(s) submitted:
•-2
•2
•-6
•-6
•6
•2
•-8
•8
•-4
•-4
•4
•-22
•6
(correct)
Correct Answers:
•-2
•2
•-6
•-6
•6
•2
•-8
•8
•-4
•-4
•4
•-22
•6
Problem 2. (1 point)
Let a=h−3,−3,−4i.
Find a unit vector in the same direction as a.
h, , i
Solution:
SOLUTION:
The vector ahas length p(−3)2+ (−3)2+ (−4)2=√34, so the
unit vector with the same direction is −3
√34 ,−3
√34 ,−4
√34 .
Answer(s) submitted:
•-(3/(sqrt(34)))
•-(3/(sqrt(34)))
•-(4/(sqrt(34)))
(correct)
Correct Answers:
•-0.514495755427526
•-0.514495755427526
•-0.685994340570035
Problem 3. (1 point)
Find the unit vector in the direction opposite to v=h2,2i.
Solution:
Solution: We first compute the unit vector evin the direction of
vand then multiply by −1 to obtain a unit vector in the opposite
direction. This gives:
ev=1
kvkv=1
p(2)2+ (2)2h2,2i=1
√8h2,2i=2
√8,2
√8
The desired vector is thus
−ev=−2
√8,2
√8=−2
√8,−2
√8.
Answer(s) submitted:
•<((-2)/(2.82842)),((-2)/(2.82842))>
(correct)
Correct Answers:
•<-0.707107,-0.707107>
1
Problem 4. (1 point)
Find the components and length of the following vectors:
−5i−5j
Components: ,
Length:
−2i+4j
Components: ,
Length:
−5i+3j
Components: ,
Length:
−3i+5j
Components: ,
Length:
Solution:
SOLUTION
Since i=h1,0iand j=h0,1i, using vector algebra we have:
−5i−5j=−5h1,0i−5h0,1i=h−5,0i+h0,−5i
=h−5+0,0−5i=−5i−5j
The length of the vector is:
k−5i−5jk=q(−5)2+ (−5)2=√50
We use vector algebra and the definition of the standard basis vec-
tor to compute the components of the vector −2i+4j:
−2i+4j=−2h1,0i+4h0,1i=h−2,0i+h0,4i
=h−2+0,0+4i=−2i+4j
The length of this vector is:
k−2i+4jk=q(−2)2+ (4)2=√20
We find the components of the vector −5i+3j:
−5i+3j=−5h1,0i+3h0,1i=h−5,0i+h0,3i
=h−5+0,0+3i=−5i+3j
The length of this vector is:
k−5i+3jk=q(−5)2+ (3)2=√34
We find the components of the vector −3i+5j, using vector alge-
bra:
−3i+5j=−3h1,0i+5h0,1i=h−3,0i+h0,5i
=h−3+0,0+5i=−3i+5j
The length of this vector is
k−3i+5jk=q(−3)2+ (5)2=√34
Answer(s) submitted:
•-5
•sqrt(50)
•-2
•sqrt(20)
•-5
•sqrt(34)
•-3
•sqrt(34)
(correct)
Correct Answers:
•−5,−5
•7.07107
•−2,4
•4.47214
•−5,3
•5.83095
•−3,5
•5.83095
Problem 5. (1 point)
If P= (2,1)and Q= (−1,3), find the components of ~
PQ
~
PQ =
Solution:
Solution: Using the definition of the components of a vector we
have ~
PQ =h−1−2,3−1i=h−3,2i
Answer(s) submitted:
•<-3,2>
(correct)
Correct Answers:
•<-3,2>
2
Problem 6. (1 point)
Determine whether the vectors ~
AB and ~
PQ are equivalent.
A= (0,0),B= (2,5),P= (2,−3),Q= (4,2)
• Select
• Equivalent
• Not Equivalent
Solution:
SOLUTION
We compute the vectors and check whether they have the same
components:
~
AB =h2−0,5−0i=h2,5i
~
PQ =h4−2,2+3i=h2,5i
so the vectors are equivalent.
Answer(s) submitted:
•Equivalent
(correct)
Correct Answers:
•Equivalent
Problem 7. (1 point)
Let R= (−5,−3). Find the point Psuch that ~
PR has components
h−3,3i.
P=
Solution:
SOLUTION
Denoting P= (x0,y0), we have:
~
PR =h−5−x0,−3−y0i=h−3,3i
Equating corresponding components yields:
−5−x0=−3
−3−y0=3⇒x0=−2,y0=−6⇒P= (−2,−6)
Answer(s) submitted:
•(-2,-6)
(correct)
Correct Answers:
•(-2,-6)
Problem 8. (1 point)
What is the terminal point of the vector a=h3,2ibased at
P= (5,1)?
Answer:
Solution:
Solution: The terminal point Qof the vector ais located 3
units to the right and 2 units up from P= (5,1). Therefore,
Q= (5+3,1+2) = (8,3)
Answer(s) submitted:
•(8,3)
(correct)
Correct Answers:
•(8,3)
Problem 9. (1 point)
Find a vector athat has the same direction as h−8,5,8ibut has
length 3.
Answer: a=
Solution:
SOLUTION
The given vector has length |h−8,5,8i|=p(−8)2+52+82=
√153, so a unit vector in the direction of h−8,5,8iis u=
1
√153 h−8,5,8i.
A vector in the same direction but with length 3 is
3u=3
√153 h−8,5,8i=−24
√153 ,15
√153 ,24
√153 .
Answer(s) submitted:
•<-((24)/(12.36931)),((15)/(12.36931)),((24)/(12.36931))>
(correct)
Correct Answers:
•<-1.94029,1.21268,1.94029>
3
Problem 10. (1 point)
A child walks due east on the deck of a ship at 1 miles per hour.
The ship is moving north at a speed of 19 miles per hour.
Find the speed and direction of the child relative to the surface of
the water.
Speed = mph
The angle of the direction from the north =
(radians)
Solution:
SOLUTION:
With respect to the water’s surface, the child’s velocity, v, is the
vector sum of the velocity of the ship with respect to the water,
and the child’s velocity with respect to the ship. If we let north be
the positive y-direction, then v=h0,19i+h1,0i=h1,19i. The
child’s speed is |v|=√12+192=√362 mph.
From the rigth triangle OPR, we have |OP|=|v|cos θ=19.
Thus the vector vmakes an angle θ=arccos19
√362 with the
north.
Answer(s) submitted:
•sqrt(362)
•1.234
(score 0.5)
Correct Answers:
•19.0262975904404
•0.0525830616109411
Problem 11. (1 point)
A horizontal clothesline is tied between 2 poles, 10 meters apart.
When a mass of 2 kilograms is tied to the middle of the clothes-
line, it sags a distance of 4 meters.
What is the magnitude of the tension on the ends of the clothes-
line?
NOTE: Use g=9.8m/s2for the gravitational acceleration.
Tension = N
Solution:
SOLUTION:
Let T1and T2represent the tension vectors in each side of the
clothesline as shown in the figure. T1and T2have equal verti-
cal components and opposite horizontal components. Let Tbe
the magnitude of T1and T2. Then T1=hTcosθ,Tsin θiand
T2=h−Tcosθ,Tsin θi, where θ=arctan4
5.
The force due to gravity acting on the mass has magnitude 2Kg ≈
(2)(9.8) = 19.6N, hence we have w=h0,−19.6i. The resultant
T1+T2of the tensile forces counterbalances w, so T1+T2=
−w. Thus 2Tsin θ=19.6 and
T=19.6
2sin θ=19.6
2sin(arctan(4/5)) =9.8
4√41
Answer(s) submitted:
•96.5387
(incorrect)
Correct Answers:
•15.6876543817105
4
Problem 12. (1 point)
The nine Ring Wraiths want to fly from Barad-Dur to Rivendell.
Rivendell is directly north of Barad-Dur. The Dark Tower reports
that the wind is coming from the west at 57 miles per hour. In
order to travel in a straight line, the Ring Wraiths decide to head
northwest. At what speed should they fly (omit units)?
Solution:
SOLITION:
Let vbe the velocity vector of the Ring Wraights and vits mag-
nitude (speed). Assume the y-axis points north and the x-axis
points east. Since the Ring Wraights is flying northwest, we have
v=hvcos(135o),vsin(135o) = D−v√2
2,v√2
2E.
Let w=h57,0ibe the wind’s velocity vector. Since v+wmust
point north, it must be v√2
2=57. Thus v=57√2.
Answer(s) submitted:
•80.6101
(correct)
Correct Answers:
•80.6101730552664
Problem 13. (1 point)
The figure shows a rectangular box in three-dimensional
space that contains several vectors. (The vector cis in the
xz-plane, and the vector eis in the xy-plane.)
Are the following statements true or false?
? 1. ~c=~
f
? 2. ~
d=~g−~c
?3. ~e=~a−~
b
? 4. ~a=~
d
? 5. ~a=−~
b
? 6. ~g=~
f+~a
(Click on graph to enlarge)
Solution:
SOLUTION
1. True.
2. False. The correct statement is ~
d=~c−~g.
3. True, by the parallelogram rule.
4. False. The true statement is ~a=−~
d.
5. False. The vectors are not parallel.
6. True, by the triangle rule.
Answer(s) submitted:
•True
•False
•True
•False
•False
•True
5
(correct)
Correct Answers:
•TRUE
•FALSE
•TRUE
•FALSE
•FALSE
•TRUE
Problem 14. (1 point)
Let a=h2,4iand b=h5,−1i.
Show that there are scalars sand tso that
sa+tb=h3,−5i
You might want to sketch the vectors to get some intuition.
s=
t=
Solution:
SOLUTION:
sa+tb=h3,−5i
⇔sh2,4i+th5,−1i=h3,−5i
⇔ h2s+5t,4s−1ti=h3,−5i
⇔2s+5t=3
4s−1t=−5
Solving the system yields
s=−1,t=1.
Answer(s) submitted:
•-1
•1
(correct)
Correct Answers:
•-1
•1
6
Problem 15. (1 point)
In the figure below the y-axis points north, the x-axis points east,
and the xy-plane corresponds to the surface of the water. Suppose
a boat is at point B, a submarine is 9 units below point S, and a
helicopter is 15 units above point H.
(1) Find the displacement vector and the distance from the
submarine to the boat.
Displacement:
Distance:
(2) Find the displacement vector and distance from the heli-
copter to the boat.
Displacement:
Distance:
(3) Find the displacement vector and distance from the sub-
marine to the helicopter.
Displacement:
Distance:
Solution:
SOLUTION
(1) The point Bhas coordinates (3,4,0)and the point Shas
coordinates (5,1,−9).
The displacement vector from the submarine to the boat
is given by
~
SB =h3−5,4−1,0−(−9)i=h−2,3,9i
The distance from the submarine to the boat is given by
the magnitude of the displacement vector:
Distance:
~
SB
=p(−2)2+32+92=√94
(2) The point Bhas coordinates (3,4,0)and the point Hhas
coordinates (1,2,15).
The displacement vector from the helicopter to the boat
is given by
~
HB =h3−1,4−2,−0−15i=h2,2,−15i
The distance from the submarine to the boat is given by
the magnitude of the displacement vector:
Distance:
~
SB
=p(2)2+ (2)2+ (−15)2=√233
(3) The point Shas coordinates (5,1,−9)and the point Hhas
coordinates (1,215).
The displacement vector from the submarine to the heli-
copter is given by
~
SH =h1−5,2−1,15 −(−9)i=h−4,1,24i
The distance from the submarine to the helicopter is given
by the magnitude of the displacement vector:
Distance:
~
SH
=q(−4)2+12+242=√593
Answer(s) submitted:
•
•sqrt(62)
•
•
•
•
(incorrect)
Correct Answers:
•<-2,3,9>
•9.69536
•<2,2,-15>
•15.2643
•<-4,1,24>
•24.3516
7
Problem 16. (1 point)
Find the following expressions using the graph below of vectors
u,v, and w.
1. u +v=
2. 2u+w=
3. 3v−6w=
4. |w|=
Note: You can click on the graph to enlarge the image.
Solution:
SOLUTION
u=h2,3i,v=h−1,1i,w=h4,1i. Thus
1. u +v=h1,4i
2. 2u+w=h4,6i+h4,1i=h8,7i
3. 3v−6w=h−3,3i−h24,6i=h−27,−3i
4. |w|=√42+12=√17
Answer(s) submitted:
•<1,4>
•<8,7>
•<-27,-3>
•sqrt(17)
(correct)
Correct Answers:
•<1,4>
•<8,7>
•<-27,-3>
•4.12311
Problem 17. (1 point)
Find vectors that satisfy the given conditions:
(1) The vector in the opposite direction to u=h−5,−2iand
of half its length is .
(2) The vector of length 5 and in the same direction as v=
h−2,4,0iis .
Solution:
SOLUTION
(1) The vector in the opposite direction to u=h−5,−2iand
of half its length is
−1
2u=h2.5,1i
(2) The vector v=h−2,4,0ihas length |v|=
p(−2)2+ (4)2+ (0)2=√20.
Thus the vector of length 5 and in the same direction as v
is
5v
|v|=5−2
√20 ,4
√20 =−10
√20 ,20
√20 ,0
√20
Answer(s) submitted:
•<2.5,1>
•<-((10)/(2sqrt(5))),((20)/(2sqrt(5))),0>
(correct)
Correct Answers:
•<2.5,1>
•<-2.23607,4.47214,0>
8
Problem 18. (1 point)
Let u=h1,−5i,v=h−5,−1i, and w=h2,1i. Find the vector x
that satisfies
2u−v+x=6x+w.
In this case, x=.
Solution:
SOLUTION
Solving the given equation for xyields
x=1
5(2u−v−w)
=1
5(2h1,−5i−h−5,−1i−h2,1i)
=1
5h5,−10i
=h1,−2i
Answer(s) submitted:
•<(5/6),-((10)/7)>
(incorrect)
Correct Answers:
•<1,-2>
Problem 19. (1 point)
Suppose u=h3,0iand v=h3,3iare two vectors that form the
sides of a parallelogram. Then the lengths of the two diagonals of
the parallelogram are .
Separate answers with a comma.
Solution:
SOLUTION
The diagonals are given by
u+v=<3,0>+<3,3>=<6,3>and u−v=<3,0>−<
3,3>=<0,−3>.
Their lengths are p(6)2+ (3)2=√45 and p(0)2+ (−3)2=√9,
respectively.
Answer(s) submitted:
•sqrt(45),sqrt(9)
(correct)
Correct Answers:
•3, 6.7082
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