1

castillo (ajc2686) – Assignment 1 – luecke – (55035) 1

This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

Which one of the following conic sectionshas graph

1−1−2−3

1

2

−1

−2

1. y2 + 2y − x − 1 = 0

2. y2 − 2y − x − 1 = 0

3. y2 + 2y + x − 1 = 0

4. x2 + 2x + y − 1 = 0 correct

5. x2 + 2x − y − 1 = 0

6. y2 − 2y + x − 1 = 0

7. x2 − 2x + y − 1 = 0

8. x2 − 2x − y − 1 = 0

Explanation:

The graph is that of a parabola open-ing downwards and having vertex at (−1, 2).Thus it is the graph of

y − 2 = −A(x + 1)2 ,

for some A > 0. But the graph intersects they-axis at y = 1 so A = 1. Consequently, the

graph is that of the conic section

(x + 1)2 + y − 2 = 0 ,

which after expansion and simplification canbe rewritten as

x2 + 2x + y − 1 = 0 .

keywords: graph, conic section, parabola,shifted parabola,

002 10.0 points


1 2−1−2−3−4

1

2

3

4

−1

−2

1. 4x2 + 8x + 9y2 + 18y = 23

2. 9x2 − 18x − 4y2 + 8y = 41

3. 4x2 + 8x + 9y2 − 18y = 23 correct

4. 9x2 + 18x − 4y2 − 8y = 41

5. 9x2 + 18x − 4y2 + 8y = 41

6. 4x2 − 8x + 9y2 − 18y = 23

Explanation:

The graph is a shifted ellipse centered at(−1, 1) whose major axis has length 6 andminor axis has length 4. Thus one equationfor this ellipse is

(x + 1)2

9+

(y − 1)2

4= 1 ,



4x2 + 8x + 9y2 − 18y = 23 .

keywords: graph, conic section, ellipse,shifted ellipse

003 10.0 points


2 4−2−4−6

2

4

−2

−4

−6

1. 4y2 − 8x − x2 − 2x = 1

2. 4y2 + 8x − x2 − 2y = 1

3. x2 − 2x + 4y2 − 8y + 1 = 0

4. x2 + 2x − 4y2 − 8y = 7

5. 4y2 + 8y − x2 − 2x = 1 correct

6. x2 + 2x − 4y2 + 8y = 7

7. x2 + 2x + 4y2 + 8y + 1 = 0

8. x2 − 2x − 4y2 + 8y = 7

Explanation:

The graph is a shifted hyperbola havingasymptotes which intersect at (−1, −1) aswell as vertices at (−1, 0) and (−1, −2). Thusone equation for this hyperbola is

(y + 1)2

1− (x + 1)2

4= 1 ,


4y2 + 8y − x2 − 2x = 1 .

keywords: graph, conic section, ellipse,shifted ellipse

004 10.0 points

Which one of the points

P (−4, −5, −8), Q(3, 7, −2), R(8, 0, −4)

in 3-space is closest to the xy-plane?

1. Q(3, 7, −2) correct

2. P (−4, −5, −8)

3. R(8, 0, −4)

Explanation:

The distance of a point (a, b, c) in 3-spacefrom the xy-plane is given by |c|. Conse-quently, of the three points

P (−4, −5, −8), Q(3, 7, −2), R(8, 0, −4)

the one closest to the xy-plane is

Q(3, 7, −2) .

keywords: plane, distance in 3-space,

005 10.0 points

Which of the following statements are true:

A. in 3-space the graph of y = 4 is a line,

B. in 2-space the graph of x = 2 is a line.

1. both of them

2. neither of them


3. B only correct

4. A only

Explanation:

A. FALSE: in 3-space the graph of y = 4 is

a plane perpendicular to the y-axis.

B. TRUE: in the plane the graph of x = 2is a vertical line.

keywords: T/F, True/False, plane, 2-space,3-space, point

006 10.0 points

A triangle ∆PQR has vertices

P (−2, 1, 1), Q(−3, 2, −1), R(−1, 0, −1) .

Use the distance formula to decide which oneof the following properties the triangle has.

1. isoceles with |PQ|= |PR| correct

2. not isoceles

3. isoceles with |RP |= |RQ|

4. isoceles with |QP |= |QR|Explanation:

Recall that for points A, B it is common todenote the line segment from A to B by AB,and to denote the length of this line segmentby |AB|. Recall also a triangle is isoceleswhen two of its sides have the same length.Thus we have to use the distance formula tocompute the side lengths of ∆PQR.

But for the vertices

P (−2, 1, 1), Q(−3, 2, −1), R(−1, 0, −1)

we see that

|PQ|2 = (−1)2 + (1)2 + (−2)2 = 6 ,

|QR|2 = (2)2 + (−2)2 + 0 = 8 ,

while

|PR|2 = (1)2 + (−1)2 + (−2)2 = 6 .

Consequently, ∆PQR is

isoceles with |PQ|= |PR| .

keywords: distance formula, isoceles triangle,3-space, 3-space geometry,

007 10.0 points

Find an equation for the sphere having cen-ter at (3, 2, −2) and radius 3.

1. x2 + y2 + z2 + 6x + 4y − 4z = 8

2. x2 + y2 + z2 − 6x − 4y + 4z + 8 = 0correct

3. x2 + y2 + z2 + 6x + 4y − 4z + 8 = 0

4. 3x2 + 2y2 − 2z2 = 9

5. 3x2 + 2y2 − 2z2 + 9 = 0

6. x2 + y2 + z2 − 6x − 4y + 4z = 8

Explanation:

The sphere consists of all points P (x, y, z)such that

dist ((x, y, z) , (3, 2,−2)) = 3 .

Thus, by the distance formula in 3-space,

(x − 3)2 + (y − 2)2 + (z + 2)2 = 9 ,

which after expansion becomes

x2 + y2 + z2 − 6x − 4y + 4z + 8 = 0 .

008 10.0 points

Find the center of the sphere given by

x2 + y2 + z2 + 8x − 6y − 2z + 25 = 0 .

1. center = (−4, −3, 2)


2. center = (−8, 6, 1)

3. center = (−4, 3, 1) correct

4. center = (−8, 6, 2)

5. center = (4, −3, −1)

6. center = (4, 3, −1)

Explanation:

After completion of squares

x2 + y2 + z2 + 8x − 6y − 2z + 25 = 0

becomes

(x+4)2+(y−3)2+(z−1)2+25−16−9−1 = 0 ,

which can then be written as

(x + 4)2 + (y − 3)2 + (z − 1)2 = 1 .

This is an equation for a sphere having

center = (−4, 3, 1)

and radius r = 1.

009 10.0 points

Which of the following sets of inequalititesdescribes the region consisting of all pointsoutside a sphere of radius 4 centered at theorigin and inside a sphere of radius 6 centeredat the origin.

1. 4 < x2 + y2 + z2 < 6

2. 16 < x2 + y2 + z2 < 36 correct

3. 4 ≤ x2 + y2 + z2 < 6

4. 4 ≤ x2 + y2 + z2 ≤ 6

5. 16 < x2 + y2 + z2 ≤ 36

6. 16 ≤ x2 + y2 + z2 ≤ 36

Explanation:

The region consists of all points whose dis-tance to the origin is greater that 4, but lessthan 6, i.e, all points satisfying the inequali-ties

4 <√

x2 + y2 + z2 < 6 ,

which can also be written as

16 < x2 + y2 + z2 < 36 .

010 10.0 points

When u, v are the displacement vectors

u =−−→AB , v =

−→AP ,

determined by the parallelogram

AB

C

D

P

Q

RS

O

express−→CR in terms of u and v.

1.−→CR = 2(u + v)

2.−→CR = 2u

3.−→CR = u + 2v

4.−→CR = 2v

5.−→CR = 2(u− v)

6.−→CR = 2v − u correct

Explanation:

By the parallelogram law for the additionof vectors we see that

−→CR = 2v − u .


keywords: vectors, linear combination, vectorsum displacement vector, parallelogram

011 10.0 points

The parallelopiped in 3-space shown in

QP

S

C

B

A

D

R

is determined by its vertices

P (−3, −4, −2), Q(2, −4, 0) ,

R(−1, 0, 1), S(−4, −1, −1) .

Find the vector v represented by the directedline segment

−−→PD.

1. v = 〈 7, 4, 5 〉

2. v = 〈 6, 7, 6 〉

3. v = 〈 4, 3, 3 〉 correct

4. v = 〈 1, 7, 4 〉

5. v = 〈 7, 8, 6 〉

6. v = 〈 0, 2, 1 〉Explanation:

As a vector sum,−−→PD =

−−→PQ +

−→PS .

But−−→PQ = 〈 5, 0, 2 〉, −→

PS = 〈−1, 3, 1 〉 .

Consequently,

v =−−→PD = 〈 4, 3, 3 〉 .

keywords: parallelopiped, 3-space, coordi-nates, vertex, directed line segment, vectorsum

012 10.0 points

Determine the vector c = a + 2b when

a = 3i + 2j + 3k , b = −3i + j + 3k .

1. c = −4i − 5j + 9k

2. c = −3i − 5j + 9k

3. c = −4i + 4j − 8k

4. c = −4i − 5j − 8k

5. c = −3i + 4j + 9k correct

6. c = −3i + 4j − 8k

Explanation:

The sum of vectors

a = a1i + a2j + a3k , b = b1i + b2j + b3k

is defined componentwise:

a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k ;

similarly, multiplication by a scalar λ also isdefined componentwise:

λa = (λa1)i + (λa2)j + (λa3)k .

When

a = 3i + 2j − 3k , b = −3i − j + 3k ,

therefore, we see that

c =(

(1)(3) + (2)(−3))

i

+(

(1)(2) + (2)(1))

j

+(

(1)(3) + (2)(3))

k .


Consequently,

c = −3i + 4j + 9k .

013 10.0 points

Determine the length of the vector −2a+b

when

a = 〈−1 ,−3 ,−1 〉 , b = 〈 1 ,−1 ,−2 〉 .

1. length =√

38

2. length = 6

3. length = 4√

2

4. length = 2√

10

5. length =√

34 correct

Explanation:

The length, |c|, of the vector

c = 〈 c1 , c2 , c3 〉

is defined by

|c| =√

c21+ c2

2+ c2

3.

Consequently, when

a = 〈−1 ,−3 ,−1 〉 , b = 〈 1 ,−1 ,−2 〉 ,

andc = −2a + b = 〈 3 , 5 , 0 〉 ,

we see that

| − 2a + b| =√

34 .

014 10.0 points

Find a unit vector n with the same directionas the vector

v = 2 i− 3 j − 6k .

1. n =1

5i +

3

10j +

3

5k

2. n =1

5i − 3

10j − 3

5k

3. n =2

9i +

1

3j +

2

3k

4. n =2

9i − 1

3j − 2

3k

5. n =2

7i +

3

7j +

6

7k

6. n =2

7i − 3

7j − 6

7k correct

Explanation:

The vector

v = 2 i − 3 j − 6k .

has length

|v| =√

22 + 32 + 62 =√

49 = 7 .

Consequently,

n =v

|v| =2

7i − 3

7j − 6

7k

is a unit vector having the same direction asv.

015 10.0 points

Find the angle between the vectors

a = 〈√

3, 2 〉 , b = 〈−5, −√

3 〉 .

1. angle =2π

3

2. angle =π

4

3. angle =π

6

4. angle =5π

6correct

5. angle =π

3


6. angle =3π

4

Explanation:

Since the dot product of vectors a and b

can be written as

a.b = |a| |b| cos θ , 0 ≤ θ ≤ π,

where θ is the angle between the vectors, wesee that

cos θ =a.b

|a| |b| , 0 ≤ θ ≤ π .

But for the given vectors,

a · b = (√

3)(−5) + (2)(−√

3) = −7√

3 ,

while

|a| =√

7 , |b| =√

28 .

Consequently,

cos θ = − 7√

3√7 · 2

√7

= −√

3

2

where 0 ≤ θ ≤ π. Thus

angle =5π

6.

016 10.0 points

A triangle ∆PQR in 3-space has vertices

P (6, 5,−2), Q(1, 6, 1), R(2, 3,−1) .

Use vectors to decide which one of the follow-ing properties the triangle has.

1. right-angled at Q

2. right-angled at P

3. right-angled at R correct

4. not right-angled at P, Q, or R

Explanation:

Vectors a and b are perpendicular whena · b = 0. Thus ∆PQR will be

(1) right-angled at P when−−→QP · −→RP = 0,

(2) right-angled at Q when−−→PQ · −−→RQ = 0,

(3) right-angled at R when−→PR · −−→QR = 0.

But for the vertices

P (6, 5,−2), Q(1, 6, 1), R(2, 3,−1)

we see that

−−→PQ = 〈−5, 1, 3 〉 ,

−−→QR = 〈 1, −3, −2 〉 ,

while −→RP = 〈 4, 2, −1 〉 .

Thus

−−→QP · −→RP = 21,

−−→PQ · −−→RQ = 14,

and −→PR · −−→QR = 0 .

Consequently, ∆PQR is

right-angled at R .

keywords: vectors, dot product, right trian-gle, perpendicular,

017 10.0 points

Find the vector projection of b onto a when

b = 〈 3, −1 〉 , a = 〈 2, −1 〉 .

1. vector proj. =7√5〈 3, −1 〉

2. vector proj. =7

5〈 2, −1 〉 correct

3. vector proj. =9

5〈 2, −1 〉

4. vector proj. =7

5〈 3, −1 〉

5. vector proj. =9√5〈 3, −1 〉


6. vector proj. =9√5〈 2, −1 〉

Explanation:

The vector projection of b onto a is givenin terms of the dot product by

projab =

(a · b|a|2

)

a .

But when

b = 〈 3, −1 〉 , a = 〈 2, −1 〉 ,

we see that

a · b = (2)(3) + (−1)(−1) = 7 ,

while

|a|2 = (2)2 + (−1)2 = 5 .

Consequently,

projab =

7

5〈 2, −1 〉 .

keywords:

018 10.0 points

Find the vector projection of b onto a when

b = i + 4 j− 2k , a = 2 i + 3 j− k .

1. vector proj. =2

3(2 i + 3 j − k)

2. vector proj. =16

21( i + 4 j − 2k)

3. vector proj. =16

21(2 i + 3 j − k)

4. vector proj. =8

7(2 i + 3 j− k) correct

5. vector proj. =2

3( i + 4 j− 2k)

6. vector proj. =8

7( i + 4 j− 2k)

Explanation:

The vector projection of b onto a is givenin terms of the dot product by

projab =

(a · b|a|2

)

a .

Now when

b = i + 4 j− 2k , a = 2 i + 3 j − k .

we see that

a · b = (2)(1) + (3)(4) + (−1)(−2) = 16 ,

while

|a|2 = (2)2 + (3)2 + (−1)2 = 14 .

Consequently,

projab =

8

7(2 i + 3 j− k) .

keywords: vector projection, vectors in space,

019 10.0 points

The box shown in

x

y

z

C

A

B

D

is the unit cube having one corner at theorigin and the coordinate planes for three ofits faces.

Find the cosine of the angle θ between CA

and CB.


1. cos θ = 0

2. cos θ =1√2

3. cos θ =

√

2

3

4. cos θ =1

2

5. cos θ =

√3

2

6. cos θ =1√3

correct

Explanation:

To use vectors we shall replace a line seg-ment with the corresponding directed line seg-ment.

Now the angle θ between any pair of vectorsu, v is given in terms of their dot product by

cos θ =u · v|u||v| .

On the other hand, since the unit cube hassidelength 1,

A = (1, 1, 0), B = (0, 1, 1) ,

while C = (0, 0, 1). In this case−→CA is a

directed line segment determining the vector

u = 〈 1, 1, −1 〉 ,

while−−→CB determines

v = 〈 0, 1, 0 〉 .

For these choices of u and v,

u · v = 1 =√

3 cos θ .

Consequently, the cosine of the angle betweenCA and CB is given by

cos θ =u · v|u| |v| =

1√3

.

keywords: vectors, dot product, unit cube,cosine, angle between vectors

020 10.0 points

The box shown in

x

y

z

A

B C

D

is the unit cube having one corner at theorigin and the coordinate planes for three ofits adjacent faces.

Determine the vector projection of−−→AB on−−→

AD.

1. vector projection =1

2(i − k)

2. vector projection = −2

3(i + j − k)


2(j − k)


3(i + j − k)


2(j − k) correct


2(i − k)

Explanation:

The vector projection of a vector b onto avector a is given in terms of the dot productby

projab =

(a · b|a|2

)

a .


On the other hand, since the unit cube hasside-length 1,

A = (0, 0, 1), B = (1, 0, 0) ,

while D = (0, 1, 0). In this case−−→AD is a

directed line segment determining the vector

a = 〈 0, 1, −1 〉 = j − k ,

while−−→AB determines the vector

b = 〈 1, 0, −1 〉 = i − k .

For these choices of a and b,

a · b = 1 , |a|2 = 2 .

Consequently, the vector projection of−−→AB

onto−−→AD is given by

projab =

1

2(j − k) .

keywords: vector projection, dot product,unit cube, component,

1

Documents

Transcript of 1