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REVIEW FOR THE EXAM 1 MATH 203 F 2014 (6330305)

Question

1.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Question Details

SEssCalc2 10.1.013. [2166860]

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SEssCalc2 10.1.015. [2166600]

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SEssCalc2 10.1.005. [2154891]

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Write the equation of the sphere in standard form. x2 + y2 + z2 + 4x − 2y − 6z = 2

Find its center and radius. center

(x, y, z) =

radius

Solution or Explanation Click to View Solution

2.

Question Details

Write the equation of the sphere in standard form. 2x2 + 2y2 + 2z2 = 4x − 20z + 1

Find its center and radius. center

(x, y, z) =

radius

Solution or Explanation Click to View Solution

3.

Question Details

Describe the surface in

3 represented by the equation x + y = 8.

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This is the set

{(x, 8 − x, z)|x

,z

} which is a vertical plane that intersects the xy-plane in the line

y = 8 − x, z = 0. This is the set

{(x, 8 − x, z)|x

,z

} which is a vertical plane that intersects the xz-plane in the line

,z

} which is a horizontal plane that intersects the xy-plane in the line

,z

} which is a horizontal plane that intersects the xz-plane in the line

y = 8 − x, z = 0. This is the set

{(x, 8 − x, z)|x

y = 8 − x, z = 0. This is the set

{(x, 8 − x, z)|x

y = 8 − x, z = 0. This is the set

{(x, y, 8 − x − y)|x

,y

} which is a vertical plane that intersects the xy-plane in the line

y = 8 − x, z = 0.

Sketch the surface.

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Solution or Explanation The equation x + y = 8 represents the set of all points in

3 whose x- and y-coordinates have a sum of 8, or equivalently

where y = 8 − x. This is the set {(x, 8 − x, z)|x

} which is a vertical plane that intersects the xy-plane in the

,z

line y = 8 − x, z = 0.

4.

Question Details

SEssCalc2 10.1.004. [2166901]

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Consider the point. (1, 4, 6) What is the projection of the point on the xy-plane? (x, y, z) =

What is the projection of the point on the yz-plane?

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(x, y, z) =

What is the projection of the point on the xz-plane? (x, y, z) =

Draw a rectangular box with the origin and (1, 4, 6) as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box.

Find the length of the diagonal of the box.

Solution or Explanation

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Click to View Solution

5.

Question Details

SEssCalc2 10.1.025. [2154143]

Describe in words the region of

-

3 represented by the inequality.

0≤z≤5 The inequality 0 ≤ z ≤ 5 represents all points z = 0 (the

---Select---

on or between the

---Select---

horizontal planes

xy -plane) and z = 5.

?

Solution or Explanation The inequality 0 ≤ z ≤ 5 represents all points on or between the horizontal planes z = 0 (the xy-plane) and z = 5.

6.

Question Details

SEssCalc2 10.2.013. [2166616]

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Find a + b, 2a + 3b, |a|, and |a − b|. a = 4, −3 ,

b = −4, 3

a+b = 2a + 3b = |a| = |a − b| =

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.2.016. [2166777]

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SEssCalc2 10.2.018. [2166827]

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Find a + b, 2a + 3b, |a|, and |a − b|. a = 2i − 4j + 3k,

b = 2j − k

a+b = 2a + 3b = |a| = |a − b| =

Solution or Explanation Click to View Solution

8.

Question Details

Find a vector that has the same direction as

−2, 6, 2

but has length 6.

Solution or Explanation −2, 6, 2

=

(−2)2 + 62 + 22 =

44 = 2

11 , so a unit vector in the direction of −2, 6, 2 is u = 2

A vector in the same direction but with length 6 is 6u = 6 ·

9.

1 2

11

−2, 6, 2 =

3 11

1 11

−2, 6, 2 .

−2, 6, 2 .

Question Details

SEssCalc2 10.2.505.XP. [2166732]

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Find a unit vector that has the same direction as the given vector. −4, 8, 8

Solution or Explanation −4, 8, 8

=

(−4)2 + 82 + 82 =

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1 −4, 8, 8 = − 1 , 2 , 2 . 144 = 12, so u = 12 3 3 3

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Question Details

SEssCalc2 10.3.035. [2166757]

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SEssCalc2 10.3.032. [2166467]

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SEssCalc2 10.3.025. [2155246]

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If a = 5, 0, −1 , find a vector b such that compab = 2. b=

Solution or Explanation Click to View Solution

11.

Question Details

Find the scalar and vector projections of b onto a. a = i + j + k,

b=i−j+k

compab =

projab =

Solution or Explanation Click to View Solution

12.

Question Details

Find the acute angle between the lines. Round your answer to the nearest degree. 3x − y = 5,

2x + y = 8

45 ° Solution or Explanation The line 3x − y = 5 2x + y = 8

y = 3x − 5 has slope 3, so a vector parallel to the line is a = 1, 3 . The line

y = −2x + 8 has slope −2, so a vector parallel to the line is b = 1, −2 . The angle between the lines is the

same as the angle θ between the vectors. Here we have a · b = (1)(1) + (3)(−2) = −5, |a| = |b| =

12 + (−2)2 =

a·b = 5 , so cos θ = |a||b|

−5 10 ·

5

=

−5 50

. Thus θ = cos−1

−5 50

12 + 32 =

10 , and

≈ 135°, and the acute

angle between the lines is 180° − 135° = 45°.

13.

Question Details

SEssCalc2 10.3.023. [2166764]

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Find a unit vector that is orthogonal to both i + j and i + k.

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.3.021. [2154522]

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Use vectors to decide whether the triangle with vertices P(0, −4, −3), Q(1, −1, −5), and R(5, −3, −6) is right-angled. Yes, it is right-angled. No, it is not right-angled.

Solution or Explanation QP = −1, −3, −2 , QR = 4, −2, −1 , and QP · QR = −4 + 6 + −2 = 0. Thus QP and QR are orthogonal, so the angle of the triangle at vertex Q is a right angle.

15.

Question Details

SEssCalc2 10.3.020. [2155351]

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Determine whether the given vectors are orthogonal, parallel, or neither. (a)

u = −3, 3, 6 ,

v = 4, −4, −8

orthogonal parallel neither

(b)

u = i − j + 3k,

v = 3i − j + k

orthogonal parallel neither

(c)

u = a, b, c ,

v = −b, a, 0

orthogonal parallel neither

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.3.011. [2154447]

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If u is a unit vector, find u · v and u · w. (Assume v and w are also unit vectors.)

u·v =

1/2

u · w=

-1/2

Solution or Explanation u, v, and w are all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60° and u · v = |u||v| cos 60° = (1)(1) 1 = 1 . If w is moved so it has the same initial point as u, we can see that the angle 2 2 between them is 120° and we have u · w = |u||w| cos 120° = (1)(1) − 1 = − 1 . 2 2

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9/17/14, 6:58 AM

Question Details

SEssCalc2 10.4.027. [2154290]

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Find the area of the parallelogram with vertices A(−3, 5), B(−1, 8), C(3, 6), and D(1, 3). 16 Solution or Explanation By plotting the vertices, we can see that the parallelogram is determined by the vectors AB = 2, 3 and AD = 4, −2 . We know that the area of the parallelogram determined by two vectors is equal to the length of the cross product of these vectors. In order to compute the cross product, we consider the vector AB as the three-dimensional vector

2, 3, 0

(and

similarly for AD), and then the area of parallelogram ABCD is

AB

× AD

=

i 2 4

j 3 −2

k 0 0

= (0)i − (0)j + (−4 − 12)k = |−16k| = 16.

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Question Details

SEssCalc2 10.4.029. [2166894]

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Consider the points below. P(1, 0, 1), Q(−2, 1, 3), R(4, 2, 5) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.

(b) Find the area of the triangle PQR.

Solution or Explanation (a) Because the plane through P, Q, and R contains the vectors PQ and PR, a vector orthogonal to both of these vectors (such as their cross product) is also orthogonal to the plane. Here PQ = −3, 1, 2 PQ Therefore,

× PR =

and PR = 3, 2, 4 , so

(1)(4) − (2)(2), (3)(2) − (−3)(4), (−3)(2) − (3)(1) = 0, 18, −9

0, 18, −9

or any nonzero scalar multiple thereof, such as

0, −18, 9

is orthogonal to the plane through P, Q,

and R. (b) Note that the area of the triangle determined by P, Q, and R is equal to half of the area of the parallelogram determined by the three points. From part (a), the area of the parallelogram is PQ

19.

× PR

=

0, 18, −9

=

0 + 324 + 81 =

405 = 9

1 ·9 5 , so the area of the triangle is 2

Question Details

5 =

9

5 2

.

SEssCalc2 10.4.032. [2166900]

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Consider the points below. P(−1, 2, 1),

Q(0, 6, 3),

R(5, 3, −1)

(a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.

(b) Find the area of the triangle PQR.

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.4.033. [2154642]

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Find the volume of the parallelepiped determined by the vectors a, b, and c. a=

1, 5, 4 ,

b=

−1, 1, 5 ,

c=

3, 1, 3

72 cubic units Solution or Explanation Recalling that the volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product, V = |a · (b a · (b

× c) =

× c)| , one obtains

1 5 4 1 5 −1 5 −1 1 −5 +4 = 1 · (3 − 5) − 5 · (−3 − 15) + 4 · (−1 − 3) = 72. −1 1 5 = 1 1 3 3 3 3 1 3 1 3

Thus the volume of the parallelepiped is 72 cubic units.

21.

Question Details

SEssCalc2 10.4.037. [2155392]

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Use the scalar triple product to determine if the vectors u = i + 5j − 2k, v = 4i − j, and w = 8i + 14j − 6k are coplanar. Yes, they are coplanar. No, they are not coplanar.

Solution or Explanation Click to View Solution

22.

Question Details

SEssCalc2 10.4.038. [2155066]

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Use the scalar triple product to determine whether the points A(1, 2, 3), B(4, −3, 7), C(6, 1, 2), and D(3, 6, −2) lie in the same plane. Yes, they lie in the same plane. No, they do not lie in the same plane.

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.4.039.MI. [2166867]

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A bicycle pedal is pushed by a foot with a 60-N force as shown. The shaft of the pedal is 18 cm long. Find the magnitude of the torque about P. (Round your answer to one decimal place.) 10.6 N · m

Solution or Explanation Click to View Solution

24.

Question Details

SEssCalc2 10.4.040. [2154363]

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Find the magnitude of the torque about P if an F = 76-lb force is applied as shown. (Round your answer to the nearest whole number.) 415 ft-lb

Solution or Explanation Click to View Solution

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SEssCalc2 10.5.014. [2166528]

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(a) Find parametric equations for the line through (2, 3, 4) that is perpendicular to the plane x − y + 2z = 6. (Use the parameter t.) (x(t), y(t), z(t)) =

(b) In what points does this line intersect the coordinate planes? xy-plane

(x, y, z) =

yz-plane

(x, y, z) =

xz-plane

(x, y, z) =

Solution or Explanation Click to View Solution

26.

Question Details

SEssCalc2 10.5.015. [2166604]

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Find a vector equation for the line segment from (3, −1, 4) to (7, 5, 3). (Use the parameter t.) r(t) =

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.5.017. [2166728]

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Determine whether the lines L1 and L2 are parallel, skew, or intersecting. L1: x = 9 + 6t, y = 12 − 3t, z = 3 + 9t L2: x = 2 + 8s, y = 6 − 4s, z = 8 + 10s parallel skew intersecting

If they intersect, find the point of intersection. (If an answer does not exist, enter DNE.)

Solution or Explanation Since the direction vectors

6, −3, 9

and

8, −4, 10

are not scalar multiples of each other, the lines aren't parallel. For the

lines to intersect, we must be able to find one value of t and one value of s that produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations: 9 + 6t = 2 + 8s, 12 − 3t = 6 − 4s, 3 + 9t = 8 + 10s. Solving the last two equations we get t = − 20 , s = − 13 and checking, we see that these values don't 3 2 satisfy the first equation. Thus the lines aren't parallel and don't intersect, so they must be skew lines.

28.

Question Details

SEssCalc2 10.5.023. [2166780]

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SEssCalc2 10.5.025. [2166693]

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Find an equation of the plane. The plane through the point (7, −1, −2) and parallel to the plane 9x − y − z = 9

Solution or Explanation Click to View Solution

29.

Question Details

Find an equation of the plane. The plane through the points (0, 9, 9), (9, 0, 9), and (9, 9, 0)

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.5.026. [2166488]

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SEssCalc2 10.5.027. [2166507]

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Find an equation of the plane. The plane through the origin and the points (2, −2, 7) and (9, 2, 4)

Solution or Explanation Click to View Solution

31.

Question Details

Find an equation of the plane. The plane that passes through (9, 0, −2) and contains the line x = 7 − 2t, y = 1 + 3t, z = 3 + 2t

Solution or Explanation Click to View Solution

32.

Question Details

SEssCalc2 10.5.029. [2166579]

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Find an equation of the plane. The plane that passes through the point (−2, 1, 2) and contains the line of intersection of the planes x + y − z = 5 and 4x − y + 5z = 2

Solution or Explanation Click to View Solution

33.

Question Details

SEssCalc2 10.5.032. [2166532]

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Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 2 and y + 4z = 1 and is perpendicular to the plane x + y − 4z = 3

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.5.033. [2377880]

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Find the point at which the line x = 5 − t, y = 4 + t, z = 4t intersects the plane x − y + 5z = 19. (x, y, z) =

Solution or Explanation Click to View Solution

35.

Question Details

SEssCalc2 10.5.035. [2166328]

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Determine whether the planes are parallel, perpendicular, or neither. 9x + 9y + 9z = 1, 9x − 9y + 9z = 1 parallel perpendicular neither

If neither, find the angle between them. (Round your answer to one decimal place. If the planes are parallel or perpendicular, enter PARALLEL or PERPENDICULAR, respectively.) °

Solution or Explanation Normal vector for the planes are n1 = 9, 9, 9 and n2 = 9, −9, 9 . The normals are not parallel, so neither are the planes. Furthermore, n1 · n2 = 81 − 81 + 81 = 81 ≠ 0, so the planes aren't perpendicular. The angle between them is given by 81 n1 · n2 cos θ = = = 81 = 1 θ = cos−1 1 ≈ 70.5°. 243 243 |n1||n2| 243 3 3

36.

Question Details

SEssCalc2 10.5.039. [2378157]

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(a) Find parametric equations for the line of intersection of the planes x + y + z = 2 and x + 3y + 3z = 2. (x(t), y(t), z(t)) =

(b) Find the angle between these planes. (Round your answer to one decimal place.) 22.0 ° Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.5.049. [2166817]

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Find the distance from the point to the given plane. (1, −3, 8),

3x + 2y + 6z = 5

Solution or Explanation By the equation D =

38.

|ax1 + by1 + cz1 + d| a2 + b2 + c2

, the distance is D =

|3(1) + 2(−3) + 6(8) − 5| 32 + 22 + 62

=

Question Details

|40| 49

= 40 . 7

SEssCalc2 10.5.050. [2166544]

-

Find the distance from the point to the given plane. (−9, 6, 5),

x − 2y − 4z = 8

Solution or Explanation By the equation D =

39.

|ax1 + by1 + cz1 + d| a2 + b2 + c2

, the distance is D =

Question Details

|1(−9) − 2(6) − 4(5) − 8| 12 + (−2)2 + (−4)2

=

|−49| 21

=

49 21

.

SEssCalc2 10.5.051. [2166516]

-

Find the distance between the given parallel planes. 5x − 5y + z = 20,

10x − 10y + 2z = 2

Solution or Explanation Click to View Solution

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Question Details

SEssCalc2 10.5.042. [2166365]

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(a) Find the point at which the given lines intersect. r = 1, 2, 0 + t 2, −2, 2 r = 3, 0, 2 + s −2, 2, 0 (x, y, z) =

(b) Find an equation of the plane that contains these lines.

Solution or Explanation Click to View Solution

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Name (AID): REVIEW FOR THE EXAM 1 MATH 203 F 2014 (6330305)

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