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Vectors Classwork & Homework Solutions | 673.7KB

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Intro to Vectors – Class Work
Draw vectors to represent the scenarios.
1. A plane flies east at 300 mph.
2. A ship sails northwest at 20 knots.
3. A river flows south at 4 mph.
Draw the following vector. Show the component forces for the given vectors.
4. π‘Žβƒ— = (6, −8)
5. 𝐢⃗ = (−3,7)
6. 𝑒
βƒ—βƒ— π‘—π‘œπ‘–π‘›π‘  (2,3) π‘‘π‘œ (4, −9)
7. v
βƒ—βƒ— joins (4, −9) to (2,3)
Referring to Questions 4-7, find the following
8. |π‘Žβƒ—|
9. |𝐢⃗|
11. |𝑣⃗|
𝟏𝟎
10. |𝑒
βƒ—βƒ—|
𝟐√πŸ‘πŸ•
√πŸ“πŸ–
𝟐√πŸ‘πŸ•
Referring to Questions 4-7, find the direction of the vector
12. π‘Žβƒ—
13. 𝐢⃗
14. 𝑒
βƒ—βƒ—
πŸ“πŸ‘. πŸπŸ‘π’ 𝑺 𝒐𝒇 𝑬
πŸ”πŸ”. πŸ–π’ 𝑡 𝒐𝒇 𝑾
15. v
βƒ—βƒ—
πŸ–πŸŽ. πŸ“π’ 𝑺 𝒐𝒇 𝑬
πŸ–πŸŽ. πŸ“π’ 𝑡 𝒐𝒇 𝑾
Intro to Vectors – Homework
Draw vectors to represent the scenarios.
16. A plane flies west at 200 mph.
17. A ship sails northeast at 10 knots.
18. A river flows north at 3 mph.
Draw the following vector. Show the component forces for the given vectors.
19. π‘Žβƒ— = (−3, −4)
20. 𝐢⃗ = (5, −12)
21. 𝑒
βƒ—βƒ— π‘—π‘œπ‘–π‘›π‘  (1,4) π‘‘π‘œ (8,6)
22. v
βƒ—βƒ— joins (−2,3) to (3, −2)
Referring to Questions 16-19, find the following
23. |π‘Žβƒ—|
24. |𝐢⃗|
25. |𝑒
βƒ—βƒ—|
26. |𝑣⃗|
πŸ“
πŸπŸ‘
πŸ“√𝟐
√πŸ“πŸ‘
Referring to Questions 16-19, find the direction of the vector
27. π‘Žβƒ—
28. 𝐢⃗
29. 𝑒
βƒ—βƒ—
𝒐
πŸ“πŸ‘. πŸπŸ‘ 𝑺 𝒐𝒇 𝑬
Pre-Calc Vectors KEY
𝒐
πŸ“πŸ‘. πŸπŸ‘ 𝑺 𝒐𝒇 𝑬
30. v
βƒ—βƒ—
𝒐
πŸ“πŸ‘. πŸπŸ‘ 𝑺 𝒐𝒇 𝑬
~1~
πŸ“πŸ‘. πŸπŸ‘π’ 𝑺 𝒐𝒇 𝑬
NJCTL.org
Converting Between Rectangular and Polar Forms – Class Work
Convert the following Polar coordinates to rectangular.
31. (4, 45°)
32. (3, 60°)
(𝟐√𝟐, 𝟐√𝟐)
πŸ‘ πŸ‘√πŸ‘
( ,
𝟐
π
34. (7, )
𝟐
35. (12, −
5
(πŸ“. πŸ”πŸ”, πŸ’. 𝟏𝟏)
7
(−πŸ‘. πŸ’πŸ, πŸ—. πŸ’)
)
4π
)
(−𝟐. πŸ”πŸ•, −𝟏𝟏. πŸ•)
Convert the following rectangular coordinates to polar.
36. (5, 8)
37. (4, -9)
(πŸ—. πŸ’πŸ‘, πŸ“πŸ•. πŸ—πŸ—π’ )
39. (-4,-2)
38. (-3,6)
(πŸ—. πŸ–πŸ“, πŸπŸ—πŸ‘. πŸ—πŸ”π’ )
(πŸ‘, πŸπŸ•πŸŽπ’ )
Converting Between Rectangular and Polar Forms – Homework
Convert the following Polar coordinates to rectangular.
41. (5, 135°)
42. (6, 30°)
44. (5,
πŸ“√𝟐 πŸ“√𝟐
𝟐
,
𝟐
)
3π
5
(πŸ”. πŸ•πŸ, πŸπŸ—πŸ”. πŸ“πŸ•π’ )
40. (0, -3)
(πŸ’. πŸ’πŸ•, πŸπŸŽπŸ”. πŸ“πŸ•π’ )
(−
33. (10, 110°)
)
(−𝟏. πŸ“πŸ“, πŸ’. πŸ•πŸ”)
(−𝟏𝟏. πŸ–πŸ, 𝟐. πŸŽπŸ–)
(πŸ‘√πŸ‘, πŸ‘)
45. (−12, −
3π
7
)
(−𝟐. πŸ”πŸ•, 𝟏𝟏. πŸ•)
Convert the following rectangular coordinates to polar.
46. (4, 2)
47. (3, -9)
(πŸ’. πŸ’πŸ•, πŸπŸ”. πŸ“πŸ•π’ )
49. (-7, 8)
(𝟏𝟎. πŸ”πŸ‘, πŸπŸ‘πŸ. πŸπŸ–π’ )
Pre-Calc Vectors KEY
43. (12,170°)
(πŸ—. πŸ’πŸ—, πŸπŸ–πŸ–. πŸ’πŸ‘π’ )
48. (12, -12)
(𝟏𝟐√𝟐, πŸ‘πŸπŸ“π’ )
50. (-4 , -5)
(πŸ”. πŸ’, πŸπŸ‘πŸ. πŸ‘πŸ’π’ )
~2~
NJCTL.org
Scalar Multiplication – Class Work
Given 𝑒
βƒ—βƒ— = (4,2) π‘Žπ‘›π‘‘ 𝑣⃗(−2,3), find the following and draw the transformation.
51. 2𝑒
βƒ—βƒ—
52. 3𝑣⃗
(πŸ–, πŸ’)
53.
1
2
𝑒
βƒ—βƒ—
(−πŸ”, πŸ—)
54. −𝑣⃗
(𝟐, 𝟏)
(𝟐, −πŸ‘)
Scalar Multiplication – Homework
Given 𝑒
βƒ—βƒ— = (5, −4) π‘Žπ‘›π‘‘ 𝑣⃗(−3, −2), find the following and draw the transformation.
55. 2𝑒
βƒ—βƒ—
56. 3𝑣⃗
(𝟏𝟎, −πŸ–)
Pre-Calc Vectors KEY
57.
(−πŸ—, −πŸ”)
1
2
𝑒
βƒ—βƒ—
(𝟐. πŸ“, −𝟐)
~3~
58. −𝑣⃗
(πŸ‘, 𝟐)
NJCTL.org
Vector Addition – Class Work
Use the vectors to draw the expression. Draw the resultant vector.
59. 𝐸⃗⃗ + 𝐹⃗
⃗⃗ + ⃗B⃗
60. βƒ—D
βƒ—βƒ—
62. 2C
βƒ—βƒ— + 𝐹⃗
63. 2𝐡
61. 𝐴⃗ + 𝐢⃗
Find the resultant vector. State the magnitude and direction of each resultant vector.
64. 𝑒
βƒ—βƒ— = (4,2) π‘Žπ‘›π‘‘ 𝑣⃗ = (3,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— + 𝑣⃗
65. 𝑒
βƒ—βƒ— = (−5,6) π‘Žπ‘›π‘‘ 𝑣⃗ = (−3,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— + 𝑣⃗
βƒ—βƒ— + 𝒗
βƒ—βƒ— = (πŸ•, πŸ’)
𝒖
|𝒖
βƒ—βƒ— + 𝒗
βƒ—βƒ—| = πŸ–. πŸŽπŸ”
𝜽 = πŸπŸ—. πŸ•π’
βƒ—βƒ— + 𝒗
βƒ—βƒ— = (−πŸ–, πŸ–)
𝒖
|𝒖
βƒ—βƒ— + 𝒗
βƒ—βƒ—| = πŸ–√𝟐
𝜽 = πŸπŸ‘πŸ“π’
66. A tug-of-war contest is made up of 2-member teams. Consider the rope to be on the x-axis with the flag at the
origin. Team A’s members pull (3,2) and (4,-1). Team B’s members pull (-4,0) and (-3,0). What is the
magnitude and direction of each team’s actions? What is the movement of the flag?
⃗𝑨⃗ = (πŸ•, 𝟏)
βƒ—βƒ—βƒ—| = πŸ“√𝟐
|𝑨
⃗𝑩
βƒ—βƒ— = (−πŸ•, 𝟎)
βƒ—βƒ—βƒ—| = πŸ•
|𝑩
𝜽 = πŸ–. πŸπŸ‘π’
𝜽 = πŸπŸ–πŸŽπ’
⃗𝑨
βƒ—βƒ— + ⃗𝑩
βƒ—βƒ— = (𝟎, 𝟏)
βƒ—βƒ— + ⃗𝑩
βƒ—βƒ—| = 𝟏
|𝑨
𝜽 = πŸ—πŸŽπ’
Pre-Calc Vectors KEY
~4~
NJCTL.org
Vector Addition – Homework
Use the vectors to draw the expression. Draw the resultant vector.
βƒ—βƒ— + 𝐹⃗
67. 𝐡
68. ⃗E⃗ + ⃗B⃗
βƒ—βƒ—βƒ—
70. 3D
71. 𝐸⃗⃗ + 2𝐹⃗
βƒ—βƒ— + 𝐢⃗
69. 𝐷
Find the resultant vector. State the magnitude and direction of each resultant vector.
72. 𝑒
βƒ—βƒ— = (5,1) π‘Žπ‘›π‘‘ 𝑣⃗ = (−5,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— + 𝑣⃗
73. 𝑒
βƒ—βƒ— = (−5,3) π‘Žπ‘›π‘‘ 𝑣⃗ = (−3,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— + 𝑣⃗
βƒ—βƒ— + 𝒗
βƒ—βƒ— = (𝟎, πŸ‘)
𝒖
|𝒖
βƒ—βƒ— + 𝒗
βƒ—βƒ—| = πŸ‘
𝜽 = πŸ—πŸŽπ’
βƒ—βƒ— + 𝒗
βƒ—βƒ— = (−πŸ–, πŸ“)
𝒖
|𝒖
βƒ—βƒ— + 𝒗
βƒ—βƒ—| = πŸ—. πŸ’πŸ‘
𝜽 = πŸπŸ’πŸ•. πŸ—πŸ—π’
74. A tug-of-war contest is made up of 2-member teams. Consider the rope to be on the x-axis with the flag at the
origin. Team A’s members pull (4,3) and (4,-2). Team B’s members pull (-5,0) and (-4,0). What is the
magnitude and direction of each team’s actions? What is the movement of the flag?
⃗𝑨⃗ = (πŸ–, 𝟏)
βƒ—βƒ—βƒ—| = πŸ–. πŸŽπŸ”
|𝑨
⃗𝑩
βƒ—βƒ— = (−πŸ—, 𝟎)
βƒ—βƒ—βƒ—| = πŸ—
|𝑩
𝜽 = πŸ•. πŸπŸ‘π’
𝜽 = πŸπŸ–πŸŽπ’
βƒ—βƒ—βƒ— + 𝑩
βƒ—βƒ—βƒ— = (−𝟏, 𝟏)
𝑨
βƒ—βƒ— + ⃗𝑩
βƒ—βƒ—| = 𝟏. πŸ’πŸ
|𝑨
𝜽 = πŸπŸ‘πŸ“π’
Pre-Calc Vectors KEY
~5~
NJCTL.org
Vector Subtraction – Class Work
Use the vectors to draw the expression. Draw the resultant vector.
βƒ—βƒ— − 𝐹⃗
75. 𝐡
βƒ—βƒ— − B
βƒ—βƒ—
76. E
βƒ—βƒ—βƒ— − βƒ—Aβƒ—
78. 3D
βƒ—βƒ—
79. 𝐸⃗⃗ − 2𝐹⃗ + 𝐡
βƒ—βƒ— − 𝐢⃗
77. 𝐷
Find the resultant vector. State the magnitude and direction of each resultant vector.
80. 𝑒
βƒ—βƒ— = (5,1) π‘Žπ‘›π‘‘ 𝑣⃗ = (−5,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— − 𝑣⃗
81. 𝑒
βƒ—βƒ— = (−5,3) π‘Žπ‘›π‘‘ 𝑣⃗ = (−3,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— − 𝑣⃗
βƒ—βƒ— − 𝒗
βƒ—βƒ— = (𝟏𝟎, −𝟏)
𝒖
|𝒖
βƒ—βƒ— − 𝒗
βƒ—βƒ—| = 𝟏𝟎. πŸŽπŸ“
𝜽 = πŸ‘πŸ“πŸ’. πŸ‘π’
βƒ—βƒ— − 𝒗
βƒ—βƒ— = (−𝟐, 𝟏)
𝒖
|𝒖
βƒ—βƒ— − 𝒗
βƒ—βƒ—| = 𝟐. πŸπŸ’
𝜽 = πŸπŸπŸ”. πŸ”π’
82. 𝑒
βƒ—βƒ— = (−3,1) π‘Žπ‘›π‘‘ 𝑣⃗ = (−4, −2), 𝑓𝑖𝑛𝑑 2𝑒
βƒ—βƒ— − 3𝑣⃗
βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—
βƒ—βƒ— = (πŸ”, πŸ–)
πŸπ’– − πŸ‘π’—
|πŸπ’–
βƒ—βƒ— − πŸ‘π’—
βƒ—βƒ—| = 𝟏𝟎
𝜽 = πŸ“πŸ‘. πŸπ’
Pre-Calc Vectors KEY
~6~
NJCTL.org
Vector Subtraction – Homework
Use the vectors to draw the expression. Draw the resultant vector.
83. 𝐴⃗ − 𝐢⃗
βƒ—βƒ—βƒ— − F
βƒ—βƒ—
84. D
βƒ—βƒ— − βƒ—Fβƒ—
86. 3C
βƒ—βƒ— − 3𝐴⃗ + 𝐢⃗
87. 𝐷
βƒ—βƒ—
85. 𝐢⃗ − 𝐡
Find the resultant vector. State the magnitude and direction of each resultant vector.
88. 𝑒
βƒ—βƒ— = (2, −3) π‘Žπ‘›π‘‘ 𝑣⃗ = (−4,2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— − 𝑣⃗
89. 𝑒
βƒ—βƒ— = (−4,6) π‘Žπ‘›π‘‘ 𝑣⃗ = (−7, −2), 𝑓𝑖𝑛𝑑 𝑒
βƒ—βƒ— − 𝑣⃗
βƒ—βƒ— − 𝒗
βƒ—βƒ— = (πŸ”, −πŸ“)
𝒖
|𝒖
βƒ—βƒ— − 𝒗
βƒ—βƒ—| = πŸ•. πŸ–πŸ
𝜽 = πŸ‘πŸπŸŽ. πŸπ’
βƒ—βƒ— − 𝒗
βƒ—βƒ— = (πŸ‘, πŸ–)
𝒖
|𝒖
βƒ—βƒ— − 𝒗
βƒ—βƒ—| = πŸ–. πŸ“πŸ’
𝜽 = πŸ”πŸ—. πŸ’π’
90. 𝑒
βƒ—βƒ— = (−5,4) π‘Žπ‘›π‘‘ 𝑣⃗ = (−8,0), 𝑓𝑖𝑛𝑑 3𝑒
βƒ—βƒ— − 2𝑣⃗
βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—
βƒ—βƒ— = (𝟏, 𝟏𝟐)
πŸ‘π’– − πŸπ’—
|𝒖
βƒ—βƒ— − 𝒗
βƒ—βƒ—| = 𝟏𝟐. πŸŽπŸ“
𝜽 = πŸ–πŸ“. πŸπ’
Pre-Calc Vectors KEY
~7~
NJCTL.org
Vector Equations of a Line – Class Work
Write the vector equation of the line and the parametric equation for the line:
91. through (7, 4) and parallel to 𝑣⃗ = (1, −2).
92. through (-3, 5) and parallel to 𝑣⃗ = (4,3).
(𝒙 − πŸ•, π’š − πŸ’) = 𝒕(𝟏, −𝟐)
𝒙= πŸ•+𝒕
{
π’š = πŸ’ − πŸπ’•
(𝒙 + πŸ‘, π’š − πŸ“) = 𝒕(πŸ’, πŸ‘)
𝒙 = −πŸ‘ + πŸ’π’•
{
π’š = πŸ“ + πŸ‘π’•
93. through (11, 0) and parallel to 𝑣⃗ = (−7,0).
94. through (-5, -8) and parallel to 𝑣⃗ = (0,8).
(𝒙 − 𝟏𝟏, π’š) = 𝒕(−πŸ•, 𝟎)
𝒙 = 𝟏𝟏 − πŸ•π’•
{
π’š=𝟎
(𝒙 + πŸ“, π’š + πŸ–) = 𝒕(𝟎, πŸ–)
𝒙 = −πŸ“
{
π’š = −πŸ– + πŸ–π’•
95. through (6, -1) and parallel to 𝑣⃗ = (−9, −10).
96. through (2, 8) and (5, 9)
(𝒙 − πŸ”, π’š + 𝟏) = 𝒕(−πŸ—, −𝟏𝟎)
𝒙 = πŸ” − πŸ—π’•
{
π’š = −𝟏 − πŸπŸŽπ’•
(𝒙 − 𝟐, π’š − πŸ–) = 𝒕(πŸ‘, 𝟏)
𝒙 = 𝟐 + πŸ‘π’•
{
π’š= πŸ–+𝒕
π‘₯ = 2 + 4𝑑
98. Write the vector equation for {
𝑦 = 3 − 7𝑑
97. through (0, 3) and (7,0)
(𝒙, π’š − πŸ‘) = 𝒕(πŸ•, −πŸ‘)
𝒙 = πŸ•π’•
{
π’š = πŸ‘ − πŸ‘π’•
(𝒙 − 𝟐, π’š − πŸ‘) = 𝒕(πŸ’, −πŸ•)
Vector Equations of a Line – Homework
Write the vector equation of the line and the parametric equation for the line:
99. through (3, 9) and parallel to 𝑣⃗ = (2, −5).
100. through (-11, 13) and parallel to 𝑣⃗ = (6,10).
(𝒙 − πŸ‘, π’š − πŸ—) = 𝒕(𝟐, −πŸ“)
𝒙 = πŸ‘ + πŸπ’•
{
π’š = πŸ— − πŸ“π’•
101.
(𝒙 + 𝟏𝟏, π’š − πŸπŸ‘) = 𝒕(πŸ”, 𝟏𝟎)
𝒙 = −𝟏𝟏 + πŸ”π’•
{
π’š = πŸπŸ‘ + πŸπŸŽπ’•
through (2, 14) and parallel to 𝑣⃗ = (−11,2).
(𝒙 − 𝟐, π’š − πŸπŸ’) = 𝒕(−𝟏𝟏, 𝟐)
𝒙 = 𝟐 − πŸπŸπ’•
{
π’š = πŸπŸ’ + πŸπ’•
103.
(𝒙 + πŸ’, π’š + πŸ—) = 𝒕(πŸ‘, πŸπŸ–)
𝒙 = −πŸ’ + πŸ‘π’•
{
π’š = −πŸ— + πŸπŸ–π’•
through (1, -3) and parallel to 𝑣⃗ = (−12, −11).
(𝒙 − 𝟏, π’š + πŸ‘) = 𝒕(−𝟏𝟐, −𝟏𝟏)
𝒙 = 𝟏 − πŸπŸπ’•
{
π’š = −πŸ‘ − πŸπŸπ’•
105.
104. through (5, 7) and (-4, 3)
(𝒙 − πŸ“, π’š − πŸ•) = 𝒕(−πŸ—, −πŸ’)
𝒙 = πŸ“ − πŸ—π’•
{
π’š = πŸ• − πŸ’π’•
106. Write the vector equation for {
through (1, 7) and (-4,7)
(𝒙 − 𝟏, π’š − πŸ•) = 𝒕(−πŸ“, 𝟎)
𝒙 = 𝟏 − πŸ“π’•
{
π’š=πŸ•
Pre-Calc Vectors KEY
102. through (-4, -9) and parallel to 𝑣⃗ = (3,18).
π‘₯ = −2 + 5𝑑
𝑦 = 3𝑑
(𝒙 + 𝟐, π’š) = 𝒕(πŸ“, πŸ‘)
~8~
NJCTL.org
Dot Product – Class Work
Find the dot product of the vectors. State whether they are perpendicular or form an obtuse or acute angle.
107.
π‘Žβƒ— = (2,4) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (3,5)
108. π‘Žβƒ— = (3, −2) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (4,6)
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = πŸπŸ”
𝒂
𝒂𝒄𝒖𝒕𝒆
109.
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = 𝟎
𝒂
π’‘π’†π’“π’‘π’†π’π’…π’Šπ’„π’–π’π’‚π’“
π‘Žβƒ— = (−2,1) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (4, −2)
110. π‘Žβƒ— = (−5,8) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (10,6)
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = −𝟏𝟎
𝒂
𝒐𝒃𝒕𝒖𝒔𝒆
111.
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = −𝟐
𝒂
𝒐𝒃𝒕𝒖𝒔𝒆
π‘Žβƒ— = (8, −4) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (3,6)
112. π‘Žβƒ— = (−4,6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (9, −6)
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = 𝟎
𝒂
π’‘π’†π’“π’‘π’†π’π’…π’Šπ’„π’–π’π’‚π’“
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = −πŸ•πŸ
𝒂
𝒐𝒃𝒕𝒖𝒔𝒆
Dot Product – Homework
Find the dot product of the vectors. State whether they are perpendicular or form an obtuse or acute angle.
113.
π‘Žβƒ— = (3,6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (2, −9)
114. π‘Žβƒ— = (8,4) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (3, −6)
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = −πŸ’πŸ–
𝒂
𝒐𝒃𝒕𝒖𝒔𝒆
115.
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = 𝟎
𝒂
π’‘π’†π’“π’‘π’†π’π’…π’Šπ’„π’–π’π’‚π’“
π‘Žβƒ— = (10,8) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (4,5)
116. π‘Žβƒ— = (3,4) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (9,12)
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = πŸ–πŸŽ
𝒂
𝒂𝒄𝒖𝒕𝒆
117.
βƒ—βƒ— βˆ™ ⃗𝒃⃗ = πŸ•πŸ“
𝒂
𝒂𝒄𝒖𝒕𝒆
π‘Žβƒ— = (0,9) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (7,5)
118. π‘Žβƒ— = (−2,8) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (4,1)
βƒ—βƒ— = πŸ’πŸ“
βƒ—βƒ— βˆ™ 𝒃
𝒂
𝒂𝒄𝒖𝒕𝒆
Pre-Calc Vectors KEY
βƒ—βƒ— = 𝟎
βƒ—βƒ— βˆ™ 𝒃
𝒂
π’‘π’†π’“π’‘π’†π’π’…π’Šπ’„π’–π’π’‚π’“
~9~
NJCTL.org
Angles Between Vectors – Class Work
Find the angle between the two given vectors.
119.
π‘Žβƒ— = (2,4) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (7,1)
120. cβƒ— = (−1,4) and βƒ—βƒ—
d = (8,2)
𝜽 = πŸ“πŸ“. πŸ‘π’
121.
𝜽 = πŸ—πŸŽπ’
𝑑⃗ = (−3,0) π‘Žπ‘›π‘‘ 𝑒⃗ = (3, −1)
122. 𝑓⃗ = (4, −3)π‘Žπ‘›π‘‘ 𝑔⃗ = (−1, −2)
𝜽 = πŸπŸ”πŸ. πŸ”π’
123.
𝜽 = πŸ•πŸ—. πŸ•π’
βƒ—βƒ— = (2, −6)π‘Žπ‘›π‘‘ 𝑖⃗ = (−1,3)
β„Ž
βƒ—βƒ— = (2,0)
124. 𝑗⃗ = (−1,4)π‘Žπ‘›π‘‘ π‘˜
𝜽 = πŸπŸ–πŸŽπ’
𝜽 = πŸπŸŽπŸ’. πŸŽπŸ‘π’
Angles Between Vectors – Homework
Find the angle between the two given vectors.
125.
π‘Žβƒ— = (3,5) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (7,2)
126. cβƒ— = (−2,4) and βƒ—βƒ—
d = (8,1)
𝜽 = πŸ’πŸ‘. πŸπ’
127.
𝜽 = πŸπŸŽπŸ—. πŸ’π’
𝑑⃗ = (−5, −1) π‘Žπ‘›π‘‘ 𝑒⃗ = (2, −1)
128. 𝑓⃗ = (5, −3)π‘Žπ‘›π‘‘ 𝑔⃗ = (−4, −7)
𝜽 = πŸπŸ’πŸ. πŸπ’
129.
𝜽 = πŸ–πŸ–. πŸ–π’
βƒ—βƒ— = (4, −8)π‘Žπ‘›π‘‘ 𝑖⃗ = (−2,3)
β„Ž
βƒ—βƒ— = (−1,0)
130. 𝑗⃗ = (−1,6)π‘Žπ‘›π‘‘ π‘˜
𝜽 = πŸπŸ•πŸ. πŸ—π’
Pre-Calc Vectors KEY
𝜽 = πŸ–πŸŽ. πŸ“π’
~10~
NJCTL.org
3-Dimensional Space – Class Work
131.
What is the distance between (1,2,3) and (4,5,6)?
𝒅 = πŸ“. 𝟐
132.
What is the distance between (-4,0,-7) and (3,-2,-9)?
𝒅 = πŸ•. πŸ“πŸ“
133.
How far is (5,3,-4) from the origin?
𝒅 = πŸ•. πŸŽπŸ•
134.
What is the length of a diagonal of a box with sides 4x4x8?
𝒍 = πŸ—. πŸ–
135.
What is radius and the center of the sphere with equation: (x-2)2 + y2 + (z-4)2= 36?
π‘ͺ = (𝟐, 𝟎, πŸ’)
𝒓=πŸ”
136.
What is radius and the center of the sphere with equation: x 2 + 6x+ y2 + (z-7)2= 16?
π‘ͺ = (−πŸ‘, 𝟎, πŸ•)
𝒓=πŸ“
137.
What is radius and the center of the sphere with equation: x 2 - 6x+ y2 +8y+ z2 -10z= -1?
π‘ͺ = (πŸ‘, −πŸ’, πŸ“)
𝒓=πŸ•
Pre-Calc Vectors KEY
~11~
NJCTL.org
3-Dimensional Space – Homework
138.
What is the distance between (10,6,2) and (3,5,7)?
𝒅 = πŸ–. πŸ”πŸ”
139.
What is the distance between (-2,1,-5) and (4,-6,-3)?
𝒅 = πŸ—. πŸ’πŸ‘
140.
How far is (-5,-3,-4) from the origin?
𝒅 = πŸ•. πŸŽπŸ•
141.
What is the length of a diagonal of a box with sides 5x7x6?
𝒍 = 𝟏𝟎. πŸ’πŸ—
142.
What is radius and the center of the sphere with equation: (x+3)2 + y2 + (z+5)2= 64?
π‘ͺ = (−πŸ‘, 𝟎, −πŸ“)
𝒓=πŸ–
143.
What is radius and the center of the sphere with equation: x 2 + 12x+ y2 + (z-8)2= -16?
π‘ͺ = (−πŸ”, 𝟎, πŸ–)
𝒓 = 𝟐√πŸ“
144.
What is radius and the center of the sphere with equation: x 2 - 10x+ y2 +4y+ z2 +20z=15?
π‘ͺ = (πŸ“, −𝟐, −𝟏𝟎)
𝒓 = 𝟏𝟐
Pre-Calc Vectors KEY
~12~
NJCTL.org
Vectors, Lines, and Planes – Class Work
Given 𝑒
βƒ—βƒ— = (1,2, −3) π‘Žπ‘›π‘‘ 𝑣⃗ = (−4,5, −6) compute the following.
145.
βƒ—βƒ— + 𝑣⃗
u
146. u
βƒ—βƒ— − 𝑣⃗
(−πŸ‘, πŸ•, −πŸ—)
148.
(πŸ“, −πŸ‘, πŸ‘)
149. |u
βƒ—βƒ—|
βƒ—βƒ— βˆ™ 𝑣⃗
u
πŸπŸ’
151.
πŸ‘. πŸ•πŸ’
the angle between u
βƒ—βƒ— π‘Žπ‘›π‘‘ 𝑣⃗
(𝒙 − 𝟏, π’š − 𝟐, 𝒛 + πŸ‘) = 𝒕(−πŸ“, πŸ‘, −πŸ‘)
154. u
βƒ—βƒ— × π‘£βƒ—
Write the equation from #152 in parametric form.
(πŸ‘, πŸπŸ–, πŸπŸ‘)
Vectors, Lines, and Planes – Homework
Given 𝑒
βƒ—βƒ— = (5,3, −4) π‘Žπ‘›π‘‘ 𝑣⃗ = (−2,6,0) compute the following.
155.
βƒ—βƒ— + 𝑣⃗
u
156. u
βƒ—βƒ— − 𝑣⃗
(πŸ‘, πŸ—, −πŸ’)
(πŸ•, −πŸ‘, −πŸ’)
πŸ–
161.
(𝟏𝟏, 𝟐𝟏, −𝟏𝟐)
160. |v
βƒ—βƒ—|
πŸ•. πŸŽπŸ•
the angle between u
βƒ—βƒ— π‘Žπ‘›π‘‘ 𝑣⃗
(𝒙 − πŸ“, π’š − πŸ‘, 𝒛 + πŸ’) = 𝒕(−πŸ•, πŸ‘, πŸ’)
Write the equation from #162 in parametric form.
𝒙 = πŸ“ − πŸ•π’•
{ π’š = πŸ‘ + πŸ‘π’•
𝒛 = −πŸ’ + πŸ’π’•
Pre-Calc Vectors KEY
πŸ”. πŸ‘πŸ
162. Write the vector equation of the line through u
βƒ—βƒ— π‘Žπ‘›π‘‘ 𝑣⃗
𝜽 = πŸ•πŸ—. πŸ•π’
163.
157. 3u
βƒ—βƒ— + 2𝑣⃗
159. |u
βƒ—βƒ—|
βƒ—βƒ— βˆ™ 𝑣⃗
u
πŸ–. πŸ•πŸ•
152. Write the vector equation of the line through u
βƒ—βƒ— π‘Žπ‘›π‘‘ 𝑣⃗
𝒙 = 𝟏 − πŸ“π’•
{ π’š = 𝟐 + πŸ‘π’•
𝒛 = −πŸ‘ − πŸ‘π’•
158.
(−πŸ“, πŸπŸ”, −𝟐𝟏)
150. |v
βƒ—βƒ—|
𝜽 = πŸ’πŸ. πŸ—πŸ•π’
153.
147. 3u
βƒ—βƒ— + 2𝑣⃗
164. u
βƒ—βƒ— × π‘£βƒ—
(πŸπŸ’, πŸ–, πŸ‘πŸ”)
~13~
NJCTL.org
Unit Review
Multiple Choice
βƒ—βƒ—|?
1. A vector has component forces of Ax= 5.2 and Ay= -4.7, what is |A
a. 0.50
b. 7.01
B
c. 10.9
d. 49.13
5
2. 𝑒
βƒ—βƒ— = (1,4), find u
βƒ—βƒ—.
4
5
a. ( , 5)
4
9 21
b. ( , )
4
c.
d.
(9,4)
25
4
3. π‘Žβƒ— = (4, −6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (−2,5), 𝑓𝑖𝑛𝑑 𝑏⃗⃗ + π‘Žβƒ—.
a. (6,11)
b. (2,-11)
c. (6,-1)
d. (2,-1)
4. π‘Žβƒ— = (4, −6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (−2,5), 𝑓𝑖𝑛𝑑 2𝑏⃗⃗ − π‘Žβƒ—.
5.
6.
7.
8.
A
4
D
a. (8,-12)
b. (2,-1)
C
c. (-8,16)
d. (-10,-17)
π‘Žβƒ— = (4, −6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (−2,5), 𝑓𝑖𝑛𝑑 𝑏⃗⃗ βˆ™ π‘Žβƒ—.
a. 16
b. -16
C
c. -38
d. 38
What is the slope of the line with vector equation (x-3,y+5)=t(2,6)
a. 2
b. 3
B
c. 5
d. 6
An example of perpendicular vectors is
a. u
βƒ—βƒ— = (4,5) and v
βƒ—βƒ— = (−2,3)
b. u
βƒ—βƒ— = (2, −6) and v
βƒ—βƒ— = (−9,3)
D
c. u
βƒ—βƒ— = (−3,4) and v
βƒ—βƒ— = (−8,6)
d. u
βƒ—βƒ— = (−4,6) and v
βƒ—βƒ— = (−9, −6)
(4,
The angle between π‘Žβƒ— =
−6) π‘Žπ‘›π‘‘ 𝑏⃗⃗ = (−2,5) is
a.
b.
c.
d.
.979
2.934
152.345
168.111
Pre-Calc Vectors KEY
D
~14~
NJCTL.org
9. What is the distance between (4,-2,-5) and (-1, 7,-6)?
a. 5.916
B
b. 10.344
c. 12.450
d. 15.067
10. Which of the following points is 12 units from the origin?
a. (3,4,5)
b. (0,12,1)
D
c. (-5,12,0)
d. (-4, 8, 8)
11. What is the radius of x2 +8x + y2 – 12y + z2 + 2z - 9=16?
a. 4
b. 5
D
c. √31
d. √78
12. Given 𝑒
βƒ—βƒ— = (5, 3, −7) π‘Žπ‘›π‘‘ 𝑣⃗ = (−2, 1, −8), find u
βƒ—βƒ— βˆ™ 𝑣⃗.
a. -59
b. -46
C
c. 49
d. 69
13. Given 𝑒
βƒ—βƒ— = (5, 3, −7) π‘Žπ‘›π‘‘ 𝑣⃗ = (−2, 1, −8), find |v
βƒ—βƒ—|
a. 3.742
b. 5.099
D
c. 7.211
d. 8.307
14. Given 𝑒
βƒ—βƒ— = (5, 3, −7) π‘Žπ‘›π‘‘ 𝑣⃗ = (−2, 1, −8), find u
βƒ—βƒ— × π‘£βƒ—
a. (-17, 54, 11)
b. (-31, -26, -1)
A
c. (-17,-54, -1)
d. (-10, 3, 56)
Extended Response
1. A plane flies northeast at 300 miles per hour.
a. Draw a vector representation of the plane. Show the component forces.
b. The wind is blowing south at 50 miles per hour, what is the result on the planes component forces?
(𝟐𝟏𝟐. πŸπŸ‘, 𝟐𝟏𝟐. πŸπŸ‘) → (𝟐𝟏𝟐. πŸπŸ‘, πŸπŸ”πŸ. πŸπŸ‘)
c.
Where will the plane be in 5 hours relative to its starting position?
𝟏, πŸŽπŸ”πŸŽ. πŸ”πŸ“ π’Žπ’Šπ’π’†π’” 𝑬𝒂𝒔𝒕
πŸ–πŸπŸŽ. πŸ”πŸ“ π’Žπ’Šπ’π’†π’” 𝑡𝒐𝒓𝒕𝒉
Pre-Calc Vectors KEY
~15~
NJCTL.org
2. A box is being slid across the floor by a person pulling a rope with component force of F x=6 and Fy=3.
a. Another person pulls a second rope with F x= -4 and Fy=3. Draw a vector diagram to model this
situation.
b. Where does the box end up?
(𝟐, πŸ”)
c.
If the second person had wanted the box to slide due north, what component forces should they
have applied? Explain.
𝑭𝒙 = −πŸ”
π‘­π’š = π’‚π’π’š # > −3
3. Three people are holding the lines of a balloon during a parade. π‘Žβƒ— = (5,7, −4), 𝑏⃗⃗ = (3, 18, −2), π‘Žπ‘›π‘‘ 𝑐⃗ =
(−2, −3, −1).
a. What is direction of the balloon?
(πŸ”, 𝟐𝟐, −πŸ•)
b. What is the angle of the lines between person A and person B?
𝜽 = πŸ‘πŸ. πŸπŸ–π’
c.
What vector would represent the effects of helium on the balloon?
(𝟎, 𝟎, 𝒛) π’˜π’‰π’†π’“π’† 𝒛 π’Šπ’” π’‚π’π’š π’‘π’π’”π’Šπ’•π’Šπ’—π’† π’π’–π’Žπ’ƒπ’†π’“
4. Line m passes through (-6, 4, -8) and (2, 7, -9).
a. Write the vector equation of line m.
(𝒙 + πŸ”, π’š − πŸ’, 𝒛 + πŸ–) = 𝒕(πŸ–, πŸ‘, −𝟏)
b. Write the parametric equation of line.
𝒙 = −πŸ” + πŸ–π’•
{ π’š = πŸ’ + πŸ‘π’•
𝒛 = −πŸ– − 𝒕
c.
Find a point, other than the ones given, that lies on the line.
π‘¨π’π’”π’˜π’†π’“π’” π’˜π’Šπ’π’ π’—π’‚π’“π’š
𝑰𝒇 𝒕 = −𝟏, 𝒕𝒉𝒆𝒏 (−πŸπŸ’, 𝟏, −πŸ•)
Pre-Calc Vectors KEY
~16~
NJCTL.org
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