1. Initial point: (0, 0); Terminal point: (3, -4)
A Lesson 13
B Lesson 13
C Lesson 13
D Lesson 13
E Lesson 13
F Lesson 13
2. Initial point: (3, 5); Terminal point: (-2, -1)
A Lesson 13
B Lesson 13
C Lesson 13
D Lesson 13
E Lesson 13
F Lesson 13
In problems 3 and 4, let u = Lesson 13, v = Lesson 13, and w = Lesson 13. Find the component form of the vector.
3. 2u + 3w
A Lesson 13
B Lesson 13
C Lesson 13
D Lesson 13
E Lesson 13
F Lesson 13
4. -2u - 3v
A Lesson 13
B Lesson 13
C Lesson 13
D Lesson 13
E Lesson 13
F Lesson 13
5. Find a unit vector in the direction of v = Lesson 13. Write your answer as a linear combination of the standard unit vectors i and j. Round each component to the nearest hundredth, if necessary.
A 0.71i - 0.71j
B 0i - 0j
C 1.41i - 1.41j
D 1i - 1j
E 2.83i - 3.83j
F -1i - 1j
6. Find the component form of v if the direction angle is 55° and the magnitude is 14. (See diagram below.) Round each component to the nearest hundredth, if necessary.
A vLesson 13
B v Lesson 13
C v Lesson 13
D v Lesson 13
E v Lesson 13
F v Lesson 13
For problems 7 and 8, find the magnitude and direction angle of the given vector.
7. Lesson 13
A magnitude: 5; direction angle: 53.13°
B magnitude: Lesson 13; direction angle: 53.13°
C magnitude: Lesson 13; direction angle: 3.40°
D magnitude: Lesson 13; direction angle: 0.80°
E magnitude:5; direction angle: 3.40°
F magnitude: 5; direction angle: 0.80°
8. -3i - 5j
A magnitude: Lesson 13; direction angle: 120.96°
B magnitude: Lesson 13; direction angle: 239.04°
C magnitude: Lesson 13; direction angle: 59.04°
D magnitude: 34; direction angle: 84.94°
E magnitude: 34; direction angle: 95.06°
F magnitude: 34; direction angle: 275.06°
For problems 9 - 11, find the dot product if u = Lesson 13, v = Lesson 13, and w = Lesson 13.
9. Lesson 13
A 2
B 8
C -10
D 10
E 4
F -3
10. Lesson 13
A -11
B 0
C 1
D -17
E 11
F -1
11. Lesson 13
A 37
B -37
C -33
D -29
E 29
F 9
For problems 12 and 13, using the theorem given in this lesson, find the angle between the given vectors.
12. Lesson 13 and Lesson 13
A 88.3°
B 55.5°
C 11°
D 74.92°
E 1.7°
F 62.4°
13. Lesson 13 and Lesson 13
A 120°
B 60°
C 315°
D 225°
E 135°
F 45°
14. An airplane is flying on a bearing of 335° at 530 miles per hour. Find the component form of the velocity of the airplane.
A v Lesson 13
B v Lesson 13
C v Lesson 13
D v Lesson 13
E v Lesson 13
F v Lesson 13
15. An airplane is flying on a bearing of 170° at 460 miles per hour. Find the component form of the velocity of the airplane.
A v Lesson 13
B v Lesson 13
C v Lesson 13
D v Lesson 13
E v Lesson 13
F v Lesson 13
16. Now, assume that the airplane from problem 12 is flying in a wind that is blowing with the bearing 200° at 80 miles per hour. Find the actual ground speed of the airplane.
A 530.79 miles per hour
B 52.52 miles per hour
C 528.19 miles per hour
D 24.09 miles per hour
E 23.08 miles per hour
F 453.01 miles per hour
17. Use the information from problem 13 to find the actual direction (angle) of the airplane. (This is the angle from the horizontal x-axis, not the bearing.)
A 174.32°
B -75.18°
C 250°
D -80°
E 80°
F 84.32°
18. A basketball is shot at a 70° angle with the horizontal with an initial velocity of 10 meters per second. Find the component form of the initial velocity.
A v Lesson 13
B v Lesson 13
C v Lesson 13
D v Lesson 13
E vLesson 13
F v Lesson 13
19. A force of 50 pounds acts on an object at an angle of 45°. A second force of 75 pounds acts on the object at an angle of -30°. Find the direction and magnitude of the resultant force.
A magnitude: 2.14 lbs; direction: -2.14°
B magnitude: 125 lbs; direction: -30°
C magnitude: 125 lbs; direction: -1.22°
D magnitude: 100.33 lbs; direction: 15°
E magnitude: 100.33 lbs; direction: -1.22°
F magnitude: 2.14 lbs; direction: 15°
20. Juana and Diego Gonzales, ages six and four respectively, own a strong and stubborn puppy named Corporal. It is so hard to take Corporal for a walk that they devise a scheme to use two leashes. If Juana pulls with a force of 23 lbs at an angle of 18° and Diego pulls with a force of 27 lbs at an angle of -15°, how hard is Corporal pulling if the puppy holds the children at a standstill?
A 50 lbs
B 4 lbs
C 3 lbs
D 47.95 lbs
E 33 lbs
F 7 lbs
See your 12-6-11,5:33am post.
To solve these problems, you will need to understand the concepts of vectors, vector components, vector operations (such as addition, subtraction, scalar multiplication), magnitude (or length) of vectors, direction angles, dot product, and vector projections.
1. To find the vector from the initial point to the terminal point, subtract the coordinates of the initial point from the coordinates of the terminal point. For problem 1, (3 - 0, -4 - 0) = (3, -4).
2. Similar to problem 1, subtract the coordinates of the initial point from the coordinates of the terminal point. For problem 2, (-2 - 3, -1 - 5) = (-5, -6).
3. Given that u = (Lesson 13), w = (Lesson 13), and you want to find 2u + 3w, multiply each component of u and w by their corresponding scalar, then add the resulting vectors together. For example, if u = (a, b) and w = (c, d), then 2u + 3w = (2a + 3c, 2b + 3d).
4. Similar to problem 3, but this time you need to subtract -2u and 3v. Subtract each component of -2u and 3v from the corresponding components of the vectors u and v.
5. To find a unit vector in the direction of v, divide each component of v by its magnitude (or length). This can be done by calculating the square root of the sum of the squares of the components (a^2 + b^2) and then dividing each component by the magnitude.
6. To find the component form of v given the direction angle (55°) and magnitude (14), use the formula v = (|v| * cos(theta), |v| * sin(theta)), where |v| is the magnitude of v and theta is the direction angle in radians.
7. To find the magnitude of a vector, use the formula magnitude = sqrt(a^2 + b^2), where a and b are the components of the vector. To find the direction angle, use the formula tan(theta) = b/a.
8. Similar to problem 7, calculate the magnitude using the formula magnitude = sqrt(a^2 + b^2), and then find the direction angle using the formula tan(theta) = b/a.
9. Given u = (Lesson 13), v = (Lesson 13), and w = (Lesson 13), the dot product of two vectors can be found by multiplying their corresponding components and adding the results. For example, for vectors u and v, the dot product is calculated as a*c + b*d.
10. Similar to problem 9, but this time find the dot product of -2u and v.
11. Similar to problems 9 and 10, find the dot product of -2u and -3v.
12. The angle between two vectors can be found using the formula cos(theta) = (u dot v) / (|u| * |v|), where u dot v is the dot product of the vectors u and v, and |u| and |v| are the magnitudes of the vectors.
13. Similar to problem 12, find the angle between the vectors u and v using the formula cos(theta) = (u dot v) / (|u| * |v|).
14. Given the bearing (direction) and velocity of the airplane, convert the bearing to an angle in radians and use the formula v = (v_1 * cos(theta), v_1 * sin(theta)), where v_1 is the velocity and theta is the angle.
15. Similar to problem 14, but this time convert the bearing to an angle in radians and use the same formula to find the component form of the velocity of the airplane.
16. To find the actual ground speed of the airplane with the wind, utilize vector addition. Add the velocity vector of the airplane with the velocity vector of the wind.
17. Given the bearing of the airplane, convert it to an angle in radians and subtract it from 180° to find the actual direction angle of the airplane.
18. Given the angle with the horizontal and initial velocity of the basketball, use the same formula as in problem 14 and problem 15 to find the component form of the initial velocity.
19. To find the resultant force, use vector addition. Add the force vectors using their magnitudes and angles, then use the formula resultant magnitude = sqrt(a^2 + b^2), and resultant angle = tan inverse(b/a).
20. Similar to problem 19, add the force vectors using their magnitudes and angles, then use the formula to find the magnitude of the resultant force.