1. Splash Screen

2. Five-Minute Check (over Lesson 8-1) Then/Now New Vocabulary
Key Concept: Component Form of a Vector
Example 1: Express a Vector in Component Form
Key Concept: Magnitude of a Vector in the Coordinate Plane
Example 2: Find the Magnitude of a Vector
Key Concept: Vector Operations
Example 3: Operations with Vectors
Example 4: Find a Unit Vector with the Same Direction as a Given Vector
Example 5: Write a Unit Vector as a Linear Combination of Unit Vectors
Example 6: Find Component Form
Example 7: Direction Angles of Vectors
Example 8: Real-World Example: Applied Vector Operations
Lesson Menu

3. Determine the magnitude and direction of the resultant of the vector sum described as 25 miles east and then 47 miles south.
A miles, N28°E
B miles, S62°E
C miles, S28°E
D. 72 miles, S28°E
5–Minute Check 1

4. An airplane heads due east at 400 miles per hour
An airplane heads due east at 400 miles per hour. If a 40-mile-per-hour wind blows from a bearing of N 25°E, what are the ground speed and direction of the plane?
A mph, N85°E
B mph, N5°E
C mph, S5.4°E
D mph, S84.6°E
5–Minute Check 2

5. Which of the following represents a vector quantity?
A. a car driving at 55 miles per hour
B. a cart pulled up a 30° incline with a force of newtons
C. the temperature of a cup of coffee
D. wind blowing at 30 knots
5–Minute Check 3

6. You performed vector operations using scale drawings. (Lesson 8-1)
Represent and operate with vectors in the coordinate plane.
Write a vector as a linear combination of unit vectors.
Then/Now

7. component form
unit vector
linear combination
Vocabulary

8. Key Concept 1

9. = x2 – x1, y2 – y1 Component form
Express a Vector in Component Form
Find the component form of with initial point A(1, –3) and terminal point B(1, 3).
= x2 – x1, y2 – y1 Component form
= 1 – 1, 3 – (–3) (x1, y1 ) = (1, –3) and ( x2, y2 ) = (1, 3)
= 0, 6 Subtract.
Answer: 0, 6
Example 1

10. Find the component form of given initial point A(–4, –3) and terminal point B(5, 3).
Example 1

11. Key Concept 2

12. Find the Magnitude of a Vector
Find the magnitude of with initial point A(1, –3) and terminal point B(1, 3).
Magnitude formula
(x1 , y1 ) = (1, –3) and ( x2 , y2 ) = (1, 3)
Simplify.
Example 2

13. CHECK From Example 1, you know that = 0, 6. 
Find the Magnitude of a Vector
Answer: 6
CHECK From Example 1, you know that = 0, 6. 
Example 2

14. Find the magnitude of given initial point A(4, –2) and terminal point B(–3, –2).
C. 7.3
D. 23
Example 2

15. Key Concept 3

16. A. Find 2w + y for w = 2, –5, y = 2, 0, and z = –1, –4.
Operations with Vectors
A. Find 2w + y for w = 2, –5, y = 2, 0, and z = –1, –4.
2w + y = 2 2, –5 + 2, 0 Substitute.
= 4, 10 + 2, 0 Scalar multiplication
= 4 + 2, –10 + 0 or 6, –10 Vector addition
Answer: 6, –10
Example 3

17. B. Find 3y – 2z for w = 2, –5, y = 2, 0, and z = –1, –4.
Operations with Vectors
B. Find 3y – 2z for w = 2, –5, y = 2, 0, and z = –1, –4.
3y – 2z = 3y + (–2z) Rewrite subtraction as addition.
= 32, 0 + (–2)–1, –4 Substitute.
= 6, 0 + 2, 8 Scalar multiplication
= 6 + 2, 0 + 8 or 8, 8 Vector addition
Answer: 8, 8
Example 3

18. Find 3v + 2w for v = 4, –1 and w = –3, 5.
B. 6, 7
C. 6, 13
D. –1, 13
Example 3

19. Find a unit vector u with the same direction as v = 4, –2.
Find a Unit Vector with the Same Direction as a Given Vector
Find a unit vector u with the same direction as v = 4, –2.
Unit vector with the same direction as v.
Substitute.
; Simplify. or
Example 4

20. Rationalize the denominator.
Find a Unit Vector with the Same Direction as a Given Vector
Rationalize the denominator.
Scalar multiplication
Rationalize denominators.
Therefore, u =
Answer: u =
Example 4

21. Find a Unit Vector with the Same Direction as a Given Vector
Check Since u is a scalar multiple of v, it has the same direction as v. Verify that the magnitude of u is 1.
Magnitude Formula
Simplify.
Simplify. 
Example 4

22. Find a unit vector u with the same direction as w = 5, –3.B. C. D.
Example 4

23. First, find the component form of .
Write a Unit Vector as a Linear Combination of Unit Vectors
Let be the vector with initial point D(–3, –3) and terminal point E(2, 6). Write as a linear combination of the vectors i and j.
First, find the component form of
= x2 – x1, y2 – y1 Component form
= 2 – (–3), 6 – (–3) (x1 , y1 ) = (–3, –3) and ( x2 , y2 ) = (2, 6)
= 5, 9 Subtract.
Example 5

24. Write a Unit Vector as a Linear Combination of Unit Vectors
Then, rewrite the vector as a linear combination of the standard unit vectors.
= 5, 9 Component form
= 5i + 9j a, b = ai + bj
Answer: 5i + 9j
Example 5

25. Let be the vector with initial point D(–4, 3) and terminal point E(–1, 5). Write as a linear combination of the vectors i and j.
A. 2i + 3j
B. 3i + 8j
C. 5i + 2j
D. 3i + 2j
Example 5

26. Component form of v in terms of |v| and θ
Find Component Form
Find the component form of the vector v with magnitude 7 and direction angle 60°.
Component form of v in terms of |v| and θ
|v| = 7 and θ = 60°
Simplify.
Answer:
Example 6

27. Find Component Form
Check Graph v = ≈ 3.5, 6.1. The measure of the angle v makes with the positive x-axis is about 60° as shown, and |v| = 
Example 6

28. Find the component form of the vector v with magnitude 12 and direction angle 300°.B. C. D.
Example 6

29. Direction angle equation
Direction Angles of Vectors
A. Find the direction angle of p = 2, 9 to the nearest tenth of a degree.
Direction angle equation
a = 2 and b = 9
Solve for .
Use a calculator.
Example 7

30. So the direction angle of vector p is about 77.5°, as shown below.
Direction Angles of Vectors
So the direction angle of vector p is about 77.5°, as shown below.
Answer: 77.5°
Example 7

31. Direction angle equation
Direction Angles of Vectors
B. Find the direction angle of r = –7i + 2j to the nearest tenth of a degree.
Direction angle equation
a = –7 and b = 2
Solve for .
Use a calculator.
Example 7

32. Since r lies in Quadrant II as shown below,  = 180 – 15.9° or 164.1°.
Direction Angles of Vectors
Since r lies in Quadrant II as shown below,  = 180 – 15.9° or 164.1°.
Answer: °
Example 7

33. Find the direction angle of p = –1, 4 to the nearest tenth of a degree.
B. 76.3°
C °
D °
Example 7

34. Applied Vector Operations
SOCCER A soccer player running forward at 7 meters per second kicks a soccer ball with a velocity of 30 meters per second at an angle of 10° with the horizontal. What is the resultant speed and direction of the kick?
Since the soccer player moves straight forward, the component form of his velocity v1 is 7, 0. Use the magnitude and direction of the soccer ball’s velocity v2 to write this vector in component form.
Example 8

35. v 2 = | v2 | cos θ, | v2 | sin θ Component form of v2
Applied Vector Operations
v 2 = | v2 | cos θ, | v2 | sin θ Component form of v2
= 30 cos 10°, 30 sin 10° |v2| = 30 and θ = 10°
≈ 29.5, 5.2 Simplify.
Add the algebraic vectors representing v1 and v2 to find the resultant velocity, r.
r = v1 + v2 Resultant vector
= 7, 0 + 29.5, 5.2 Substitution
= 36.5, 5.2 Vector Addition
Example 8

36. Applied Vector Operations
The magnitude of the resultant is |r| = or about Next find the resultant direction θ.
Example 8

37. Applied Vector Operations
a, b = 36.5, 5.2
Therefore, the resultant velocity of the kick is about 36.9 meters per second at an angle of about 8.1° with the horizontal.
Answer: 36.9 m/s; 8.1°
Example 8

38. SOCCER A soccer player running forward at 6 meters per second kicks a soccer ball with a velocity of 25 meters per second at an angle of 15° with the horizontal. What is the resultant speed and direction of the kick?
A m/s; 15.1°
B m/s; 8.1°
C m/s; 15.1°
D m/s; 12.1°
Example 8

39. End of the Lesson