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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

Determine the magnitude and direction of the resultant of the vector sum described as 25 miles east and then 47 miles south. A. 53.2 miles, N28°E B. 53.2 miles, S62°E C. 53.2 miles, S28°E D. 72 miles, S28°E 5–Minute Check 1

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. 418.5 mph, N85°E B. 418.5 mph, N5°E C. 384.8 mph, S5.4°E D. 384.8 mph, S84.6°E 5–Minute Check 2

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 40 newtons C. the temperature of a cup of coffee D. wind blowing at 30 knots 5–Minute Check 3

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

component form unit vector linear combination Vocabulary

Key Concept 1

= 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

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

Key Concept 2

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

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

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

Key Concept 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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: 164.1° Example 7

Find the direction angle of p = –1, 4 to the nearest tenth of a degree. B. 76.3° C. 104.5° D. 166.7° Example 7

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

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

Applied Vector Operations The magnitude of the resultant is |r| = or about 36.9. Next find the resultant direction θ. Example 8

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

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. 25.0 m/s; 15.1° B. 25.0 m/s; 8.1° C. 30.8 m/s; 15.1° D. 30.8 m/s; 12.1° Example 8

End of the Lesson