9.7 Vectors

Geometry

Mrs. Spitz

Spring 2005

Objectives:

• This lesson is worth 1/3 of your test grade on Thursday.

• Find the magnitude and the direction of the vector

• Add vectors• Assignment: pp. 576-577 #13-20, 35-40 all. Also

complete Ch. 9 Review: pp. 582-584 #1-24 all.• Reminder: Ch. 9 Test is next week Wednesday.

IF you are leaving early, please take the exam early during study hall

Pay attention:

• I’m only giving you exactly what is on the test. There are 4 questions that have to do with graphing the vector, putting it into component form, and then finding the magnitude. The next three examples are good practice and/or notes to have on the test Thursday.

Finding the magnitude of a vector

• You begin with an initial point to a terminal point given in terms of points, usually P and Q. You graph it as you would a ray. Initial point is P(0, 0). Terminal point is Q(-6, 3).

4

2

-2

-4

-6

-5 5 10

Q(-6, 3)

P(0, 0)

Write the component form

• Here you write the following Component Form =‹x2 – x1, y2 – y1›<-6 – 0, 3 – 0><-6, 3> is the component form.Next use the distance formula to find the

magnitude.

|PQ| = √(-6 – 0)2 + (3 – 0)2 = √62 + 32

= √36 + 9 = √45 ≈ 6.7

4

2

-2

-4

-6

-5 5 10

Q(-6, 3)

P(0, 0)

Graph Initial/Terminal points

• Initial point is P(0, 2). Terminal point is Q(5, 4).

• Reminder that Q is the second point. P is the initial point. Graph the ray starting at P and going through Q as to the right. Then you can start looking for component form and magnitude.

8

6

4

2

-2

5 10 15

Q(5, 4)

P (0, 2)

Write the component form

• Here you write the following Component Form =‹x2 – x1, y2 – y1›<5 – 0, 4 – 2><5, 2> is the component form.Next use the distance formula to find the

magnitude.

|PQ| = √(5 – 0)2 + (4 – 2)2 = √52 + 22

= √25 + 4 = √29 ≈ 5.4

8

6

4

2

-2

5 10 15

Q(5, 4)

P (0, 2)

Graph Initial/Terminal points

• Initial point is P(3, 4). Terminal point is Q(-2, -1).

• Reminder that Q is the second point. P is the initial point. Graph the ray starting at P and going through Q as to the right. Then you can start looking for component form and magnitude.

4

2

-2

-4

-6

5 10

P(3, 4)

Q(-2, -1)

Write the component form

• Here you write the following Component Form =‹x2 – x1, y2 – y1›<-2 – 3, -1 – 4><-5, -5> is the component form.Next use the distance formula to find the

magnitude.

|PQ| = √-2 – 3)2 + (-1– 4)2 = √(-5)2 + (-5)2

= √25 + 25 = √50 ≈ 7.1

4

2

-2

-4

-6

5 10

P(3, 4)

Q(-2, -1)

Adding Vectors

• Two vectors can be added to form a new vector. To add u and v geometrically, place the initial point of v on the terminal point of u, (or place the initial point of u on the terminal point of v). The sum is the vector that joins the initial point of the first vector and the terminal point of the second vector. It is called the parallelogram rule because the sum vector is the diagonal of a parallelogram. You can also add vectors algebraically.

What does this mean?

• Adding vectors:

Sum of two vectors

The sum of u = <a1,b1> and v = <a2, b2> is

u + v = <a1 + a2, b1 + b2>

In other words: add your x’s to get the coordinate of the first, and add your y’s to get the coordinate of the second.

Example:

• Let u = <3, 5> and v = <-6, 1>• To find the sum vector u + v, add the x’s and

add the y’s of u and v.

u + v = <3 + (-6), 5 + (-1)>

= <-3, 4>

There are 6 of these on the test Wednesday!!!

Reminders:

• Test Wednesday before you leave

• Binder Check Wednesday

• HW: 9.7 is due Friday

• HW: Chapter 9 Review is due Monday.

• HW Worksheets 9.5A & B and 9.6A are due also Monday.