The Dot Product Angles Between Vectors Orthogonal Vectors Section 6.2a HW: p. 520 13-27 odd.

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The Dot Product The Dot Product Angles Between Vectors Angles Between Vectors Orthogonal Vectors Orthogonal Vectors Section 6.2a HW: p. 520 13-27 odd

Transcript of The Dot Product Angles Between Vectors Orthogonal Vectors Section 6.2a HW: p. 520 13-27 odd.

The Dot ProductThe Dot ProductAngles Between VectorsAngles Between VectorsOrthogonal VectorsOrthogonal Vectors

Section 6.2aHW: p. 520 13-27 odd

Definition: Dot Product

The dot product or inner product of u = u , uand v = v , v is

1 21 2

u v = u v + u v1 21 2

The sum of two vectors is a… vector!vector!

The product of a scalar and a vector is a… vector!vector!

The dot product of two vectors is a… scalar!scalar!

Properties of the Dot Product

Let u, v, and w be vectors and let c be a scalar.

1. u v = v u

2. u u = |u| 2

3. 0 u = 0

4. u (v + w) = u v + u w

(u + v) w = u w + v w

5. (cu) v = u (cv) = c(u v)

Finding the Angle BetweenTwo Vectors (Law of cosines)

vu

v – u

0

2 2v v v u u v u u u v 2 u v cosθ

2 2v u v u u v 2 u v cosθ

2u v 2 u v cosθ

2 2 2 2v 2u v u u v 2 u v cosθ

2 2 2v u u v 2 u v cosθ

u vcosθ

u v

1 u v

θ cosu v

Theorem:Angle Between Two Vectors

vu

v – u

01 u v

θ cosu v

If 0 is the angle between nonzerovectors u and v, then

u vcosθ

u v

and

Definition: Orthogonal Vectors

The vectors u and v are orthogonal ifand only if u v = 0.

The terms “orthogonal” and “perpendicular”The terms “orthogonal” and “perpendicular”are nearly synonymous (with the exceptionare nearly synonymous (with the exception

of the zero vector)of the zero vector)

Guided Practice

Find each dot product

1. 3, 4 5, 2

2. 1, –2 –4, 3

3. (2i – j) (3i – 5j)

= 23= 23

= –10= –10

= 11= 11

Guided Practice

Use the dot product to find the length ofvector v = 4, –3 ((hint: use property 2!!!)hint: use property 2!!!)

Length = 5Length = 5

2v v v

v v v 4 4 3 3

25

Guided Practice

Find the angle between vectors u and v

0 = 55.4910 = 55.491

u = 2, 3 , v = –2, 5

u v 2 2 3 5 112 2u 2 3 13 2 2v 2 5 29

1 11θ cos

13 29

Guided Practice

Find the angle between vectors u and v

u cos i sin j3 3

5 5v 3cos i 3sin j

6 6

1 3,2 2

3 3 3,

2 2

01 3 3 3 3

u v2 2 2 2

3 3 3 3

4 4

Guided Practice

Find the angle between vectors u and v

u cos i sin j3 3

5 5v 3cos i 3sin j

6 6

1 3,2 2

3 3 3,

2 2

0 = 900 = 90

Is there an easier way to solve this???Is there an easier way to solve this???

1θ cos 0

Guided Practice

Prove that the vectors u = 2, 3 andv = –6, 4 are orthogonal

u v = 0!!!u v = 0!!!

u v 2 6 3 4 12 12 0 Check the dot product:

Graphical Support???Graphical Support???

First, let’s look at a brain exercise…Page 520, #30:

Find the interior angles of the triangle with vertices (–4,1),(1,–6), and (5,–1).

Start with a graph…

A(–4,1)

B(5,–1)

C(1,–6)

First, let’s look at a brain exercise…

A(–4,1)

B(5,–1)

C(1,–6)

4 1,1 6CA 55555555555555

5 1, 1 6CB 55555555555555

5,7

4,5

5 4 7 5CA CB 5555555555555555555555555555

15

2 25 7CA 55555555555555

74 2 24 5CB 55555555555555

41

First, let’s look at a brain exercise…

A(–4,1)

B(5,–1)

C(1,–6)

1 15cos

74 41C

74.197

First, let’s look at a brain exercise…

A(–4,1)

B(5,–1)

C(1,–6)

9,2BA 55555555555555

4, 5BC 55555555555555

26BA BC 5555555555555555555555555555

85BA 55555555555555

41BC 55555555555555

1 26cos

85 41B

63.869

180 74.197 63.869A 41.934

Definition: Vector Projection

The vector projection of u = PQ onto a nonzero vector v = PSis the vector PR determined by dropping a perpendicular fromQ to the line PS.

u

P

Q

SR

Thus, u can be broken into components PR and RQ:

u = PR + RQ

v

Definition: Vector Projection

u

P

Q

SR

Notation for PR, the vector projection of u onto v:

PR = proj uv

The formula:

proj u = vvu v|v| 2

Some Practice Problems

Find the vector projection of u = 6, 2 onto v = 5, –5 .Then write u as the sum of two orthogonal vectors, oneof which is proj u.v Start with a graph… v 2u proj u u

u v 6 5 2 5 20

22v 5 5 50

v 2

u vproj u v

v

205, 5

50 2, 2

Some Practice Problems…

Find the vector projection of u = 6, 2 onto v = 5, –5 .Then write u as the sum of two orthogonal vectors, oneof which is proj u.v

u = proj u + u = 2, –2 + 4, 4v 2

Start with a graph…v 2u proj u u

2 vu u proj u

6,2 2, 2

4,4

Some Practice Problems…

Find the vector projection of u = 3, –7 onto v = –2, –6 .Then write u as the sum of two orthogonal vectors, oneof which is proj u.v

u = proj u + u = –1.8,–5.4 + 4.8,–1.6v 2

Some Practice Problems…

Find the vector v that satisfies the given conditions:2u = –2,5 , u v = –11, |v| = 10

1 2v ,v v

1 22 5 11v v 2 21 2 10v v

A system to solve!!!

1 2

5 11

2 2v v

22

2 2

5 1110

2 2v v

2 22 2 2

25 110 12110

4 4 4v v v

Some Practice Problems…

Find the vector v that satisfies the given conditions:2u = –2,5 , u v = –11, |v| = 10

1 2v ,v v

22 2

29 110 810

4 4 4v v

2 22 2 2

25 110 12110

4 4 4v v v

22 229 110 81 0v v

2 21 29 81 0v v 2

811,29

v

v 3, 1 OR 43 81

,29 29

Some Practice Problems…

Now, let’s look at p.520: 34-38 even:

What’s the plan???What’s the plan??? If u v = 0 If u v = 0 orthogonal! orthogonal!If u = If u = kkv v parallel! parallel!

34) Neither

36) Orthogonal

38) Parallel