6.4

Vectors and Dot Products

The Definition of the Dot Product of Two Vectors

The dot product of u = and v = is

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u1,u2

€

v1,v2

€

u ⋅v = u1v1 + u2v2

Ex.’s Find each dot product.

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a. 4,5 ⋅ 2,3

b. 2,−1 ⋅ 1,2

c. 0,3 ⋅ 4,−2€

=4 2( ) + 5 3( ) = 23

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=2 1( ) + −1( ) 2( ) = 0

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=0 4( ) + 3 −2( ) = −6

Properties of the Dot Product

Let u, v, and w be vectors in the plane or in space and let c be a scalar.

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u ⋅v = v ⋅u

0 ⋅v = 0

u ⋅ v + w( ) = u ⋅v + u ⋅w

v ⋅v = v2

c u ⋅v( ) = cu ⋅v = u ⋅cv

Let

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u = −1,3 , v = 2,−4 , and w = 1,−2

Find

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u ⋅v( )w First, find u . v

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u ⋅v = −1( ) 2( ) + 3 −4( ) = −14

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u ⋅v( )w

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=−14 1,−2 = −14,28

Find u . 2v = 2(u . v) = 2(-14) = -28

The Angle Between Two Vectors

If is the angle between two nonzero vectors u and v, then

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θ

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cosθ =u ⋅vu v

Find the angle between

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u = 4,3 & v = 3,5€

v = 3,5

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u = 4,3

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cosθ =4,3 ⋅ 3,5

4,3 3,5

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=27

5 34

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θ =arccos27

5 34≈ 22.2°

Definition of Orthogonal Vectors (90 degree angles)

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

Are the vectors orthogonal?

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u = 2,−3 & v = 6,4

Find the dot product of the two vectors.

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u ⋅v = 2,−3 ⋅ 6,4 = 2 6( ) + −3( ) 4( ) = 0

Because the dot product is 0, the two vectors are orthogonal.

End of notes.

Finding Vector Components

Let u and v be nonzero vectors such that u = w1 + w2

where w1 and w2 are orthogonal and w1 is parallel to(or a scalar multiple of) v. The vectors w1 and w2 are called vector components of u. The vector w1 is the projection of u onto v and is denoted by w1 = projvu.The vector w2 is given by w2 = u - w1.

w2

v

u

w1

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θ

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θ is acute

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{w1

uw2

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{

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θ€

θ is obtuse

v

Projection of u onto v

Let u and v be nonzero vectors. The projection of u onto v is

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projvu =u ⋅v

v2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟v

Find the projection of onto Then write u as the the sum of two orthogonal vectors, one which is projvu.

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u = 3,−5

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v = 6,2 .

w1 = projvu =

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u ⋅v

v2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟v =

8

40

⎛

⎝ ⎜

⎞

⎠ ⎟ 6,2 =

6

5,2

5

w2 = u - w1 =

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3,5 −6

5,2

5=

9

5,−

27

5

So,

u = w1 + w2 =

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6

5,2

5+

9

5,−

27

5= 3,−5