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Dot product and vector projections (Sect. 12.3), Study Guides, Projects, Research of Physics

Definition. The dot product of the vectors v and w in Rn, with n = 2,3, having magnitudes |v|, |w| and angle in between θ, where.

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Dot product and vector projections (Sect. 12.3)
ITwo definitions for the dot product.
IGeometric definition of dot product.
IOrthogonal vectors.
IDot product and orthogonal projections.
IProperties of the dot product.
IDot product in vector components.
IScalar and vector projection formulas.
Two main ways to introduce the dot product
Geometrical
definition Properties Expression in
components.
Definition in
components Properties Geometrical
expression.
We choose the first way, the textbook chooses the second way.
pf3
pf4
pf5
pf8
pf9
pfa

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Dot product and vector projections (Sect. 12.3)

I (^) Two definitions for the dot product.

I (^) Geometric definition of dot product.

I (^) Orthogonal vectors.

I (^) Dot product and orthogonal projections.

I (^) Properties of the dot product.

I (^) Dot product in vector components.

I (^) Scalar and vector projection formulas.

Two main ways to introduce the dot product

Geometrical

definition

→ Properties →

Expression in

components.

Definition in

components

→ Properties →

Geometrical

expression.

We choose the first way, the textbook chooses the second way.

Dot product and vector projections (Sect. 12.3)

I (^) Two definitions for the dot product.

I (^) Geometric definition of dot product.

I (^) Orthogonal vectors.

I (^) Dot product and orthogonal projections.

I (^) Properties of the dot product.

I (^) Dot product in vector components.

I (^) Scalar and vector projection formulas.

The dot product of two vectors is a scalar

Definition

The dot product of the vectors v and w in Rn, with n = 2, 3, having magnitudes |v |, |w| and angle in between θ, where 0 ≤ θ ≤ π, is denoted by v · w and given by

v · w = |v | |w| cos(θ).

O

V

W

Initial points together.

Perpendicular vectors have zero dot product.

Definition

Two vectors are perpendicular, also called orthogonal, iff the angle in between is θ = π/2.

0 = / 2

V W

Theorem

The non-zero vectors v and w are perpendicular iff v · w = 0.

Proof.

0 = v · w = |v| |w| cos(θ)

|v| 6 = 0, |w| 6 = 0

cos(θ) = 0

0 6 θ 6 π

⇔ θ =

π

2

The dot product of i, j and k is simple to compute

Example

Compute all dot products involving the vectors i, j , and k.

Solution: Recall: i = 〈 1 , 0 , 0 〉, j = 〈 0 , 1 , 0 〉, k = 〈 0 , 0 , 1 〉.

x

i j

k

z

y

i · i = 1, j · j = 1, k · k = 1,

i · j = 0, j · i = 0, k · i = 0,

i · k = 0, j · k = 0, k · j = 0. C

Dot product and vector projections (Sect. 12.3)

I (^) Two definitions for the dot product.

I (^) Geometric definition of dot product.

I (^) Orthogonal vectors.

I (^) Dot product and orthogonal projections.

I (^) Properties of the dot product.

I (^) Dot product in vector components.

I (^) Scalar and vector projection formulas.

The dot product and orthogonal projections.

Remark: The dot product is closely related to orthogonal

projections of one vector onto the other.

Recall: v · w = |v| |w| cos(θ).

O W

V

|v| cos(θ) =

v · w

|w|

O W

V

|w| cos(θ) =

v · w

|v|

Remark: If |u| = 1, then v · u is the projection of v along u.

Properties of the dot product.

(c) u · (v + w) = u · v + u · w, is non-trivial. The proof is:

W

w

V+W

|V+W| cos( 0 )

V

0 (^0) W

|V| cos( 0 )V

|W| cos( 0 )W

U

(^0) V

|v + w| cos(θ) =

u · (v + w) |u|

,

|w| cos(θw ) =

u · w |u|

,

|v| cos(θv ) =

u · v |u|

,

    

   

⇒ u · (v + w) = u · v + u · w

Dot product and vector projections (Sect. 12.3)

I (^) Two definitions for the dot product.

I (^) Geometric definition of dot product.

I (^) Orthogonal vectors.

I (^) Dot product and orthogonal projections.

I (^) Properties of the dot product.

I (^) Dot product in vector components.

I (^) Scalar and vector projection formulas.

The dot product in vector components (Case R

2

Theorem

If v = 〈vx , vy 〉 and w = 〈wx , wy 〉, then v · w is given by

v · w = vx wx + vy wy.

Proof.

Recall: v = vx i + vy j and w = wx i + wy j. The linear property of the dot product implies

v · w = (vx i + vy j ) · (wx i + wy j )

v · w = vx wx i · i + vx wy i · j + vy wx j · i + vy wy j · j.

Recall: i · i = j · j = 1 and i · j = j · i = 0. We conclude that

v · w = vx wx + vy wy.

The dot product in vector components (Case R

3

Theorem

If v = 〈vx , vy , vz 〉 and w = 〈wx , wy , wz 〉, then v · w is given by

v · w = vx wx + vy wy + vz wz.

I (^) The proof is similar to the case in R^2.

I (^) The dot product is simple to compute from the vector

component formula v · w = vx wx + vy wy + vz wz.

I (^) The geometrical meaning of the dot product is simple to see

from the formula v · w = |v| |w| cos(θ).

Scalar and vector projection formulas.

Theorem

The scalar projection of v along w is the number pw (v ),

pw (v ) =

v · w

|w|

The vector projection of v along w is the vector pw (v ),

pw (v ) =

v · w

|w|

w

|w|

| W |

O

| W |

w

V

W

P ( V ) = ( V W ) W | W |

O

| W |

w

V

W

P ( V ) = ( V W ) W

Example

Find the scalar projection of b = 〈− 4 , 1 〉 onto a = 〈 1 , 2 〉.

Solution: The scalar projection of b onto a is the number

pa(b) = |b| cos(θ) =

b · a

|a|

12 + 2^2

We therefore obtain pa(b) = −

x

p (b)

a

b

a

y

Example

Find the vector projection of b = 〈− 4 , 1 〉 onto a = 〈 1 , 2 〉.

Solution: The vector projection of b onto a is the vector

pa(b) =

b · a

|a|

a

|a|

We therefore obtain pa(b) = −

x

p (b)

a

b

a

y

Example

Find the vector projection of a = 〈 1 , 2 〉 onto b = 〈− 4 , 1 〉.

Solution: The vector projection of a onto b is the vector

pb(a) =

a · b

|b|

b

|b|

We therefore obtain pa(b) =

x

b

a

p (a)

b

y