Dot Product Academic Resource Center

In This Presentation…

• We will give a definition

• Look at properties

• See the relationship in projections

• Look at vectors in different coordinate systems

• Do example problems

Dot Product

Definition:

If a = <a1, a2> and b = <b1, b2>, then the dot product of a and b is number a · b given by

a · b = a1b1 + a2b2

Likewise with 3 dimensions,

Given a = <a1, a2, a3> and b = <b1, b2 , b3 >

a · b = a1b1 + a2b2 + a3b3

Dot Product

The result of a dot product is not a vector, it is a real number and is sometimes called the scalar product or the inner

product.

Orthogonal Vectors

Two vectors a and b are orthogonal (perpendicular) if and only if a · b = 0

Example: The vectors i, j, and k that correspond to the x, y, and z components are all orthogonal to each other

Examples

Find a · b:

1. Given a = <-4, -5> and b = <2, -9>

a · b = a1b1 + a2b2

a · b = (-4)(2) + (-5)(-9) = -8 + 45

a · b = 37

2. Given a = 2j + 7k and b = -7i + 4j - k

a · b = (0)(-7) + (2)(4) + (7)(-1) = 0 + 8 + (-7)

a · b = 1

Dot Product

Properties of the dot product

1. a · a = |a|2

2. a · b = b · a

3. a · (b + c) = a · b + a · c

4. (ca) · b = c(a · b) = a · (cb)

5. 0 · a = 0

(Note that 0 (bolded) is the zero vector)

Dot Product

If the angle between the two vectors

a and b is θ, then

a · b = |a||b|cos θ

or

θ =

Examples

Find a · b:

1. Given |a| = 8, |b| = 4 and θ = 60°

a · b = |a||b|cos θ = (8)(4)cos(60°) = 12(1/2)

a · b = 6

2. Given |a| = 3, |b| = 2 and θ = π/4

a · b = (3)(2)cos(π/4) = 6(√(2)/2)

a · b = 3√(2)

Example

Find the angle between the two vectors:

Given a = <-1, -3> and b = <3, -3>

a · b = (-1)(3) + (-3)(-3) = -3 + 9 = 6

|a| = √(1+9) = √(10)

|b| = √(9+9) = √(18) = 3√(2)

θ = 6/(√(10)*3√(2)) = 6/(6√(5)) = 1/√(5)

θ = 1/√(5) = √(5)/5 rad

Direction Angles

Given a = <a1, a2, a3>

cos α =

cos β =

cos γ =

Example

Find the direction angles of the vector a:

Given a = 3i – j + 4k

|a| = √(9+1+16) = √(26)

cos α = 3/√(26) α = cos-1(3/√(26)) = 0.94 rad

cos β = -1/√(26) β = cos-1(-1/√(26)) = 1.8 rad

cos γ = 4/√(26) γ = cos-1(4/√(26)) = 0.67 rad

Projections

Scalar projection of b onto a:

compa b =

Vector projection of b onto a:

proja b = ( ) =

Example

Find the scalar and vector projection of b onto a:

Given a = <-1, -3> and b = <3, -3>

a · b = (-1)(3) + (-3)(-3) = -3 + 9 = 6

|a| = √(1+9) = √(10)

|b| = √(9+9) = √(18) = 3√(2)

compa b = a · b/|a| = 6/√(10)

proja b = a · b/|a|2*a = 6/10*a = <-3/5, -9/5>

Application Example 1

Problem: A cart is pulled a distance of 50m along a horizontal path by a constant force of 25 N. The handle of the cart is pulled at an angle of 60° above the horizontal. Find the work done by the force.

Solution:

F and d are force and displacement vectors

W = F · d = |F||d|cosθ = (25)(50)cos(60°)

W = 625 J

Application Example 2

Problem: Given a constant vector field F = 7i + 3j – k find the work done from point P(5,3,-4) to the point Q(1,4,-7)

Solution:

d = <1 – 5, 4 – 3, -7 – (-4)> = <-4, 1, -3>

W = F · d = <7,3,-1> · <-4,1,-3>

= (7)(-4) + (3)(1) + (-1)(-3) = -28 + 4 +3

W = -21 J

Vectors

Given a vector in any coordinate system, (rectangular, cylindrical, or spherical) it is possible to obtain the corresponding vector in either of the two other coordinate systems

Given a vector A = Axax + Ayay + Azaz we can obtain A = Aρaρ + AΦaΦ + Azaz and/or

A = Arar + AΦaΦ + Aθaθ

Rectangular Coordinate System

Define A = Axax + Ayay + Azaz as a rectangular vector where each component is a function of x, y, and z

Cylindrical Coordinate System

Define A = Aρaρ + AΦaΦ + Azaz as a rectangular vector where each component is a function of ρ, Φ, and z

Spherical Coordinate System

Define A = Arar + AΦaΦ + Aθaθ as a rectangular vector where each component is a function of r, Φ, and θ

Rectangular to Cylindrical

Dot products of unit vectors in cylindrical and rectangular coordinate systems

x = ρ cosΦ

y = ρ sinΦ

z = z

ax ax ax

ax cos Φ -sin Φ 0

ay sin Φ cos Φ 0

az 0 0 1

Rectangular to Spherical

Dot products of unit vectors in spherical and rectangular coordinate systems

x = r sinθ cosΦ

y = r sinθ sinΦ

z = r cosθ

ar Aθ aΦ

ax sin θ cos Φ cos θ cos Φ -sin Φ

ay sin θ sin Φ cos θ sin Φ cos Φ

az cos θ -sin θ 0

Conversion

Given a rectangular vector A = Axax + Ayay + Azaz , we want to find the vector in cylindrical coordinates A = Aρaρ + AΦaΦ + Azaz

To find any desired component of a vector, we take the dot product of the vector and a unit vector in the desired direction.

Aρ = A · aρ and A Φ = A · aΦ

Example 1

Express the vector F = 4ax -2ay +8az in cylindrical coordinates:

Fρ = F · aρ = 4(ax · aρ) – 2(ay · aρ) + 8(az · aρ)

= 4(cosΦ) – 2(sinΦ) + 8(0)

= 4cosΦ – 2sinΦ

FΦ = F · aΦ = 4(ax · aΦ) – 2(ay · aΦ) + 8(az · aΦ)

= 4(-sinΦ) – 2(cosΦ) + 8(0)

= -4sinΦ – 2cosΦ

Fz = 8

F = (4cosΦ – 2sinΦ)aρ + (-4sinΦ – 2cosΦ)aΦ + 8az

Example 1 (cont.)

Evaluate F given ρ = 2.5, Φ = 0.7, z = 1.5:

F = (4cosΦ – 2sinΦ)aρ + (-4sinΦ – 2cosΦ)aΦ + 8az

Fρ = 4cosΦ – 2sinΦ = 4cos(0.7) – 2sin(0.7) = 1.77

FΦ = -4sinΦ – 2cosΦ = -4sin(0.7) – 2cos(0.7) = -4.1

Fz = 8

F = 1.77aρ – 4.1aΦ + 8az

Example 2

Given a vector field E = xax + yay + zaz, convert to cylindrical and spherical coordinates:

Eρ = E · aρ = x(ax · aρ) + y(ay · aρ) + z(az · aρ)

= ρcosΦ(cosΦ) – ρsinΦ(sinΦ) + z(0)

= ρcos2Φ + ρsin2Φ = ρ(cos2Φ + sin2Φ) = ρ

EΦ = E · aΦ = x(ax · aΦ) + y(ay · aΦ) + z(az · aΦ)

= ρcosΦ(-sinΦ) + ρsinΦ(cosΦ) + z(0)

= -ρcosΦsinΦ + ρcosΦsinΦ = 0

Ez = z

E = ρaρ + zaz (cylindrical)

Example 2 (cont.)

Express E in spherical coordinates:

Er = E · ar = x(ax · ar) + y(ay · ar) + z(az · ar)

= x(sinθcosΦ) + y(sinθsinΦ) + z(cosθ)

= rsin2θcos2Φ + rsin2θsin2Φ + rcos2θ

= rsin2θ(cos2Φ + sin2Φ) + rcos2θ = rsin2θ + rcos2θ

= r(sin2θ + cos2θ) = r

Eθ = E · aθ = x(ax · aθ) + y(ay · aθ) + z(az · aθ)

= x(cosθcosΦ) + y(cosθsinΦ) + z(-sinθ)

= rsinθcosθcos2Φ + rcosθsinθsin2Φ – rsinθcosθ

= rsinθcosθ(cos2Φ + sin2Φ) – rsinθcosθ

= rsinθcosθ – rsinθcosθ = 0

Example 2 (cont.)

EΦ = E · aΦ = x(ax · aΦ) + y(ay · aΦ) + z(az · aΦ)

= x(-sinΦ)+ y(cosΦ)

= -rsinθsinΦcosΦ + rsinθsinΦcosΦ = 0

E = r ar (spherical)

References

• Calculus – Stewart 6th Edition, Section 13.3 “The Dot Product”