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 =
a · b = a1b1 + a2b2
Likewise with 3 dimensions,
Given a =
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 =
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) 2 proja b = a · b/|a| *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
a a a x x x ax cos Φ -sin Φ 0
ay sin Φ cos Φ 0 x = ρ cosΦ az 0 0 1 y = ρ sinΦ z = z Rectangular to Spherical
Dot products of unit vectors in spherical and rectangular coordinate systems
a A a r θ Φ ax sin θ cos Φ cos θ cos Φ -sin Φ
ay sin θ sin Φ cos θ sin Φ cos Φ x = r sinθ cosΦ az cos θ -sin θ 0 y = r sinθ sinΦ z = r cosθ 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”