Calculus(12.3 Dot Product)
Dot Product Properties 0•a=
0
Dot product of (i + 2j -3k) •(2j - k)
1(0) +2(2) - 3(-1) = 7
What are the two products of vectors?
1. Dot Product 2. Cross Product
Dot product of <2,4> • <3,1>
2(3) + 4(-1) =2
a•b>0
Acute Θ
a•b<0
Obtuse Θ
Find Vector projection of b=<1,1,2> and a=<-2,3,1>
Proj(a)b=(a→•b→/|a|)(a→/|a|)= =3/√14(<-2,3,1>/√14) =<-3/7,9/14,3/14>
To find orthogonal vectors you must:
Take the dot product of the 2 vectors to see if it equals 0
Angle ß is what?
The Angle between a→ and j→
What is a dot product?
The angle Between two vectors ( a and b) and a scalar value
Angle α is what?
The angle between a→ and i→
Angle Γ is what ?
The angle between a→ and k→
Direction angles(α,ß,Γ) are related to the unit vector a in what way?
These direction angles in the domain of [0,π] are the angles that the unit vector a→ makes with the x, y and z axis.
Can you multiply two vectors?
Yes
Dot Product Properties a•(b+c)=
a•b+a•c
Two vectors a&b are orthogonal(perpendicular) if and only if:
a•b=0
a→=<a1,a2,a3> , b→=<b1b2,b3> Dot product of a &b is given by:
a•b=a1b1+a2b2+a3b3
If Θ is the angle between vectors a &b then:
a•b=|a||b|cosΘ 0≤Θ≤π
Dot Product Properties a•b=
b•a
Dot Product Properties (ca)•b=
c(a•b)=a•(cb)
Scalar Projection of b onto a:
comp(a) b=a•b/|a|
Find scalar projection of b=<1,1,2> and a=<-2,3,1>
comp(a)b=a→•b→/|a|= =-2+3+2/√4+9+1 = 3/√14
Vector Projection of b onto a:
proj(a)b=(a•b/|a|²)(a)
If vectors a &b have lengths 4 &6, and the angle between is π/3 then find a•b
|a|=4 , |b|=6 , Θ=π/3 a•b=|a||b|cosΘ=4*6*cos(π/3) =24*1/2=12
Find the angle between a=<2,2,-1> & b=<5,-3,2>
|a|=√4+4+1=√9=3 |b|=√25+9+4=√38 a•b=2(5)+2(-3)+(-1)(2)=2 Cos(Θ)= 2/(3√(38)) Θ=cos-¹(2/(3√(38)))
Dot Product Properties a•a=
|a|²
Find the direction angles of vector a=<1,2,3>
|a→|=√1+4+9=√14 α=cos-¹(1/√14) ß=cos-¹(2/√14) Γ=cos-¹(3/√14)
How do you find ß?
ß=cos-¹(a2/|a→|)
How do you find Γ?
Γ=cos-¹(a3/|a→|)
If Θ is the angle between the nonzero vectors a & b, then
Θ= cos-¹(a•b/(|a||b|)) 0≤Θ≤π
a•b=0
Θ=π/2
How do you find α?
α = cos-¹(a1/|a→|)
