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→|)