Dot product

Dot product



Dot Product
Let u=<a,b,c>u=<a,b,c> and v=<d,e,f>v=<d,e,f>. Then the dot product of these two vectors will be:

uv=ad+be+dfu \cdot v = ad + be + df

Dot Product Property
If the dot product of two vectors uu and vv gives 0, then the vectors are perpendicular. In other words,

uv=0u \cdot v=0 \to perpendicular vectors

Scalar and Vector Projection
Suppose we have two vectors aa and bb. Suppose they create an angle θ\theta such that we get the following picture:

v=abb |v| = \frac{a \cdot b } {|b|}

To find the vector projection aa onto bb (which is v), we use the formula:

v=abbbb v = \frac{a \cdot b}{b \cdot b}b

Additional Dot Product Properties
Let u,v,wu, v, w be vectors and cc be a scalar. Then the properties of dot products are:
  1. uu=u2u \cdot u = |u|^2
  2. uv=vuu \cdot v = v \cdot u
  3. u(v+w)=uv+uwu \cdot (v+w) = u \cdot v + u \cdot w
  4. (cu)v=u(cv)=c(uv) (cu) \cdot v = u \cdot (cv) = c(u \cdot v)
  • Introduction
    Dot Product Overview:
    Dot Product and its Special Property
    • Multiplying the corresponding entries, and adding
    • Dot product = 0 \to vectors are perpendicular

  • 1.
    Using the Dot Product
    Find the dot product of u=<1,2,7> u = <-1, -2 , 7> and v=<2,1,2>v = <-2,1,-2> .

  • 2.
    Find the dot product of u=<1,5,3>u= <1, -5, -3> and v=<1,1,2>v= <-1, 1, 2>.

  • 3.
    Using the Dot Product Property
    Suppose two vectors u=<a,4,3> u = < a, 4, -3> and v=<1,2,3> v=<1, 2, 3> are perpendicular. Find aa.

  • 4.
    Finding Scalar and Vector Projections
    Find the scalar and vector projection of BA\vec{BA} onto CA\vec{CA} if A=(1,0,2)A=(1, 0, 2), B=(3,2,1)B=(3, -2, 1) and C=(4,1,5)C=(-4, 1, 5).

  • 5.
    Verifying Properties of Dot Product
    Use the two vectors u=<3,1,5>u=<3, 1, 5> and v=<1,4,6>v=<1,4,-6> to show that:

    uu=u2u \cdot u = |u|^2

  • 6.
    Use the 3 vectors u=<3,1,5>u=<3, 1, 5>, v=<1,4,6>v=<1,4,-6>, and w=<1,0,3>w=<1, 0, 3> to show that:

    u(v+w)=uv+uwu \cdot (v+w)=u \cdot v+ u \cdot w