Published on: Wed Sep 01 2010

We reviewed Vectors today, how vectors can be used to describe a line in 2D and 3D space. <1,3> + t<1,1> or (1+t)i + (3+t)j and in 3D 2si + (2+2s)j Why the s and the t? They are just variables describing the scaling factor of your place in the system.

For describing planes, this type of generalization works as well. There are two direction vectors and one initial vector pointing to the zero-zero location of the plane and two direction vectors describing the tilt of the plane.

For example, the Equation of three points, as a plane is:

POINTS: (1,2,3) , (5,5,5) , (-1,6,9)

Form a line with a starting zero zero location of <1,2,3> plus a scaling in t direction and s direction.

<1,2,3> + t<4,3,2> + s<-2,4,6>**Vector Rules**

u · u > 0 unless u = 0

u · v = v · u

(u + v) ·w = (u ·w) + (v ·w)

r(u · v) = (ru) · v = u ·(rv)

*****r is a scalar

DOT PRODUCT

u · v = |u| |v| cos θ

u · v = u_{1}v_{1}+u_{2}v_{2}+...u_{n}v_{n}

Why does the dot product work this way?? Because of the Law of Cosines, which states:

c^{2} = a^{2} + b^{2} - 2ab cos θ

and if we work out a triangle, eventually we will see

a = u

b = v

c = u - v

|u - v|^{2} = |u|^{2} + |v|^{2} - 2|u||v| cos θ

If we want to find the angle between two vectors, we can use the dot product.

arccos (a∙b / (|a|*|b|) )

That is regular multiplication of the length of a and b.

Now we can project b onto A by using |b| cos θ

And we can describe a plane, by saying (x - p) ∙n = 0

So we are describing the plane with a normal n and a point (is that p?)

Tags: Dot product, find the angle, project B onto A