Vector Angles: Prove A.B = |A||B|cosA

Date: 07/13/98 at 02:37:31
From: Bryan J. 
Subject: Vector angles

Hi Doctors,

I just finished my pre-calc class, and at the end we covered vectors, 
including this theorem:
  
             A.B = |A||B|cosA
     dot product = length(length)cos(angle between them).

If the proof isn't too complicated, what is the basis for that 
theorem? Any help would be appreciated.

Bryan

Date: 07/22/98 at 10:51:35
From: Doctor Rick
Subject: Re: Vector angles

Hi, Bryan. 

No, I don't think this proof is very complicated. Let's see if you 
agree. 

To start with, let's draw a vector A with length a, making an angle 
alpha with the x-axis:
  
y
|
+....... + A
|       /:
|      / :
|     /  :
|   a/   :
|   /    : x = a*sin(alpha)
|  /     :
| /      :
|/)alpha :
+--------+------x
 y = a*cos(alpha)

The x component of the vector is a*cos(alpha), and the y component is
a*sin(alpha), from the right triangle I have drawn. Now make another 
vector B, of length b and making angle beta with the x-axis:

  A = (a*cos(alpha), a*sin(alpha))
  B = (b*cos(beta), b*sin(beta))

The dot product is now (just multiplying the x components and the y
components, and adding them together)

  A . B = a*b*cos(alpha)*cos(beta) + a*b*sin(alpha)*sin(beta)
        = a * b * (cos(alpha)*cos(beta)+sin(alpha)*sin(beta))

All that's left is to remember the trigonometric identity,

  cos(alpha - beta) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta)

and we have

  A . B = |A| * |B| * cos(alpha - beta)

But (alpha - beta) is the angle between the two vectors.

Does that explain it? Write back if you want more explanation.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 07/16/98 at 03:23:28
From: Bryan J. 
Subject: vector theorem proof

Hi Doctors,

In determing some angles in 3-D figures, I used the theorem:
  
   A.B = |A||B|cosA    - pertaining to vectors
	
I was thinking this looked like the law of cosines, but I couldn't 
maket the mental connection. Could you please help me to understand 
how the above theorem was derived?

Thanks.

Date: 07/16/98 at 07:08:18
From: Doctor Anthony
Subject: Re: vector theorem proof

This is not so much a 'theorem' as a 'definition'. You DEFINE the 
scalar products of two vectors to be the product:

 a.b = |a| x |b| x cos of angle between a and b.

Following this definition you get the very convenient result that two 
vectors given in component form such as:

   a = (x1,y1,z1)   b = (x2,y2,z2)  then the scalar product is:

    a.b = x1.x2 + y1.y2 + z1.z2

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/