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Inner Product

Date: 12/10/97 at 03:31:38
From: Goga
Subject: Inner product


How can I find the angles of the triangle formed by the points (8,2), 
(-20,7), (45,33), using the inner product?

Thank you for your time, Dr. Math.

Date: 12/11/97 at 01:20:24
From: Doctor Pete
Subject: Re: Inner product


Recall that the inner product of two vectors A = (x,y), B = (z,w) in 
the plane is

     <A,B> = A.B = xz + yw.


     A.B = |A||B| Cos[T],

where |A| = Sqrt[x^2 + y^2], |B| = Sqrt[z^2 + w^2], are the lengths 
of the vectors A, B, and Cos[T] is the cosine of the angle between 
A and B. (Sqrt [x^2 + y^2] is simply the square root of x squared 
plus y squared, etc.)

So combining these two formulas, we find

                        xz + yw
     Cos[T] = ---------------------------- .
              Sqrt[(x^2 + y^2)(z^2 + w^2)]

From this we can easily solve for T.  Be careful to note that if 
A = (8,2), B = (-20,7), C = (45,33), you do not want to find 
A.B, B.C, C.A, but rather, (A-B).(C-B), (B-A).(C-A), (B-C).(A-C).  
But an easier way to do this is to use the Law of Cosines,

     c^2 = a^2 + b^2 - 2ab Cos[C],

where a is the distance between B and C, b is the distance between A 
and C, and c is the distance between A and B.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Trigonometry

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