Date: 12/10/97 at 03:31:38 From: Goga Subject: Inner product Hi, 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 Hi, Recall that the inner product of two vectors A = (x,y), B = (z,w) in the plane is <A,B> = A.B = xz + yw. Alternatively, 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 Check out our web site! http://mathforum.org/dr.math/
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