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Finding the Angle between Two Vectors

Date: 01/15/2006 at 19:39:55
From: Victor
Subject: vector math

There's two vectors A and B, which both have equal magnitudes.  In 
order for the magnitude of A+B to be 120 times larger than the 
magnitude of A-B, what must the angle between them be?

Date: 01/16/2006 at 04:55:56
From: Doctor Luis
Subject: Re: vector math

Hi Victor,

Using vector norm notation, the problem informs us that A and B are
two vectors such that 

    |A| = |B|

Further, they want us to determine the angle T (between A and B) such that

  |A+B| = 120 * |A-B|

Ok.  Now that we have expressed the requirements in concise 
mathematical notation, let's solve the problem.

The easiest way to find T is probably to use the dot product between A
and B (denoted A.B).  I'm sure you'll recognize the identity

  A.B = |A| * |B| * cos(T)

Solving for cos(T) we get

  cos(T) = A.B/(|A| * |B|)

if we use |A|=|B|, then

  cos(T) = (A.B)/|A|^2

Now, it is clear that the problem will be easier if we find the value
(A.B) in terms of |A|^2. We can do that from the following 
relationship between the dot product and vector norm:

  v.v = |v|^2

(which is actually an instance of the identity above, applied to the
same vector v, so that T=0, or cos(T)=1).

Well the important thing is to realize that we can apply v.v = |v|^2
to |A+B|^2 and to |A-B|^2 (and then applying the distributive rule of
the dot product),

  |A + B|^2 = (A + B).(A + B)
            = A.(A+B) + B.(A+B)
            = (A.A + A.B) + (B.A + B.B)
            = |A|^2 + 2(A.B) + |B|^2
            = 2|A|^2 + 2(A.B)           (using |A|=|B|)

  |A - B|^2 = (A - B).(A - B)
            = A.(A-B) - B.(A-B)
            = (A.A - A.B) - (B.A - B.B)
            = |A|^2 - 2(A.B) + |B|^2
            = 2|A|^2 - 2(A.B)            (using |A|=|B|)

Now, we'll use that second equation that the problem gave us:

    |A+B| = 120 |A-B|


  |A+B|^2 = 120^2 * |A-B|^2

  (2|A|^2 - 2(A.B)) = 120^2 * (2|A|^2 - 2(A.B))

You can use this last equation to solve for A.B in terms of |A|^2,
which will allow you to find the ratio A.B/|A|^2 = cos(T), from which
you can finally determine the value of T.

As a sanity check, you should notice that the answer is small (at
least relative to 180 degrees), which means that A and B are pointing
in almost the same direction.  This makes sense, since they'll 
reinforce each other when added, but almost cancel out when 
subtracted.  This is how |A+B| can manage to be 120 times larger than 
|A-B|, even though the two vectors A and B have the same magnitude.

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum 
Associated Topics:
High School Linear Algebra

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