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Homework Help: Angle between two vectors (Only by having their lengths)

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Ok guys, I'm studying Physics and I started, of course, by learning about vectors.

    So I have this problem:

    There are two vectors: A (its length is 3 m) and B (its length is 4 m). I have not other informations about them.
    I want to know how I can dispose them to get a vector S (which is A + B) such as:
    a) its length is 7 m.
    b) its length is 1 m.
    c) its length is 5 m.

    Now, I know that the points a and b are obvious but I want a rigorous solution which is valid for point c and for every other length of the vector S.

    2. Relevant equations
    : From now on when I write a I mean the length of the vector A. When I write b I mean the length of the vector B. Let's call o the agle between them.

    Maybe this equation will be useful.

    cos(o) = (A . B) / (a b)

    3. The attempt at a solution
    I tried but I cannot evaluate the scalar product A . B because it is equals to (a b cos(o)). And if I replace it in the equation I have:

    cos(o) = (a b cos(o)) / (a b) = cos(o)

    I need your help.
  2. jcsd
  3. Oct 28, 2015 #2


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    Can you think of a way of writing the length of the composite vector using only the inner product and the lengths of A and B?
  4. Oct 28, 2015 #3


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    You have ##|A||B|\cos(\theta) =A \cdot B##. I am not sure that is exactly what you want to do, since C= A+B.
    In theory, let's say you make A point along the x axis. A + B would be ##( |A| +B_x)\hat x + B_y \hat y## which is adding the part of B in the x direction to A and the remaining component of B is in the y direction.
    ##B_x## can also be thought of as ## B\cdot \hat x ## and similarly ##B_y = B\cdot \hat y = B - B_x##.
    In more general terms, you don't even need to align A with the x axis, you could just make ##\hat a## a unit vector in the A direction.

    If that still doesn't help, check your work with triangles and geometry...in that way (c) should at least be very straightforward.
  5. Oct 28, 2015 #4
    Thanks for your reply. I have not specified it but I want to know the angle between the vectors A and B. For this reason I mentioned the equation: ##|A||B|\cos(\theta) =A \cdot B##.

    Even if i read your reply I am still not able to evaluate the angle between them (I want a rigorous solution which is valid for every length of S).

    And for evaluating I mean that I want to know how many degrees (or radians) is the angle between the two vectors. With no unknown, I want to be able to precisely evalute the angle. In (a) the angle measures 0°, in (b) the angle measures 180°, in (c) it is 90° because 5 = sqrt(a^2 + b^2).
  6. Oct 28, 2015 #5
    Remember that by definition, ##s^2 = \vec S.\vec S = (\vec A +\vec B).(\vec A + \vec B) ##. You are almost there !
  7. Oct 28, 2015 #6


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    As you note, parts a) and b) are obvious, and part c) relates to a well-known right triangle.
    If the only information you have are the magnitudes of vectors A and B and the magnitude of S, you don't have enough to go on for all possible values of |S|. For example, if |S| = 0 or if |S| > 7, there is no solution.

    If S is such that it could be the third side of a triangle, you can use the Law of Cosines to find one of the angles of the triangle formed by A, B, and A + B. With that information, you could find the vector A + B.
  8. Oct 28, 2015 #7
    Ok. Thanks for your replies, I solved.
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