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Homework Help: Angle between vectors:

  1. Jan 26, 2008 #1
    [SOLVED] Angle between vectors:

    1. The problem statement, all variables and given/known data
    What is the angle between A and B ? Answer in units of degree.

    2. Relevant equations
    From the previous problem:
    A = 1.02 units long in positive y direction [0i + 1.02j]
    B = [-6.18i + 3.64j]

    3. The attempt at a solution

    I got the question from the previous part right, finding scalar product of the 2 vectors. So I know the vectors are set up correctly.

    I am under the impression that [tex]\theta[/tex] = [tex]\theta[/tex]b - [tex]\theta[/tex]a

    Arctan of (-1.02/0 ) gives an error ( obviously ) so i just did arctan of (-1.02) = 45.567 degrees

    Arctan of (3.64 / -6.18 ) gives a degree of -30.497

    When doing the answer, I got [tex]\theta[/tex] = 76.064. I tried this as both a negative and a positive and I was wrong.

    What am I missing here? Is one angle supposed to change sign? Is my order wrong? Maybe my arctan of (1.02) is wrong?

  2. jcsd
  3. Jan 26, 2008 #2
    So you already know there are two ways to calculate scalar product right?

    [tex]\vec a \cdot \vec b = a_xb_x + a_yb_y + a_zb_z = |\vec a| |\vec b| \cos \theta[/tex].

    Where [itex]\theta[/itex] is the angle in which you seek.
  4. Jan 26, 2008 #3
    but I already used one method to compute the angle, so lets build on that one please.

    According to my math, it would be -30.497deg - (+ 45.567deg) = -76.064 deg

    Why is this wrong? I asked if I forgot something with the quadrants or if there was some kind of sign changing error I'm unaware of.

    Thanks again for your second opinion
  5. Jan 26, 2008 #4
    Well, your instinct about the arctan(1.02) is correct, i.e. that's where you got into trouble.

    Think about that arctan. Why does that give you the angle, and for that matter, what angle does it give you? Now look at your first vector - do you really even need to use arctan to determine theta for it? There's a good reason why the formula arctan([tex]v_y/v_x[/tex]) breaks down for this case.

    Is that making it clear enough, or do you need a little more help?
    Last edited: Jan 26, 2008
  6. Jan 26, 2008 #5
    yea I started getting a headache.

    Ive been stumped on the arctan thing for a while, its been a while and honestly dont even remember what the purpose of inverse tan is.... i understand that tan gives you opposite over adjacent, but I really am truly stumped on what I did wrong with this angle.

    Using normal Tan wont give me an angle, i know this, so Im left clueless and still need your help.

    I basically did the whole problem by myself, just stumped on this part - if you helped me with it you wouldn't necessarily be " doing the problem for me " and that is what we are trying to achieve here at the forums. With an explanation, you would do more help than harm.

    Thank you for considering this
  7. Jan 26, 2008 #6
    Okay ... the tangent, as you say, is the ratio of the side opposite an angle (in a rt. triangle) to the side adjacent to it. This helps with vectors, since you can generally draw the x and y components of a vector and complete a right triangle. The tangent of the angle that the vector makes with the x axis is then the x component (the side opposite) over the y component (the side adjacent).

    The arctan function is just the inverse of the tangent. You give it a number and it will give you the angle whose tangent is that number. In other words, if you have the two sides, you can take their ratio, apply the arctan function, and get the angle.

    Now for vector B this should all make sense - you can draw the triangle exactly as I described. What about for vector A? Can you draw this triangle? Keep in mind that one side is the x component and the other is the y component (with the vector itself generally being the hypotenuse). That's why you got (-1.02/0) as the ratio of the sides. So ... what does that tell you about the angle?

    Keep in mind what angle it is that we're looking for here ...
  8. Jan 26, 2008 #7
    eh still confused even tho I shouldnt be... Im thinking now that since there is no ratio for vector A, and that it has only one component ( straight up on the y-axis ) that maybe Vector A's angle is really 360 deg?

    Which would make our answer angle 360 - 30.497 ... or something similar to that?
  9. Jan 26, 2008 #8
    Why don't you try taking a visual or geometric approach to this?

    Draw an x-y plane and plot each vector:

    A=up 1.02
    B=left 6.18 and up 3.64

    Clearly A makes a 90 degree angle with the x axis
    B makes a 30.49 degree angle with the negative x- axis

    Now use your original idea; [itex]\theta_f=\theta_B-\theta_A[/itex]
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