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Homework Help: Angle between 2 vectors given dot product and cross product

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A dot B=0.707m^2, A cross B=4.950m^2 k^. If |A|=2.500m and B makes an angle of 135° with the positive x-axis, what are A and B in component form?

    2. Relevant equations
    A*B = |A||B|cos(θ)
    A X B = |A||B|sin(θ)@RHR

    3. The attempt at a solution
    I have no prior physics experience and my professor seemed to glance over some fundamentals so I'm really at a loss.

    I've tried several things but can't seem to make anything work. I tried solving as a system of equations, but it just didn't look right. If I solved like I would expect I'd end with a magnitude for B that had no unit which made no sense because A has a unit of meters.

    I really want to LEARN this so if someone could give me a clue about which way to take it, I'll be happy to pound it out. Not asking for a handout but rather some guidance for a complete physics newb.

  2. jcsd
  3. Jun 11, 2012 #2


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    You know that [itex]|A||B|cos(\theta)= 0.707[/itex] and [itex]|A||B|sin(\theta)= 4.95[/itex] so, dividing the second equation by the first, [itex]sin(\theta)/cos(\theta)= tan(\theta)= 4.94/.707= 6.99[/itex]. What is [itex]\theta[/itex]? (there are two possibilities- knowing that AxB points toward the positive z axis tells you which one.) Once you know that, and know that "B makes an angle of 135 degrees with the x-axis" what angle does A make with the x-axis .

    Of course, for any vector, v, with given length, |v|, and knowing that it makes angle [itex]\theta[/itex] with the x-axis, the x component it [itex]|v|cos(\theta)[/itex] and the y-component is [itex]|v|sin(\theta)[/itex]. (Of course, the z-component is 0.)
  4. Jun 11, 2012 #3

    What does k^ at the end of the above sentence mean?

    Lets see your system of equations. Looks like it should be 4 equations with 4 unknowns.

  5. Jun 11, 2012 #4


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    Hello herbally. Welcome to PF !

    Compile all the information given and use the various ways of expressing the dot product & cross product.

    In general, what is the direction of the cross product [itex]\vec{C}\times\vec{D}[/itex] (Which is a vector) in relation to the direction of vectors [itex]\vec{C}[/itex] and [itex]\vec{D}\ ?[/itex]
  6. Jun 11, 2012 #5


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    I assumed it was the k vector- the unit vector in the direction of the positive z-axis.

  7. Jun 11, 2012 #6
    @HallsofIvy - Had no idea I could divide the two equations. Is that something you just do any time you have both cross product and dot product? Theta=81.9. It's positive k hat direction so if I remember correctly its a z-out vector.

    Sadly, I haven't quite figured out how to determine which non-unique answer is correct based on pos or neg direction. Is it relative to the sign of the x and/or y components of vector B?

    @Ratch - Kind of embarrassed to even post the system of equations. I don't think it was getting me anywhere. Plus I had two equations and two variables so I was clearly off.

    Thanks tons for your responses. Been going crazy trying to figure this stuff out and feel a thousand percent better already.
  8. Jun 11, 2012 #7


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    Do you know how to make dot product and cross product if the vectors are given in component form?
    You also must know that the cross product AxB is a vector normal to both A and B. As the cross product is given parallel with the unit vector k (pointing along the z axis), both A and B must lie in the x,y plane. A=axi+ayj, B=bxi+byj.

    You also know the magnitude of A and the angle of B with the positive x axis. Write up the components with the magnitude and the angle.

  9. Jun 12, 2012 #8
    @ehild - I taught myself how to do dot product and cross product and have a pretty good grasp of it. According to my professor, cross product produces a vector with no parallel components (i.e. z-axis). I'm not really sure what you mean by "normal to both A and B".

    The main thing troubling me at this point is determining which tan vector is the correct one. I assume that once I know which vector is right, then I can plot it along with vector B which is 135deg from x-axis and get the components from there using pythagorean & sohcahtoa.

    Can you offer some insight in regards to non-unique tan answers?

    Thanks again.
  10. Jun 12, 2012 #9


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    Then you can write both products with the components of A and B.

    The cross product AxB is perpendicular (normal) to both vectors A and B.

    As both the dot product and the cross product are positive, both the sine and the cosine of theta are positive. In what quadrant is theta then?

    If you know the sign rule of cross product you must know what is the relative position of B with respect to A.

  11. Jun 13, 2012 #10
    Thanks a bunch for your help folks! I went to tutoring yesterday and before I got help I heard the professor talking to a student about an unrelated issue and had an epiphany. I plugged the value for dot product as well as theta and the magnitude of A into the dot product formula and got the magnitude for B.

    Then I plotted them and solved for components using sohcahtoa.

    One more question though....ehild, I know you said that since dot/cross products are both positive then the vector must be in quadrant I which makes perfect sense. My professor pointed out in a different problem that when you take inverse tan and the denominator is negative, you always add 180 degrees. Can someone add some clarity to that perhaps?

    Thanks again for all the help. Really feel like a turned a corner yesterday.
  12. Jun 13, 2012 #11
    I'm not sure what you exactly mean here. In which context was this done??

    If [itex]x=tan\theta[/itex] and only the denominator is negative, then [itex]cos\theta[/itex] is negative. Now if you have, [itex]cos(\pi + \theta)[/itex] that will be equal to [itex]-cos\theta[/itex], giving you a positive cosine, and hence a positive tangent.
  13. Jun 13, 2012 #12
    He's referring to sohcahtoa for getting angle between vectors.


    According to him, if the denominator is negative then the angle it produces is 180° out from the correct angle.
  14. Jun 13, 2012 #13

    Since the defined principal value of the tan is between -90° and 90°, you add 180° to the value of the tan when it falls into the second or third quadrant to get into the first or second quadrant. Adding 180° to the tan does not change its value.

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