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Cross Product Questions

  1. Aug 22, 2012 #1
    I have two problems:


    1. The problem statement, all variables and given/known data
    In general, what can be said about the vector product x×(x×y)

    3. The attempt at a solution
    I thought the result of this would be parallel to y. However the answer suggests it is orthogonal to x. Can anyone explain how I could approach this question? I tried to visualize it in my head but it was very difficult.


    1. The problem statement, all variables and given/known data
    Given two vectors of length 2 and 3 separated by an angle of 30 degrees, what is the cross product of the two vectors?

    3. The attempt at a solution
    I know that the cross product = (length of vector a)*(length of vector b)*sin(theta)
    This gives us 3.0, however the answer key suggests there is not enough information to answer the question.
     
  2. jcsd
  3. Aug 22, 2012 #2

    vela

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    You seem to be making an assumption about the relative orientations of ##\vec{x}## and ##\vec{y}##. Try look at a few specific cases and see if it clears up any misconceptions you have. Concentrate on the direction of the cross product. Don't worry about the magnitudes for now. Draw pictures!

    First case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (0, 1, 0)##. That is the two vectors lie along the x-axis and y-axis respectively. What is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get?

    Second case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (1, 1, 0)##. This time, the second vector still lies in the xy-plane, but it's no longer aligned to the y-axis. Again, what is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get? What effect did changing ##\vec{y}## have on the direction of the final answer?


    This isn't quite right. The cross product gives you a vector, but quantity on the righthand side is a number. The two sides of your equation can't be equal. What you mean is
    $$|\vec{a}\times\vec{b}| = |a||b|\sin\theta.$$ You found the magnitude of the cross product, but that's only half the answer. You still have to give its direction.
     
  4. Aug 22, 2012 #3
    Wow, thanks for clearing everything up! Makes sense now.
     
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