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Cross Product of vectors in vector mechanics by beer and johnston

  1. Jul 18, 2013 #1

    I was reading vector mechanics by beer and jhonston. I came across the equation wherein the cross prodcut of two vectors P and Q is given. It says P x Q = P x Q` . I am not bale to understand how dis is possible. Because as the vector Q changes even the angle teetha will change then how are both equal?

    I am going by the procedure that if Q changes even teetha changes and thus the rhs can not be equal to LHS. I am attaching the two pages 77 and 78 of the book. The equation is given on page 78. Can someone please explain this to me

    Attached Files:

  2. jcsd
  3. Jul 18, 2013 #2
    You are correct that the angle between Q and P is not the same as the angle between Q' and P. However, the length of Q is also not the same as the length of Q'. The cross product depends on both of these things. It might not be immediately obvious why these two changes exactly compensate for one another and leave the cross product unchanged, and it is possible to actually show it mathematically, but I think the intuitive description given in the text regarding the area of the parallelogram generated by P and Q (or P and Q') is a much better way of thinking about it. Keep in mind that Q and Q' are not two arbitrary vectors; they are related to each other in that their difference is parallel to P. That relationship implies that the parallelogram generated by P and Q has the same area as the one generated by P and Q'; this is depicted in Fig. 3.7.

    By the way, if you still want to work it out mathematically, let [itex]\theta[/itex] be the angle between P and Q and [itex]\theta'[/itex] the angle between P and Q'. The aforementioned relationship between Q and Q' implies [itex]Q \sin \theta = Q' \sin \theta'[/itex]. If you can work that out, then it's obvious that the cross products with P are equal since you just have to multiply both sides of the previous equation by [itex]P[/itex].
    Last edited: Jul 18, 2013
  4. Jul 19, 2013 #3
    The vector Q' is larger in magnitude than the vector Q, but the angle between Q' and P is smaller than the angle between Q and P. For the parallelogram geometry that they discuss, the two changes exactly cancel out so that P x Q' = P x Q.

  5. Jul 19, 2013 #4
    Thanks guys... but how can we be sure that the increase in magnitude and decrease in angle is exactly dat much to make it equal to the orignal cross product
  6. Jul 19, 2013 #5
    This is a geometry problem involving parallelograms that you should be able to work out on your own. Just drop a normal from each of the points Q and Q' to the line of action of P.

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