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Vector Cross Product

  1. Sep 12, 2006 #1
    This is the question:

    Two vectors A and B have magnitude A = 3 and B = 3. Their vector product is A X B = -5k+2i. What is the angle Between A and B.

    OK so I'll start with what I do know.

    I do know that the cross product is the magnitude of A times magnitude of B times sin theta of B.
    I end up with

    3*3sinTHETA = 5.4 ( i got 5.4 from finding the magnitude with the components that they gave me )

    eventually getting an angle of 37 degrees by dividing by 9 and using arcsin

    im not sure what this angle is though..I think it is the angle of B but if it is how does that help me to find the angle between A and B ?
  2. jcsd
  3. Sep 13, 2006 #2


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    The [itex]38^o[/itex] angle is the actual angle between the two vectors [itex]\vec{A},\ \vec{B}[/itex].
  4. Sep 13, 2006 #3
    This is not true.

    The cross product is:
    [tex] \vec A \times \vec B = \hat n |AB \sin \theta_{AB}| [/tex]

    where [itex] \hat n [/itex] is a unit vector with direction found by the right hand rule. It is orthogonal to the plane formed by [itex] \vec A [/itex] and [itex] \vec B [/itex].

    The MAGNITUDE of the cross product however, can be written as:

    [tex] |\vec A \times \vec B| = |AB \sin \theta_{AB}| [/tex]

    Notice that [itex] \hat n [/itex] disappears because it's magnitude is unity (equal to one).

    Also note that [itex] \theta_{AB} [/itex] is the smaller angle between vectors [itex] \vec A [/itex] and [itex] \vec B [/itex].

    Does that help?
    Last edited: Sep 13, 2006
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