1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Why is A cross B= ABsin(theta) false?

  1. Sep 30, 2016 #1
    1. The problem statement, all variables and given/known data

    problem 2.2(b)
    2. Relevant equations
    A cross B = ABsin(theta)

    3. The attempt at a solution
    I believe that the answer is that you cannot multiply two vectors but you can dot or cross them. In order to make this a true statement you have to take the magnitudes of A and B.

    Attached Files:

  2. jcsd
  3. Sep 30, 2016 #2


    User Avatar

    Staff: Mentor

    Note the change in font:
    \mathbf{A} \times \mathbf{B} = A B \sin \theta
    The author is using a common convention that ##A = | \mathbf{A} |##, so it is the magnitude that is considered.
  4. Sep 30, 2016 #3

    Buzz Bloom

    User Avatar
    Gold Member

  5. Sep 30, 2016 #4
    From my readings
    When you cross two vectors you get a new vector that is orthogonal to them and that the vectors form a parallelogram.
    The new vector c pointing perpendicular to them gives us the height. But the height is defined by by C(sin(theta)).
    In order to find the area its Height times base. The length of the base is |B| and the height is |A|(sin(theta)). It's sin(theta) because we broke up the A vector into its two components and the height of vector A is opposite the angle which gives us sin.

    To continue and apply this knowledge

    If we want to find the volume it's height times the area of the parallelogram. Now I know the height of the new vector C is Csin(theta) so now I just multiply that by the area I got from above. Is this right?
  6. Sep 30, 2016 #5
    Plus A cross B is a vector and we should get another vector back. ABsin(theta) is not a vector so it would make the statement false. We have a vector equal to a scalar in this equation.
  7. Sep 30, 2016 #6


    User Avatar

    Staff: Mentor

    That's it.

    But you see that the equation is almost correct.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted