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Homework Help: Vectors - basic help

  1. Mar 9, 2005 #1
    vectors -- basic help

    Two vectors are given by A=-3i+4j and B=2i+3j. I need to find the angle between A and B

    I know AxB is 17k because:
    -3i+3j + 4j*2i = -9k-8k = 17k

    I'm not sure how to find the vector
     
  2. jcsd
  3. Mar 9, 2005 #2

    BobG

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    Do you know how to find the dot product? (It's very similar to the cosine difference identity in trigonometry)

    Edit: By the way, AxB is -18k, not 17k. You have to cross multiply (-3*2)-(4*3) to find the cross product.
     
    Last edited: Mar 9, 2005
  4. Mar 9, 2005 #3
    my book says -17k for an answer. It might be wrong or something.

    To find the angle between A and B...
    sin^(-1)[AXB]/[AB]
    where AXB is the cross product... and AB is ??

    not sure what the equation is says
     
  5. Mar 9, 2005 #4

    BobG

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    Yeah, it's -17k. What ever gave you the idea it wasn't? (Okay, I admit, the second time I looked at it, I looked at your -3i + 3j and thought I was looking at your two vectors).

    Yes, that would work, since your cross product only has one component - otherwise you'd need the norm. A dot B over the product of the norms also works, but gives you the cosine of the angle, instead.

    In other words, you'd wind up with (-3*2)+(4*3)/(5*√13) = cos (θ)
     
  6. Mar 9, 2005 #5

    HallsofIvy

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    The first question you said was " I need to find the angle between A and B"
    then "my book says -17k for an answer." You do understand that "-17k" is not an angle, don't you?

    Yes, it is true that the length of uxv is |u||v|sin(θ) and you can use that to find the angle. But the cross product is much more complicated than the dot product and u.v is |u||v|cos(θ) that's a much simpler way to find the angle.
     
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