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Cross product help (I think)

  1. Jun 24, 2008 #1
    Hey guys, I normally don't use forums for help but I've been stuck on these questions for some time now and I just want to see if there's anyone out there who can set me in the right direction. These questions are from a grade 12 Calculus and Vectors class and they relate to the cross product.

    1. How much work is done by a force, in Newtons, represented by the vector F = (4, 7, 1) if it results in a displacement, in metres, represented by the vector delta(I'm assuming it's delta but the symbol is really small) S = (5, 3, 8).

    2.If vectors A and B are perpendicular and A = (4, – 4, 8) and
    B = (5, k, 2k), find k.

    For the first one I tried applying the cross product to make a new vector until I remembered that finding the Work for F and the new vector would result in a dot product of 0, because of the fact that cross products are perpendicular to both vectors. After realizing it wouldn't work I tried calculating the distance between the vectors but this didn't help either. I'm stumped on both of these and any help would be greatly appreciated, Thanks alot!
  2. jcsd
  3. Jun 24, 2008 #2


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    Actually they are related to the dot product.

    Work is F.s

    No need to use the cross-product
  4. Jun 24, 2008 #3
    So if I multiplied [4, 7, 1]*[5,3,8] and got 49 as my dot product, then that's how much work is needed to result delta S = [5, 3, 8]? Why would they even mention the delta in the first place, because that's the only reason I assumed this question was difficult...
  5. Jun 24, 2008 #4


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    No idea about the delta, but the dot product gives a scalar, while cross-pdt gives a vector. Work is a scalar so, the only way to get the amt of work done would be to find the dot-product.
  6. Jun 25, 2008 #5


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    What do you mean "mention the delta"? I presume what you had was [itex]\delta S= [5, 3, 8][/itex] or [itex]\Delta S= [5, 3, 8][/itex]. They use the notation [itex]\delta S[/itex] or [itex]\Delta S[/itex] because it implies a change in S which is what is necessary to do work: Force applied to move something- to change its position.
  7. Jun 25, 2008 #6
    Sorry, I didn't know how to include the triangle in the question so I just called it delta, but what I was having trouble with was finding the work to move from the initial displacement to the delta displacement. So basically I figured there was an initial displacement vector that I had to find, but the only way I knew of creating a vector given two vectors is through the cross product. Does anyone know how to find the initial displacement?
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