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A quick question about scalar product of vectors

  1. Apr 14, 2008 #1
    Attached is a .jpg of my problem.

    I know how to find the scalar product of B*C (I think... 5, right?), but I don't really know where the 2 and 3 come into play. I've tried multiplying the values of C by 3 and then finding the scalar product, then multiplying the quantity by two, but that was incorrect.

    I couldn't find it in my physics text. I guess it's probably something I should know, but I don't, so that's why I'm here! Any help would be greatly appreciated.

    Attached Files:

  2. jcsd
  3. Apr 14, 2008 #2


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    Your attachment couldn't be displayed. Try something else.
  4. Apr 14, 2008 #3
    Oops! Here it is:

  5. Apr 14, 2008 #4
    Well this was correct. Unless you made a mistake in carrying out the calculations ...
    Remember when you multiply the vector C by the number 3 you have to multiply each component of C by this number 3, giving you

    3C = 3(-1,-1,2)=(-3,-3,6)

    I suggest double-checking your calculations and if this doesn't help....show us what you have done and we can most likely find your mistake.:smile:

    For the scalar product of B and C, five is correct.

    B.C = (-3,0,1).(-1,-1,2)=3+0+2=5, well done.
  6. Apr 14, 2008 #5
    So for my work...

    B = (-3, 0, 1) and
    C = (-3, -3, 6)

    So... 9 + 0 + 6 = 15
    15 * 2 = 30

    ...I could have sworn thats what I was doing all along, but for some reason I kept getting 60 for my answer. Hmm.

    Anyways, thanks greatly for any and all help!
  7. Apr 14, 2008 #6
    So are you content with 30 now? It seems corect to me.
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