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A quick question about scalar product of vectors... |
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| Apr14-08, 03:09 PM | #1 |
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A quick question about scalar product of vectors...
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. |
| Apr14-08, 04:39 PM | #2 |
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| Apr14-08, 04:41 PM | #3 |
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Oops! Here it is:
img87.imageshack.us/img87/2452/vectorqqp8.jpg |
| Apr14-08, 06:32 PM | #4 |
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A quick question about scalar product of vectors...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. ![]() 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. |
| Apr14-08, 07:56 PM | #5 |
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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! |
| Apr14-08, 07:58 PM | #6 |
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So are you content with 30 now? It seems corect to me.
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