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A quick question about scalar product of vectors... 
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#1
Apr1408, 03:09 PM

P: 13

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. 


#2
Apr1408, 04:39 PM

Sci Advisor
P: 6,077

Your attachment couldn't be displayed. Try something else.



#3
Apr1408, 04:41 PM

P: 13

Oops! Here it is:
img87.imageshack.us/img87/2452/vectorqqp8.jpg 


#4
Apr1408, 06:32 PM

P: 587

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 doublechecking 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. 


#5
Apr1408, 07:56 PM

P: 13

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! 


#6
Apr1408, 07:58 PM

P: 587

So are you content with 30 now? It seems corect to me.



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