# A quick question about scalar product of vectors...

by TA1068
Tags: product, scalar, vectors
 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. Attached Thumbnails
 Sci Advisor P: 5,772 Your attachment couldn't be displayed. Try something else.
 P: 13 Oops! Here it is: img87.imageshack.us/img87/2452/vectorqqp8.jpg
P: 588

## A quick question about scalar product of vectors...

 Quote by TA1068 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.
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.

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.
 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!
 P: 588 So are you content with 30 now? It seems corect to me.

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