Scalar product of 2B and 3C given vectors B and C

[TA1068]

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.

 

[mathman]

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

 

[TA1068]

Oops! Here it is:

img87.imageshack.us/img87/2452/vectorqqp8.jpg

 

[Pere Callahan]

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.

 

[TA1068]

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 that's 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!

 

[Pere Callahan]

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