Calculate 2C*(3AxB) for Vectors A, B, and C

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To calculate 2C*(3AxB) for the vectors A, B, and C, first, recognize that AxB represents the cross product of vectors A and B. The user initially confused the problem by calculating 2A instead of focusing on the required cross product. The correct approach involves calculating the cross product of A and B, then multiplying the result by 3, and finally multiplying by 2C. Clarification on the properties of cross products and unit vectors is necessary for accurate calculations. Understanding these concepts will help resolve the confusion and complete the problem correctly.
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Homework Statement


For the following three vectors, what is 2C*(3AxB)
A=3i+3j-4k
B=-3i+2j+3k
C=-6i-8j

**All of the capital letters are vectors**

Homework Equations





The Attempt at a Solution


2A=2(3i+3j-4k)
=6i+6j-8k

2AxB=(6i+6j-8k)*(-3i+2j+3k)

(6i)(-3i)+(6i)(2j)+(6i)(2k)+(6i)(-3i)+(6j)(2j)+(6j)(3k)+(-8k)(-3i)+(-8k)(2j)+(-8k)(3k)

(12k)+(-12j)+(18k)+(18i)+24j)+(16i)

34i+12j+30k

I don't really know what to do next, honestly I don't really understand what I did to get that far. I found an example problem and just kind of plugged stuff in... If you could explain to me how to work this problem that would be great!
 
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I'm confused. You ask what is 2C*(3AxB) and then you calculate 2AxB.
 
jdawg said:

Homework Statement


For the following three vectors, what is 2C*(3AxB)
A=3i+3j-4k
B=-3i+2j+3k
C=-6i-8j

**All of the capital letters are vectors**

The Attempt at a Solution


2A=2(3i+3j-4k)
=6i+6j-8k
Question, as written, does not call for 2A.

2AxB=(6i+6j-8k)*(-3i+2j+3k)

(6i)(-3i)+(6i)(2j)+(6i)(2k)+(6i)(-3i)+(6j)(2j)+(6j)(3k)+(-8k)(-3i)+(-8k)(2j)+(-8k)(3k)

(12k)+(-12j)+(18k)+(18i)+24j)+(16i)

34i+12j+30k

I don't really know what to do next, honestly I don't really understand what I did to get that far. I found an example problem and just kind of plugged stuff in... If you could explain to me how to work this problem that would be great!

Hmmm ... OK: You need quick primer lessons:

AxB indicates a cross product:


What the example you used did was start by multiplying out the brackets as usual ... then used the rules for the cross product of the unit vectors. so, using "x" to indicate cross product:
then ixi=jxj=kxk=0
and ixj=k, jxk=i, kxi=j
and kxj=-i, jxi=-k, ixk=-j
 
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SteamKing said:
I'm confused. You ask what is 2C*(3AxB) and then you calculate 2AxB.

Ohhh that might be part of my problem, I got this question mixed up with the example I was looking at.
 
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