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

[jdawg]

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!

 

[SteamKing]

I'm confused. You ask what is 2C*(3AxB) and then you calculate 2AxB.

 

[Simon Bridge]

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

 

[jdawg]

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.

 

[HalfLostAndSearching]

First, let's clarify the notation used in the problem. A, B, and C are vectors represented by their components in the form of <x,y,z>. So, A = <3,3,-4>, B = <-3,2,3>, and C = <-6,-8,0>.

Now, let's rewrite the expression 2C*(3AxB) using vector operations:

2C*(3AxB) = 2C*(3A x B) = 2C * (3 * (A x B))

We know that the cross product of two vectors A and B is given by A x B = <AyBz - AzBy, AzBx - AxBz, AxBy - AyBx>. So, substituting the values of A and B, we get:

A x B = <(3)(3) - (-4)(2), (-4)(-3) - (3)(3), (3)(2) - (3)(-4)> = <17, -9, 18>

Now, substituting this value in the expression 2C * (3 * (A x B)), we get:

2C * (3 * (A x B)) = 2 * <-6,-8,0> * (3 * <17,-9,18>) = <-12,-16,0> * <51,-27,54> = <-612, 324, -864>

Therefore, the final result is -612i + 324j - 864k.