Question

For the following three vectors, what is 3C·(2A x B )?
A = 2.00 + 3.00 - 7.00
B = -3.00 + 3.00 + 2.00
C = 6.00 - 8.00

Answer 1

\[3\vec C\cdot(2\vec A \times\vec B)\]\[3\vec C\cdot\left(2(2.00\hat{\imath}+3.00\hat\jmath-7.00\hat k) \times (-3.00\hat\imath +3.00\hat\jmath+2.00\hat k)\right)\]\[=3\vec C\cdot\left|\begin{array}\ \hat{\imath} &\hat \jmath&\hat k\\4.00&6.00&-14.00\\-3.00&3.00&2.00 \end{array}\right| \]\[=3\vec C\cdot\left(\left((6\times2)-(-14\times3)\right)\hat\imath-((4\times2)-(-14\times-3))\hat\jmath+((4\times 3)-(-3\times6))\hat k\right)\]\[=3\vec C\cdot\left(56\hat\imath+34\hat\jmath +30\hat k\right)\]

Answer 2

i think you have left off a component of \(\vec C\)

Answer 3

From the given information, the solution proceeds like this:
2A x B = 54i + 34j + 30k................(I don't know how to type it in determinant form. The calculation of the same is as Rhaukus showed above. Just that, instead of 56i, it ought to be 54i)
Then,
3C.(2A x B) = (18i -24j + 0k) . (54i + 34j + 30k).....[assuming that you have not missed a component of C and that it simply is 0]
= (18x54)i - (24x34)j + (0x30)k
= 972i - 816j + 0k
3C.(2A x B) = 972i -816j
This should be the final answer. Hope I helped!

Answer 4

you are quite right \((6×2)−(−14×3)=54\neq56\)

but i am not sure how you can assume the \(\hat k\) direction is the empty direction

Answer 5

well, since Oj has not specified the k component of C, doesn't that mean that it is 0?
Like how in math, we can re-write the polynomial ax^3 + bx^2 + c as
ax^3 + bx^2 + 0x + c. I used the same approach here.