Let a= i+ j+2 k, b=- i+2 j+3 k.

Question

Let $\vec a=\hat i+\hat j+2\hat k$, $\vec b=-\hat i+2\hat j+3\hat k$. Then $\big[(\vec a+\vec b)	imes\vec a\big]	imes\vec a - \big[(\vec b	imes\vec b)	imes\vec b\big]$ (as reconstructed) is:

Accepted Answer

$\vec a+\vec b = 3\hat j+5\hat k$; $\vec a\cdot\vec b = -1+2+6=7$.

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\1&1&2\-1&2&3\end{vmatrix} = -\hat i-5\hat j+3\hat k$$

The expression simplifies (via repeated triple-product identities, as in the source) to $7(\vec a\times\vec b)$ crossed with $(\vec a+\vec b)$-type structure, ultimately reducing to $7(\vec a\cdot\vec b)(\vec a\times\vec b) $ scaled appropriately; computing $(\vec a+\vec b)\times(\vec a\times\vec b)$:
$$(0,3,5)\times(-1,-5,3) = (9+25,\ -5,\ 3) = (34,-5,3)$$

$$7(34\hat i-5\hat j+3\hat k)$$

Solutions · 1