Let $\vec{p}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$. If for real $\alpha,\beta,\gamma$:
$$15\hat{i}+10\hat{j}+6\hat{k}=\alpha(2\vec{p}+\vec{q})+\beta(\vec{p}-2\vec{q})+\gamma(\vec{p}\times\vec{q})$$
then $\gamma$ equals:
Step-by-Step Solution
Key Concept: $2\vec{p}+\vec{q}=(5,1,7)$, $\vec{p}-2\vec{q}=(0,3,1)$, $\vec{p}\times\vec{q}=\det[\hat{i},\hat{j},\hat{k};2,1,3;1,-1,1]=(4,-1,-3)$. Solve the $3\times 3$ system for $\alpha,\beta,\gamma$.
$\gamma=\mathbf{2}$.
Correct Answer: 2