Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ ．If for some real nu

Question

Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ ．If for some real numbers $\alpha, \beta$ and $\gamma$ ，we have\ $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} 	imes \vec{q})$ ，then the value of $\gamma$ is $\_\_\_\_$ ．\

Professor PC Thomas Classes · Answer

(2) : $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+ \beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} 	imes \vec{q}) $\ Taking dot product with $\vec{p} 	imes \vec{q}$, we have\ $\left|\begin{array}{ccc}15 & 10 & 6 \ 2 & 1 & 3 \ 1 & -1 & 1\end{array}ight|=0+0+\gamma\left\{p^{2} q^{2}-(\vec{p} \cdot \vec{q})^{2}ight\}$\ $\Rightarrow \quad 52=\gamma(14 \cdot 3-16)=26 \gamma$\ $	herefore \quad \gamma=2$\