Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}, \vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ a

Question

Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}, \vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$ ，where $\alpha$ and $\beta$ are integers．If $\vec{a} \cdot \vec{b}=-1$ and $\vec{b} \cdot \vec{c}=10$ ，then $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$ is equal to\

Professor PC Thomas Classes · Answer

(9) \ $\Rightarrow-3 \alpha-2 \beta-\alpha=10$\ $\Rightarrow 4 \alpha+2 \beta+10=0$\ $\Rightarrow 4 \alpha^{2}+4+10 \alpha=0$\ (Using (i))\ $\Rightarrow \quad(\alpha+2)(4 \alpha+2)=0$\ $\Rightarrow \quad \alpha=-2, \alpha=\frac{-1}{2}$\ (Not possible, as $\alpha \in Z$ )\ ∴ From (i), $\beta=-1$\ Now, $\vec{a}=\hat{i}+2 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,\ $\vec{c}=2 \hat{i}-2 \hat{j}+\hat{k}$\ $	herefore(\vec{a} 	imes \vec{b}) \cdot \vec{c}=\left|\begin{array}{ccc}1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 1\end{array}ight|$\ $=1(-1+4)-2(3-4)-1(-6+2)$\ $=3+2+4=9$\