MCQ
Let $\vec{a}=\vec{i}-\alpha \vec{j}+\beta \hat{k}, \vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ 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 $\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=10$, then $(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}$ is equal to $.....$
  • A
    $8$
  • B
    $5$
  • $9$
  • D
    $1$

Answer

Correct option: C.
$9$
c
$\vec{a}=(1,-\alpha, \beta)$

$\vec{b}=(3, \beta,-\alpha)$

$\vec{c}=(-\alpha,-2,1) ; \alpha, \beta \in I$

$\vec{a} \vec{b}=-1 \Rightarrow 3-\alpha \beta-\alpha \beta=-1$

$\Rightarrow \alpha \beta=2$

$\vec{b} \cdot \vec{c}=10$

$\Rightarrow-3 \alpha-2 \beta-\alpha=10$

$\Rightarrow 2 \alpha+\beta+5=0$

$\therefore \alpha=-2 ; \beta=-1$

$\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\left|\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 1\end{array}\right|$

$=1(-1+4)-2(3-4)-1(-6+2)$

$=3+2+4=9$

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