- A$8$
- B$5$
- ✓$9$
- D$1$
$\vec{b}=(3, \beta,-\alpha)$
$\vec{c}=(-\alpha,-2,1) ; \alpha, \beta \in I$
$\vec{a} \vec{b}=-1 \Rightarrow 3-\alpha \beta-\alpha \beta=-1$
$\Rightarrow \alpha \beta=2$
$\vec{b} \cdot \vec{c}=10$
$\Rightarrow-3 \alpha-2 \beta-\alpha=10$
$\Rightarrow 2 \alpha+\beta+5=0$
$\therefore \alpha=-2 ; \beta=-1$
$\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\left|\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 1\end{array}\right|$
$=1(-1+4)-2(3-4)-1(-6+2)$
$=3+2+4=9$
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$f(x)=\max \{\sin t: 0 \leq t \leq x\}, \quad 0 \leq x \leq \pi$
$\quad \quad \quad \quad \quad \quad 2+\cos x,\quad \quad \quad \quad x>\pi$
Then which of the following is true?