- A$15$
- B$9$
- ✓$12$
- D$6$
$\overrightarrow{ b } \cdot \overrightarrow{ c }=12$
$\overrightarrow{ a } \cdot \overrightarrow{ c }=\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c })-\overrightarrow{ n } \cdot \overrightarrow{ c }$
$\overrightarrow{ a } \cdot \overrightarrow{ c }=\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c })$
$|\overrightarrow{ c } \times(\overrightarrow{ a } \times \overrightarrow{ b })|=|(\overrightarrow{ c } \cdot \overrightarrow{ b }) \overrightarrow{ a }-(\overrightarrow{ c } \cdot \overrightarrow{ a }) \overrightarrow{ b }|$
$=|(\overrightarrow{ c } \cdot \overrightarrow{ b }) \overrightarrow{ a }-\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c }) \overrightarrow{ b }|$
$=|(\overrightarrow{ c } \cdot \overrightarrow{ b })||\overrightarrow{ a }-\alpha \overrightarrow{ b }|$
$=12 \times(|\overrightarrow{ n }|)$
$=12 \times 1$
$=12$
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$\left| {\begin{array}{*{20}{c}}
{\left[ \pi \right]}&{amp(1 + i\sqrt 3 )}&1 \\
1&0&2 \\
{\operatorname{sgn} ({{\cot }^{ - 1}}x)}&1&{\{ \pi \} }
\end{array}} \right|$ is-