- A$\frac{\pi }{3}$
- ✓$\frac{2\pi }{3}$
- C$\frac{5\pi }{6}$
- D$\frac{\pi }{6}$
${\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right]^2} = \left| {\begin{array}{*{20}{c}}
1&{\cos \theta }&{\cos \theta }\\
{\cos \theta }&1&{\cos \theta }\\
{\cos \theta }&{\cos \theta }&1
\end{array}} \right|$
${\left[ {\begin{array}{*{20}{l}}
{\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c }
\end{array}} \right]^2} = {(1 - \cos \theta )^2}(1 + 2\cos \theta ) \ge 0$
${\cos \theta \geq-\frac{1}{2}} $
${\theta=\frac{2 \pi}{3}}$
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If $\text{P}(\text{A})=0.4,\text{P}(\text{B})=0.8$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$ then $\text{P}(\text{A}\cup\text{B})$ is equal to:
$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$