a)cho tam giác ABC ,AD BE CF là các đường trung tuyến CMR:vtoAD.vtoBC +vtoCA.vtoBE+vtoAB .vtoCF=0 b)sin a+cos a=1/3 tính P=căn tan^2 a+cot ^2 a

Đáp án:

$\begin{array}{l}
a)\overrightarrow {AD} .\overrightarrow {BC}  + \overrightarrow {CA} .\overrightarrow {BE}  + \overrightarrow {AB} .\overrightarrow {CF} \\
 = \overrightarrow {BC} .\frac{1}{2}.\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right) + \overrightarrow {CA} .\frac{1}{2}.\left( {\overrightarrow {BC}  + \overrightarrow {BA} } \right)\\
 + \overrightarrow {AB} .\frac{1}{2}.\left( {\overrightarrow {CB}  + \overrightarrow {CA} } \right)\\
 = \frac{1}{2}\left( \begin{array}{l}
.\overrightarrow {BC} .\overrightarrow {AB}  + \overrightarrow {BC} .\overrightarrow {AC}  + \overrightarrow {CA} .\overrightarrow {BC}  + \\
\overrightarrow {CA} .\overrightarrow {BA}  + \overrightarrow {AB} .\overrightarrow {CB}  + \overrightarrow {AB} .\overrightarrow {CA} 
\end{array} \right)\\
 = \frac{1}{2}.\overrightarrow 0  = \overrightarrow 0 \\
b)\sin a + \cos a = \frac{1}{3}\\
 \Rightarrow \sqrt 2 \sin \left( {a + \frac{\pi }{4}} \right) = \frac{1}{3}\\
 \Rightarrow \sin \left( {a + \frac{\pi }{4}} \right) = \frac{1}{{3\sqrt 2 }}\\
 \Rightarrow a = \arcsin \frac{1}{{3\sqrt 2 }} - \frac{\pi }{4}\\
 \Rightarrow P = \sqrt {{{\tan }^2}a + {{\cot }^2}a}  = \frac{{49}}{{16}}
\end{array}$