Bài tập 2.62 trang 105 SBT Hình học 10

a) 

\(\begin{array}{l}
\overrightarrow {AB} .\overrightarrow {AC}  = AB.AC.\cos A = 4.6.\frac{1}{2} = 12\\
\overrightarrow {AB} .\overrightarrow {BC}  = \overrightarrow {AB} .\left( {\overrightarrow {AC}  - \overrightarrow {AB} } \right) = \overrightarrow {AB} .\overrightarrow {AC}  - A{B^2} = 12 - 16 =  - 4\\
B{C^2} = {\left( {\overrightarrow {AC}  - \overrightarrow {AB} } \right)^2} = A{C^2} - 2\overrightarrow {AB} .\overrightarrow {AC}  + A{B^2}\\
 = 36 - 2.12 + 16 = 28 \Rightarrow BC = 2\sqrt 7 \\
R = \frac{{BC}}{{2\sin A}} = \frac{{2\sqrt 7 }}{{2.\frac{{\sqrt 3 }}{2}}} = \frac{{2\sqrt {21} }}{3}
\end{array}\)

b) 

\(\begin{array}{l}
2\overrightarrow {AM}  + 3\overrightarrow {MC}  = \overrightarrow 0  \Leftrightarrow 2\overrightarrow {AM}  + 3\left( {\overrightarrow {AC}  - \overrightarrow {AM} } \right) = 0\\
 \Rightarrow \overrightarrow {AM}  = 3\overrightarrow {AC}  \Rightarrow \overrightarrow {AB}  + \overrightarrow {BM}  = 3\overrightarrow {AC} \\
 \Rightarrow \overrightarrow {BM}  = 3\overrightarrow {AC}  - \overrightarrow {AB} 
\end{array}\)

\(\begin{array}{l}
\overrightarrow {NB}  + x\overrightarrow {NC}  = \overrightarrow 0  \Rightarrow \overrightarrow {AB}  - \overrightarrow {AN}  + x\left( {\overrightarrow {AC}  - \overrightarrow {AN} } \right) = \overrightarrow 0 \\
 \Rightarrow \overrightarrow {AN}  = \frac{1}{{x + 1}}\left( {\overrightarrow {AB}  + x\overrightarrow {AC} } \right)
\end{array}\)

\(\begin{array}{l}
AN \bot BM \Leftrightarrow \overrightarrow {AN} .\overrightarrow {BM}  = 0\\
 \Leftrightarrow \left( {\overrightarrow {AB}  + x\overrightarrow {AC} } \right)\left( {3\overrightarrow {AC}  - \overrightarrow {AB} } \right) = 0\\
 \Leftrightarrow \left( {3 - x} \right)\overrightarrow {AB} .\overrightarrow {AC}  - A{B^2} + 3xA{C^2} = 0\\
 \Leftrightarrow \left( {3 - x} \right).12 - 16 + 3x.36 = 0\\
 \Leftrightarrow 96x + 20 = 0 \Leftrightarrow x =  - \frac{5}{{24}}
\end{array}\)

-- Mod Toán 10 HỌC247