Bài tập 11 trang 96 SGK Hình học 11 NC

a) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\overrightarrow {AB} .\overrightarrow {CD}  = \overrightarrow {AB} \left( {\overrightarrow {AD}  - \overrightarrow {AC} } \right)\\
 = \overrightarrow {AB} .\overrightarrow {AD}  - \overrightarrow {AB} .\overrightarrow {AC} 
\end{array}\\
\begin{array}{l}
 = AB.AD.\cos \widehat {BAD} - AB.AC\\
.\cos \widehat {BAC} = 0
\end{array}\\
{ \Rightarrow AB \bot CD.}
\end{array}\)

b)

Ta có:

\(\begin{array}{*{20}{l}}
{\overrightarrow {IJ}  = \overrightarrow {IA}  + \overrightarrow {AJ} }\\
{ = \frac{1}{2}\overrightarrow {BA}  + \frac{1}{2}\left( {\overrightarrow {AD}  + \overrightarrow {AC} } \right)}\\
{ = \frac{1}{2}\left( {\overrightarrow {AD}  + \overrightarrow {BC} } \right)}\\
{ = \frac{1}{2}\left( {\overrightarrow {AD}  + \overrightarrow {AC}  - \overrightarrow {AB} } \right)}\\
\begin{array}{l}
 \Rightarrow \overrightarrow {AB} .\overrightarrow {IJ} \\
 = \frac{1}{2}\left( {\overrightarrow {AB} .\overrightarrow {AD}  + \overrightarrow {AB}  - \overrightarrow {AC}  - A{B^2}} \right)
\end{array}\\
\begin{array}{l}
 = \frac{1}{2}(AB.AD.\cos {60^0} + AB.AC\\
.\cos {60^0} - A{B^2}) = 0
\end{array}\\
{ \Rightarrow AB \bot IJ}
\end{array}\)

Mặt khác:

\(\begin{array}{*{20}{l}}
{\overrightarrow {CD} .\overrightarrow {IJ}  = \frac{1}{2}\left( {\overrightarrow {CA}  + \overrightarrow {AD} } \right).\left( {\overrightarrow {AD}  + \overrightarrow {BA}  + \overrightarrow {AC} } \right)}\\
\begin{array}{l}
 = \frac{1}{2}( - \overrightarrow {AC} .\overrightarrow {AD}  + {\overrightarrow {AD} ^2} + \overrightarrow {CA} .\overrightarrow {BA} \\
 + \overrightarrow {AD} .\overrightarrow {BA}  - {\overrightarrow {AC} ^2} + \overrightarrow {AD} .\overrightarrow {AC} )
\end{array}\\
{ =  - \frac{1}{2}\overrightarrow {AB} .\left( {\overrightarrow {CA}  + \overrightarrow {AD} } \right) =  - \frac{1}{2}\overrightarrow {AB} .\overrightarrow {CD}  = 0}\\
{ \Rightarrow CD \bot IJ}
\end{array}\)

-- Mod Toán 11 HỌC247