Giải thích các bước giải:
Giả sử $AB=BC=AC=a$
Ta có:
$\overrightarrow {AP} = \overrightarrow {AB} + \overrightarrow {BP} = \overrightarrow {AB} + x\overrightarrow {BC} $
Và $\overrightarrow {MN} = \dfrac{5}{6}\overrightarrow {AB} - \overrightarrow {AC} = \dfrac{5}{6}\overrightarrow {AB} - \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) = \dfrac{{ - 1}}{6}\overrightarrow {AB} - \overrightarrow {BC} $
Như vậy:
$\begin{array}{l}
\overrightarrow {AP} .\overrightarrow {MN} = \left( {\overrightarrow {AB} + x\overrightarrow {BC} } \right)\left( {\dfrac{{ - 1}}{6}\overrightarrow {AB} - \overrightarrow {BC} } \right)\\
= \dfrac{{ - 1}}{6}\overrightarrow {AB} .\overrightarrow {AB} - \left( {1 + \dfrac{x}{6}} \right)\overrightarrow {AB} .\overrightarrow {BC} - x\overrightarrow {BC} .\overrightarrow {BC} \\
= \dfrac{{ - 1}}{6}A{B^2} + \left( {1 + \dfrac{x}{6}} \right)\overrightarrow {BA} .\overrightarrow {BC} - xB{C^2}\\
= \dfrac{{ - 1}}{6}{a^2} + \left( {1 + \dfrac{x}{6}} \right).{a^2}.\cos {60^0} - x.{a^2}\\
= \dfrac{{ - 1}}{6}{a^2} + \dfrac{1}{2}\left( {1 + \dfrac{x}{6}} \right){a^2} - x{a^2}\\
= {a^2}\left( {\dfrac{{ - 1}}{6} + \dfrac{1}{2} + \dfrac{x}{{12}} - x} \right)\\
= {a^2}\left( {\dfrac{{ - 11}}{{12}}x + \dfrac{1}{3}} \right)
\end{array}$
Để $AP \bot MN \Leftrightarrow \overrightarrow {AP} .\overrightarrow {MN} = 0$
$\begin{array}{l}
\Leftrightarrow \dfrac{{ - 11}}{{12}}x + \dfrac{1}{3} = 0\\
\Leftrightarrow x = \dfrac{4}{{11}}
\end{array}$
Vậy $x = \dfrac{4}{{11}}$ thỏa mãn đề
