Lời giải

a) Sai b) Đúng c) Sai d) Đúng

Ta có: \(\overrightarrow {AB} \cdot \overrightarrow {AC} = AB \cdot AC \cdot \cos A = 4\sqrt 2 \cdot 6 \cdot \cos 45^\circ = 24\).

Ta có: \(\overrightarrow {BC} = \overrightarrow {AC} - \overrightarrow {AB} ,\,\,\overrightarrow {AD} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \).

Khi đó:

\(\begin{array}{*{20}{l}}{{{\overrightarrow {BC} }^2}}&{ = {{\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)}^2} = {{\overrightarrow {AC} }^2} - 2\overrightarrow {AC} \cdot \overrightarrow {AB} + {{\overrightarrow {AB} }^2} = {6^2} - 2 \cdot 24 + {{\left( {4\sqrt 2 } \right)}^2} = 20}\\{}&{ \Rightarrow BC = 2\sqrt 5 .}\\{{{\overrightarrow {AD} }^2}}&{ = {{\left( {\frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} } \right)}^2} = \frac{1}{4}\left( {{{\overrightarrow {AB} }^2} + 2\overrightarrow {AB} \cdot \overrightarrow {AC} + {{\overrightarrow {AC} }^2}} \right)}\\{}&{ = \frac{1}{4}\left[ {{{\left( {4\sqrt 2 } \right)}^2} + 2 \cdot 24 + {6^2}} \right] = 29 \Rightarrow AD = \sqrt {29} .}\end{array}\)

Ta có: \(\overrightarrow {BE} = \overrightarrow {AE} - \overrightarrow {AB} = k\overrightarrow {AC} - \overrightarrow {AB} \). Từ đó, ta có:

\(\overrightarrow {AD} \cdot \overrightarrow {BE} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) \cdot \left( {k\overrightarrow {AC} - \overrightarrow {AB} } \right)\)

\(\begin{array}{l} = \frac{1}{2}\left( {k\overrightarrow {AB} \cdot \overrightarrow {AC} + k{{\overrightarrow {AC} }^2} - {{\overrightarrow {AB} }^2} - \overrightarrow {AB} \cdot \overrightarrow {AC} } \right) = \frac{1}{2}\left[ {24k + {6^2} \cdot k - {{\left( {4\sqrt 2 } \right)}^2} - 24} \right]\\ = 30k - 28.\end{array}\)

Khi đó \(AD \bot BE \Leftrightarrow \overrightarrow {AD} \cdot \overrightarrow {BE} = 0 \Leftrightarrow 30k - 28 = 0 \Leftrightarrow k = \frac{{14}}{{15}}\).

Cho tam giác ABC thoả mãn: b^2 + c^2 - a^2 = căn bậc hai của 3 bc. Khi đó: