Đáp án:
\[\begin{array}{l}
\left| {\overrightarrow {AC} } \right| = 2a.\\
\left| {\overrightarrow {AI} } \right| = a\\
\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = a\sqrt 7 .\\
\left| {\overrightarrow {AB} - \overrightarrow {AC} } \right| = a.
\end{array}\]
Giải thích các bước giải:
\[\begin{array}{l}
\left| {\overrightarrow {AC} } \right| = AC = \frac{{BC}}{{\sin {{30}^0}}} = \frac{a}{{\frac{1}{2}}} = 2a.\\
\Rightarrow AB = \sqrt {B{C^2} - A{C^2}} = \sqrt {4{a^2} - {a^2}} = a\sqrt 3 .\\
\left| {\overrightarrow {AI} } \right| = AI = \frac{1}{2}AC = a.\\
{\left( {\left| {\overrightarrow {AB} + \overrightarrow {AC} } \right|} \right)^2} = A{B^2} + 2\overrightarrow {AB} .\overrightarrow {AC} + A{C^2}\\
= B{C^2} + 2.\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\,\,\overrightarrow {AC} } \right) = 4{a^2} + 2a.a\sqrt 3 .\cos {30^0}\\
= 4{a^2} + 3{a^2} = 7{a^2}.\\
\Rightarrow \left| {\overrightarrow {AB} + \overrightarrow {AC} } \right| = a\sqrt 7 .\\
{\left( {\left| {\overrightarrow {AB} - \overrightarrow {AC} } \right|} \right)^2} = A{B^2} - 2\overrightarrow {AB} .\overrightarrow {AC} + A{C^2}\\
= B{C^2} - 2.\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\,\,\overrightarrow {AC} } \right) = 4{a^2} - 2a.a\sqrt 3 .\cos {30^0}\\
= 4{a^2} - 3{a^2} = {a^2}.\\
\Rightarrow \left| {\overrightarrow {AB} - \overrightarrow {AC} } \right| = a.
\end{array}\]
