Đáp án:
$P=-4a^2$
Lời giải chi tiết:
$\begin{array}{l} P=\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\left( {\overrightarrow {BC} + \overrightarrow {BD} + \overrightarrow {BA} } \right)\\ = \overrightarrow {AB} .\overrightarrow {BC} + \overrightarrow {AC} .\overrightarrow {BC} + \overrightarrow {AB} .\overrightarrow {BD} + \overrightarrow {AC} .\overrightarrow {BD} + \overrightarrow {AB} .\overrightarrow {BA} + \overrightarrow {AC} .\overrightarrow {BA} \\ = 0 + AC.BC.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right) + AB.BD.\cos \left( {\overrightarrow {AB} ,\overrightarrow {BD} } \right) + 0 + AB.BA.\cos \left( {\overrightarrow {AB} ,\overrightarrow {BA} } \right) + AC.BA.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BA} } \right)\\ = a\sqrt 2 .a.\cos {135^0} + a.a\sqrt 2 .\cos {135^0} + a.a.\cos {180^0} + a\sqrt 2 .a{.135^0}\\ = {a^2}\sqrt 2 .\left( { - \frac{{\sqrt 2 }}{2}} \right) + {a^2}\sqrt 2 .\left( { - \frac{{\sqrt 2 }}{2}} \right) + {a^2}.\left( { - 1} \right) + {a^2}\sqrt 2 .\left( { - \frac{{\sqrt 2 }}{2}} \right)\\ = - 4{a^2} \end{array}$
