Đáp án:
a) $- 3{a^2}$
b) $- 2{a^2}$
Giải thích các bước giải:
$\begin{array}{l}
a)\overrightarrow {AC} .\left( {\overrightarrow {CD} + \overrightarrow {CA} } \right) = \left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\left( { - \overrightarrow {AB} - \overrightarrow {AB} - \overrightarrow {AD} } \right)\\
= \left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\left( { - 2\overrightarrow {AB} - \overrightarrow {AD} } \right) = - 2A{B^2} - 2\overrightarrow {AD} .\overrightarrow {AB} - \overrightarrow {AD} .\overrightarrow {AB} - A{D^2}\\
= - 2{a^2} - 2.0 - 0 - {a^2} = - 3{a^2}\\
b)\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\left( {\overrightarrow {BC} + \overrightarrow {BD} + \overrightarrow {BA} } \right)\\
= \left( {\overrightarrow {AB} + \overrightarrow {AB} + \overrightarrow {AD} } \right)\left( {\overrightarrow {AD} + \overrightarrow {BA} + \overrightarrow {BC} + \overrightarrow {BA} } \right)\\
= \left( {2\overrightarrow {AB} + \overrightarrow {AD} } \right)\left( {\overrightarrow {AD} + 2\overrightarrow {BA} + \overrightarrow {AD} } \right)\\
= \left( {2\overrightarrow {AB} + \overrightarrow {AD} } \right)\left( {2\overrightarrow {AD} - 2\overrightarrow {AB} } \right)\\
= 2\overrightarrow {AB} .\overrightarrow {AD} + 2A{D^2} - 4A{B^2} - 2\overrightarrow {AD} .\overrightarrow {AB} \\
= 2.0 + 2.{a^2} - 4.{a^2} - 2.0\\
= - 2{a^2}
\end{array}$
