Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
AB \bot AD \Rightarrow \overrightarrow {AB} .\overrightarrow {AD} = 0\\
a,\\
\overrightarrow {AC} .\left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\\
= \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)\left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\\
= \left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\\
= {\overrightarrow {AB} ^2} + 2\overrightarrow {AB} .\overrightarrow {AD} + {\overrightarrow {AD} ^2} = {a^2} + 0 + {a^2} = 2{a^2}\\
b,\\
\left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\left( {\overrightarrow {BD} + \overrightarrow {BC} } \right)\\
= \left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\left( {\overrightarrow {BA} + \overrightarrow {AD} + \overrightarrow {AD} } \right)\\
= \left( {\overrightarrow {AD} + \overrightarrow {AB} } \right)\left( {2\overrightarrow {AD} - \overrightarrow {AB} } \right)\\
= 2{\overrightarrow {AD} ^2} + \overrightarrow {AB} .\overrightarrow {AD} - {\overrightarrow {AB} ^2} = {a^2}\\
c,\\
\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right).\left( {2\overrightarrow {AD} - \overrightarrow {AB} } \right)\\
= \left( {\overrightarrow {AD} + \overrightarrow {AB} - \overrightarrow {AB} } \right)\left( {2\overrightarrow {AD} - \overrightarrow {AB} } \right)\\
= \overrightarrow {AD} .\left( {2\overrightarrow {AD} - \overrightarrow {AB} } \right)\\
= 2{\overrightarrow {AD} ^2} - \overrightarrow {AD} .\overrightarrow {AB} = 2{a^2}
\end{array}\)
