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a: AB=BC=CD=DA=6a
\(AC=BD=\sqrt{\left(6a\right)^2+\left(6a\right)^2}=6a\sqrt{2}\)
\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=6a\)
\(\left|\overrightarrow{BC}+\overrightarrow{BD}\right|=\sqrt{BC^2+BD^2+2\cdot BC\cdot BD\cdot cos45}\)
\(=\sqrt{36a^2+72a^2+\sqrt{2}\cdot6a\cdot6a\sqrt{2}}\)
\(=6a\sqrt{5}\)
b: \(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=6a\cdot6a\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}\)
\(=36a^2\)
\(AC=\sqrt{AB^2+BC^2}=a\sqrt{5}\)
\(BD=\sqrt{AD^2+AB^2}=a\sqrt{2}\)
\(\overrightarrow{AC}.\overrightarrow{BD}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\left(\overrightarrow{BA}+\overrightarrow{AD}\right)\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AB}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BA}+\overrightarrow{BC}.\overrightarrow{AD}\)
\(=-\overrightarrow{AB}^2+\overrightarrow{AD}.2\overrightarrow{AD}=-\overrightarrow{AB}^2+2\overrightarrow{AD}^2\)
\(=-a^2+2a^2=a^2\)
\(cos\left(\overrightarrow{AC};\overrightarrow{BD}\right)=\dfrac{\overrightarrow{AC}.\overrightarrow{BD}}{AC.BD}=\dfrac{a^2}{a\sqrt{2}.a\sqrt{5}}=\dfrac{1}{\sqrt{10}}\)
