Cho hình vuông ABCD có cạnh = 6a
a) tính độ dài các vecto sau \(\overrightarrow{u}=\overrightarrow{AB}-\overrightarrow{AC}\) ; \(\overrightarrow{v}=\overrightarrow{BC}+\overrightarrow{BD}\)
b) tính các tích vô hương sau : \(\overrightarrow{AB}.\overrightarrow{AC}\); \(\overrightarrow{BD}.\overrightarrow{AC}\);\(\overrightarrow{AB}.\overrightarrow{CD}\)
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: \(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=6a\cdot6a\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}\)
\(=36a^2\)
