a) \(A\in\left(ABC\right)\Rightarrow d\left(A;\left(ABC\right)\right)=0\)
b) Chọn hệ trục tọa độ \(Oxyz:\)
\(A\left(0;0;0\right);B\left(a;0;0\right);C\left(\dfrac{a}{2};\dfrac{a\sqrt{3}}{2};0\right);A'\left(0;0;a\sqrt{2}\right);B'\left(a;0;a\sqrt{2}\right)\)
\(I\) là trung điểm \(BB'\Rightarrow I\left(a;0;\dfrac{a\sqrt{2}}{2}\right)\)
\(\overrightarrow{CI}=\left(\dfrac{a}{2};-\dfrac{a\sqrt{3}}{2};\dfrac{a\sqrt{2}}{2}\right);\overrightarrow{IA'}=\left(-a;0;\dfrac{a\sqrt{2}}{2}\right)\)
\(\Rightarrow\left(CIA'\right):\overrightarrow{n_1}=\left[\overrightarrow{CI}.\overrightarrow{IA'}\right]=\left(-\dfrac{a^2\sqrt{6}}{4};-\dfrac{3a^2\sqrt{2}}{4};\dfrac{a^2\sqrt{3}}{2}\right)\)
\(\overrightarrow{BI}=\left(0;0;\dfrac{a\sqrt{2}}{2}\right)\)
\(\Rightarrow\left(BIA'\right):\overrightarrow{n_2}=\left[\overrightarrow{BI}.\overrightarrow{IA'}\right]=\left(0;-\dfrac{a^2\sqrt{2}}{2};0\right)\)
\(cos\left[C;IA';B\right]=\dfrac{\left|\overrightarrow{n_1}.\overrightarrow{n_2}\right|}{\left|\overrightarrow{n_1}\right|.\left|\overrightarrow{n_2}\right|}=\dfrac{\dfrac{3a^2}{4}}{\left(\dfrac{3a^2\sqrt{3}}{4}\right).\left(\dfrac{a^2\sqrt{2}}{2}\right)}=\dfrac{1}{2}\)
\(\Rightarrow Sđ\left[C;IA';B\right]=60^o\)
