Cho tam giác \(MNP\) có \(M\left( { - 1;3} \right),N\left( {2;2} \right),P\left( { - 1;1} \right).\) Trọng tâm \(G\) của tam giác \(MNP\) có tọa độ là 

a) Đúng.

b) Đúng.

Ta có: \(\overrightarrow {A'A}  + \overrightarrow {A'B'}  - \overrightarrow {CM}  = \overrightarrow {A'A}  + \overrightarrow {AB}  + \overrightarrow {BM}  = \overrightarrow {A'B}  + \overrightarrow {BM}  = \overrightarrow {A'M} \)

c) Sai.

Ta có: \(\overrightarrow {A'M} .\overrightarrow {AC}  = \left( {\overrightarrow {A'A}  + \overrightarrow {AM} } \right).\overrightarrow {AC}  = \overrightarrow {A'A} .\overrightarrow {AC}  + \overrightarrow {AM} .\overrightarrow {AC}  = \overrightarrow {AM} .\overrightarrow {AC}  = \frac{{a\sqrt 3 }}{2}.a.\cos 30^\circ  = \frac{{3{a^2}}}{4}\)

d) Đúng.

Ta có \(\overrightarrow {AB'} .\overrightarrow {BC'}  = \left( {\overrightarrow {AB}  + \overrightarrow {BB'} } \right)\left( {\overrightarrow {BC}  + \overrightarrow {CC'} } \right)\)\( = \overrightarrow {AB} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CC'}  + \overrightarrow {BB'} .\overrightarrow {BC}  + \overrightarrow {BB'} .\overrightarrow {CC'} \)

\( = \overrightarrow {AB} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CC'}  + \overrightarrow {BB'} .\overrightarrow {BC}  + \overrightarrow {BB'} .\overrightarrow {CC'} \)\( =  - \frac{{{a^2}}}{2} + 0 + 0 + 2{a^2} = \frac{{3{a^2}}}{2}\)

Suy ra \(\cos \left( {\overrightarrow {AB'} ,\overrightarrow {BC'} } \right) = \frac{{\overrightarrow {AB'} .\overrightarrow {BC'} }}{{\left| {\overrightarrow {AB'} } \right|.\left| {\overrightarrow {BC'} } \right|}}\)\( = \frac{{\frac{{3{a^2}}}{2}}}{{a\sqrt 3 .a\sqrt 3 }} = \frac{1}{2} \Rightarrow \left( {\overrightarrow {AB'} ,\overrightarrow {BC'} } \right) = 60^\circ \)

      Cạnh hình lập phương là \(a \Rightarrow A'D\, = \,a\sqrt 2 .\)

      \[\begin{array}{l}\overrightarrow {IG}  = \overrightarrow {IB'}  + \overrightarrow {B'G} \, = \frac{1}{2}\left( {\overrightarrow {AA'}  + \overrightarrow {AB} } \right) + \frac{1}{3}\left( {\overrightarrow {B'D'}  + \overrightarrow {B'C'} } \right) = \frac{1}{2}\left( {\overrightarrow {AA'}  + \overrightarrow {AB} } \right) + \frac{1}{3}\left( {\overrightarrow {BD}  + \overrightarrow {AD} } \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\overrightarrow {AA'}  + \overrightarrow {AB} } \right) + \frac{1}{3}\left( {2\overrightarrow {AD}  - \overrightarrow {AB} } \right) = \frac{1}{2}\overrightarrow {AA'}  + \frac{1}{6}\overrightarrow {AB}  + \frac{2}{3}\overrightarrow {AD} \\{\overrightarrow {IG} ^2} = {\left( {\frac{1}{2}\overrightarrow {AA'}  + \frac{1}{6}\overrightarrow {AB}  + \frac{2}{3}\overrightarrow {AD} } \right)^2} = \frac{{13{a^2}}}{{18}} \Rightarrow IG = \frac{{a\sqrt {26} }}{6}.\end{array}\]

      Ta có: \(\overrightarrow {A'D}  = \overrightarrow {AD}  - \overrightarrow {AA'} \) và \(A'D = a\sqrt 2 \) .

\[\begin{array}{l}\overrightarrow {A'D} .\overrightarrow {IG}  = \left( {\frac{1}{2}\overrightarrow {AA'}  + \frac{1}{6}\overrightarrow {AB}  + \frac{2}{3}\overrightarrow {AD} } \right).\left( {\overrightarrow {AD}  - \overrightarrow {AA'} } \right) = \frac{{{a^2}}}{6}.\\cos\left( {\overrightarrow {A'D} \,,\,\overrightarrow {IG} } \right) = \frac{{\overrightarrow {A'D} .\overrightarrow {IG} }}{{A'D.IG}} = \frac{{\frac{{{a^2}}}{6}}}{{\frac{{a\sqrt {26} }}{6}.a\sqrt 2 }} = \frac{{\sqrt {13} }}{{26}}.\end{array}\]