Đề thi thử THPT môn Toán lần 1 năm 2026 liên trường THPT Nghệ An

a) Sai. \(BO = \frac{{\sqrt 3 }}{2}.6 = 3\sqrt 3  \Rightarrow B\left( {3\sqrt 3 ;0;0} \right)\);

\(AO = CO = \frac{{AC}}{2} = \frac{6}{2} = 3 \Rightarrow \left\{ \begin{array}{l}C\left( {0;3;0} \right)\\A\left( {0; - 3;0} \right)\end{array} \right.\)

\(M = \left( {\frac{{3\sqrt 3  + 0}}{2};\frac{{0 + 3}}{2};\frac{{0 + 0}}{2}} \right) = \left( {\frac{{3\sqrt 3 }}{2};\frac{3}{2};0} \right)\).

\(A'O = \sqrt {A'{A^2} - A{O^2}}  = \sqrt {{5^2} - {3^2}}  = 4 \Rightarrow A'\left( {0;0;4} \right)\).

\(\overrightarrow {A'E}  = \frac{1}{2}\overrightarrow {AB}  = \left( {\frac{{3\sqrt 3 }}{2};\frac{3}{2};0} \right) \Rightarrow E\left( {\frac{{3\sqrt 3 }}{2};\frac{3}{2};4} \right)\).

\(\overrightarrow {ME}  = \left( {\frac{{3\sqrt 3 }}{2} - \frac{{3\sqrt 3 }}{2};\frac{3}{2} - \frac{3}{2};4 - 0} \right) = \left( {0;0;4} \right)\).

b) Đúng. \(BO = \frac{{\sqrt 3 }}{2}.6 = 3\sqrt 3  \Rightarrow B\left( {3\sqrt 3 ;0;0} \right)\).

c) Sai. \(\overrightarrow {A'B}  = \left( {3\sqrt 3 ;0; - 4} \right) \Rightarrow A'B = \sqrt {43} \); \(\overrightarrow {BC}  = \left( { - 3\sqrt 3 ;3;0} \right)\).

\(\cos \left( {\overrightarrow {A'B} ,\overrightarrow {BC} } \right) = \frac{{3\sqrt 3 .\left( { - 3\sqrt 3 } \right) + 0.3 - 4.0}}{{\sqrt {43} .6}} = \frac{{ - 27}}{{6\sqrt {43} }} < 0\).

d) Đúng. \(\overrightarrow {AA'}  = (0;3;4)\), \(\overrightarrow {AM}  = \left( {\frac{{3\sqrt 3 }}{2};\frac{9}{2};0} \right)\).

\(d\left( {ME,AA'} \right) = \frac{{\left| {\left[ {\overrightarrow {AA'} ,\overrightarrow {ME} } \right].\overrightarrow {AM} } \right|}}{{\left| {\left[ {\overrightarrow {AA'} ,\overrightarrow {ME} } \right]} \right|}} = \frac{{3\sqrt 3 }}{2}\).