Gọi \(I\) là trung điểm của\(AB\). Khẳng định nào sau đây sai?

Ta có:

$\begin{array}{l}\overrightarrow{AM}.\overrightarrow{SB}=\left(\overrightarrow{SM}-\overrightarrow{SA}\right).\overrightarrow{SB}=\overrightarrow{SM}.\overrightarrow{SB}-\overrightarrow{SA}.\overrightarrow{SB}=SM.SB.\cos \widehat{BSM}-SA.SB.\cos \widehat{ASB}\\ =\frac{a\sqrt{3}}{2}.a.\cos {30}^{°}-a.a.\cos {60}^{°}\\ =\frac{a^2}{4}\end{array}$

Suy ra: $\cos \left(\overrightarrow{AM},\overrightarrow{SB}\right)=\frac{\overrightarrow{AM}.\overrightarrow{SB}}{AM.SB}=\frac{\frac{a^2}{4}}{\frac{a\sqrt{3}}{2}.a}=\frac{\sqrt{3}}{6}\approx 0,29$

Do \(SA \bot \left( {ABCD} \right)\) nên \(SA \bot AB,SA \bot AD\). Do \(ABCD\) là hình chữ nhật nên \(AB \bot AD\).

Do đó, đặt \(\overrightarrow {AS}  = \overrightarrow a ,\overrightarrow {AB}  = \overrightarrow b ,\overrightarrow {AD}  = \overrightarrow c \) thì \(\left| {\overrightarrow a } \right| = 4,\left| {\overrightarrow b } \right| = 1,\left| {\overrightarrow c } \right| = 2\) và \(\overrightarrow a .\overrightarrow b  = \overrightarrow b .\overrightarrow c  = \overrightarrow c .\overrightarrow a  = 0\)

Ta có: \(\cos \left( {\overrightarrow {SC} ,\overrightarrow {DM} } \right) = \frac{{\overrightarrow {SC} .\overrightarrow {DM} }}{{\left| {\overrightarrow {SC} } \right|.\left| {\overrightarrow {DM} } \right|}}\).

Do \(\overrightarrow {SC}  = \overrightarrow {AC}  - \overrightarrow {AS}  = \left( {\overrightarrow {AB}  + \overrightarrow {AD} } \right) - \overrightarrow {AS}  =  - \overrightarrow a  + \overrightarrow b  + \overrightarrow c \) và \(\overrightarrow {DM}  = \overrightarrow {AM}  - \overrightarrow {AD}  = \frac{1}{2}\overrightarrow {AB}  - \overrightarrow {AD}  = \frac{1}{2}\overrightarrow b  - \overrightarrow c \) nên:

\[\overrightarrow {SC} .\overrightarrow {DM}  = \left( { - \overrightarrow a  + \overrightarrow b  + \overrightarrow c } \right)\left( {\frac{1}{2}\overrightarrow b  - \overrightarrow c } \right) =  - \frac{1}{2}\overrightarrow a .\overrightarrow b  + \overrightarrow a .\overrightarrow c  + \frac{1}{2}{\overrightarrow b ^2} - \overrightarrow b .\overrightarrow c  + \frac{1}{2}.\overrightarrow c .\overrightarrow b  - {\overrightarrow c ^2} = \frac{1}{2}{\overrightarrow b ^2} - {\overrightarrow c ^2} = \frac{1}{2}{\left| {\overrightarrow b } \right|^2} - {\left| {\overrightarrow c } \right|^2}\]

\[ = \frac{1}{2}{.1^2} - {2^2} =  - \frac{7}{2}\];

\({\left| {\overrightarrow {SC} } \right|^2} = {\left( { - \overrightarrow a  + \overrightarrow b  + \overrightarrow c } \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} + {\overrightarrow c ^2} - 2\overrightarrow a \overrightarrow b  - 2\overrightarrow a \overrightarrow c  + 2\overrightarrow b \overrightarrow c  = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} = {4^2} + {1^2} + {2^2} +  = 21\)

\({\left| {\overrightarrow {DM} } \right|^2} = {\overrightarrow {DM} ^2} = {\left( {\frac{1}{2}\overrightarrow b  - \overrightarrow c } \right)^2} = \frac{1}{4}{\overrightarrow b ^2} - \overrightarrow b \overrightarrow c  + {\overrightarrow c ^2} = \frac{1}{4}{\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} = \frac{1}{4}.1 + {2^2} = \frac{{17}}{4}\).

Suy ra:

\(\cos \left( {\overrightarrow {SC} ,\overrightarrow {DM} } \right) = \frac{{ - \frac{7}{2}}}{{\sqrt {21} .\sqrt {\frac{{17}}{4}} }} =  - \frac{{\sqrt {357} }}{{51}} \Rightarrow \left( {\overrightarrow {SC} ,\overrightarrow {DM} } \right) \approx 111,75^\circ \).

Vậy góc giữa hai vectơ \(\overrightarrow {SC} \) và \(\overrightarrow {DM} \) là khoảng \(111,75^\circ \).