Cho tam giác ABC vuông tại A. Khẳng định nào sau đây là sai?

Đáp án đúng là D

Do AB ⊥ AC nên $\overrightarrow{\text{AB}}$. $\overrightarrow{\text{AC}}$ = 0.

Ta lại có $\overrightarrow{BA}$. $\overrightarrow{BC}$ = BA.BC.cos$\widehat{\text{B}}$> 0 (vì $\widehat{\text{B}}$ là góc nhọn nên cos$\widehat{\text{B}}$ > 0). Do đó $\overrightarrow{\text{AB}}.\overrightarrow{\text{AC}}<\overrightarrow{\text{BA}}.\overrightarrow{\text{BC}}\text{}$.

Khẳng định A đúng.

$\left(\overrightarrow{\text{AC}},\overrightarrow{\text{CB}}\right)=\left(-\overrightarrow{\text{CA}},\overrightarrow{\text{CB}}\right)=-\left(\overrightarrow{\text{CA}},\overrightarrow{\text{CB}}\right)$ là góc tù nên $\overrightarrow{\text{AC}}.\overrightarrow{\text{CB}}=\overrightarrow{\left|\text{AC}\right|}.\left|\overrightarrow{\text{CB}}\right|.\cos \left(\overrightarrow{\text{AC}}.\overrightarrow{\text{CB}}\right)$ < 0;

$\left(\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}\right)$ là góc nhọn nên $\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}=\left|\overrightarrow{\text{AC}}\right|.\overrightarrow{\left|\text{BC}\right|}.\cos \left(\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}\right)$ > 0. Suy ra $\overrightarrow{\text{AC}}.\overrightarrow{\text{CB}}<\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}\text{}$. Khẳng định B đúng.

$\left(\overrightarrow{\text{AB}},\overrightarrow{\text{BC}}\right)=\left(-\overrightarrow{BA},\overrightarrow{\text{BC}}\right)=-\left(\overrightarrow{BA},\overrightarrow{\text{BC}}\right)$ là góc tù nên $\overrightarrow{\text{AB}}.\overrightarrow{\text{BC}}$ < 0; $\left(\overrightarrow{\text{CA}}.\overrightarrow{\text{CB}}\right)$ là góc nhọn nên $\left(\overrightarrow{\text{CA}}.\overrightarrow{\text{CB}}\right)$ > 0. Suy ra $\overrightarrow{\text{AB}}.\overrightarrow{\text{BC}}<\overrightarrow{\text{CA}}.\overrightarrow{\text{CB}}$. Khẳng định C đúng.

$\left(\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}\right)$ là góc nhọn nên $\left(\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}\right)$ > 0; $\left(\overrightarrow{\text{BC}}.\overrightarrow{\text{AB}}\right)$ là góc tù nên $\left(\overrightarrow{\text{BC}}.\overrightarrow{\text{AB}}\right)$ < 0. Suy ra $\overrightarrow{\text{AC}}.\overrightarrow{\text{BC}}>\overrightarrow{\text{BC}}.\overrightarrow{\text{AB}}$.

Khẳng định D sai.

Vậy chọn đáp án D.