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Câu hỏi: Cho tam giác \(ABC\) vuông tại \(A\). Khẳng định nào dưới đây là sai?
A. \(\overrightarrow {AB} .\overrightarrow {AC} < \overrightarrow {BA} .\overrightarrow {BC} \)
B. \(\overrightarrow {AC} .\overrightarrow {CB} < \overrightarrow {AC} .\overrightarrow {BC} \)
C. \(\overrightarrow {AB} .\overrightarrow {BC} < \overrightarrow {CA} .\overrightarrow {CB} \)
D. \(\overrightarrow {AC} .\overrightarrow {BC} < \overrightarrow {BC} .\overrightarrow {AB} \)
A. \(\overrightarrow {AB} .\overrightarrow {AC} < \overrightarrow {BA} .\overrightarrow {BC} \)
B. \(\overrightarrow {AC} .\overrightarrow {CB} < \overrightarrow {AC} .\overrightarrow {BC} \)
C. \(\overrightarrow {AB} .\overrightarrow {BC} < \overrightarrow {CA} .\overrightarrow {CB} \)
D. \(\overrightarrow {AC} .\overrightarrow {BC} < \overrightarrow {BC} .\overrightarrow {AB} \)
Lời giải chi tiết
$+\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{AC}}$ nên $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}=0$
$(\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ABC}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=\mathrm{BA.BC} \cdot \cos (\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}<\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}$
$+(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}})=180^{\circ}-\widehat{\mathrm{ACB}}>90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}})<0 \Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{CB}}<0$
$(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>0$
Do đó $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{CB}}<\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}$
$+\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}=0$
$(\overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{CB}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{CB}})>0$
$\Rightarrow \overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}>0$
Do đó $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}<\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}$
$+(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>0$
$(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{AB}})=180^{\circ}-\widehat{\mathrm{ABC}}>90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{AB}})<0$
$\Rightarrow \overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{AB}}<0$
Do đó $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>\overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{AB}}$
$+\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{AC}}$ nên $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}=0$
$(\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ABC}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=\mathrm{BA.BC} \cdot \cos (\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}<\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}$
$+(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}})=180^{\circ}-\widehat{\mathrm{ACB}}>90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}})<0 \Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{CB}}<0$
$(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>0$
Do đó $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{CB}}<\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}$
$+\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}=0$
$(\overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{CB}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{CB}})>0$
$\Rightarrow \overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}>0$
Do đó $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}<\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}$
$+(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})=\widehat{\mathrm{ACB}}<90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{BC}})>0$
$\Rightarrow \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>0$
$(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{AB}})=180^{\circ}-\widehat{\mathrm{ABC}}>90^{\circ}$
$\Rightarrow \cos (\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{AB}})<0$
$\Rightarrow \overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{AB}}<0$
Do đó $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BC}}>\overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{AB}}$
Đáp án D.