Câu hỏi:

Ta có $\overrightarrow{CA}.\overrightarrow{BC} = CA.BC.cos(\overrightarrow{CA},\overrightarrow{BC}) = CA.BC.cos(\widehat{C})$.
Trong tam giác $ABC$ vuông tại $A$, ta có:
$BC = \sqrt{AB^2 + AC^2} = \sqrt{2^2 + 5^2} = \sqrt{4+25} = \sqrt{29}$.
$cos(\widehat{C}) = \frac{AC}{BC} = \frac{5}{\sqrt{29}}$.
Do đó, $\overrightarrow{CA}.\overrightarrow{BC} = 5.\sqrt{29}.\frac{5}{\sqrt{29}} = 25$.
Vì $\overrightarrow{CA}.\overrightarrow{BC} = |CA|.|BC|.cos(\widehat{(CA,BC)})$,
Suy ra $\overrightarrow{CA}.\overrightarrow{BC} = CA.(BA + AC) = \overrightarrow{CA}.\overrightarrow{BA} + CA^2 = CA.BA.cos(\widehat{BAC}) + CA^2 = 0 + 5^2 = 25$.
Ta có $\overrightarrow{CA}.\overrightarrow{BC} = -\overrightarrow{AC}.\overrightarrow{BC} = -|AC||BC|cos(\widehat{ACB}) = -5\sqrt{29} \frac{5}{\sqrt{29}} = -25$.
$ \overrightarrow{CA}.\overrightarrow{BC} = |CA|.|BC|.cos(\widehat{(\overrightarrow{CA},\overrightarrow{BC})})$
$\overrightarrow{AC}.\overrightarrow{BC} = |AC|.|BC|.cos(\widehat{(\overrightarrow{AC},\overrightarrow{BC})})$
$\overrightarrow{AC}.\overrightarrow{BC} = AC \cdot BC \cdot cos(\widehat{ACB}) = AC \cdot BC \cdot \frac{AC}{BC} = AC^2 = 5^2 = 25$.
$\overrightarrow{CA}.\overrightarrow{BC} = - \overrightarrow{AC}.\overrightarrow{BC} = -25 $
Vì $cos(\widehat{ACB})=\frac{AC}{BC}=\frac{5}{\sqrt{29}}$ nên suy ra $cos(\widehat{BCA})=\frac{AC}{BC}$.
$\overrightarrow{CA}.\overrightarrow{CB} = |CA||CB|cos(\widehat{ACB}) = |5|\sqrt{29}.\frac{5}{\sqrt{29}}=25$
Do đó $\overrightarrow{CA}.\overrightarrow{BC} = \overrightarrow{CA}.(\overrightarrow{BA} + \overrightarrow{AC}) = \overrightarrow{CA}.\overrightarrow{BA} + \overrightarrow{CA}.\overrightarrow{AC} = CA.BA.cos(90^\circ) - CA^2 = 0 - 25 = -25 $
$\overrightarrow{CA}.\overrightarrow{BC} = -\overrightarrow{AC}.\overrightarrow{BC} = -AC.BC.cos(C) = -5.\sqrt{29}.\frac{5}{\sqrt{29}} = -25$.
$\overrightarrow{CA}.\overrightarrow{BC} = (0 - 5)(2 - 0) + (0 - 0)(0 - 0) = (-5)(2) + 0 = -10 $.
$\overrightarrow{CA}.\overrightarrow{BC} = -AC.BC.cos(\widehat{ACB})$
$BC=\sqrt{AB^2+AC^2}=\sqrt{4+25}=\sqrt{29}$
$cos(\widehat{ACB})=\frac{AC}{BC}=\frac{5}{\sqrt{29}}$
$\overrightarrow{CA}.\overrightarrow{BC}=-5.\sqrt{29}.\frac{5}{\sqrt{29}}=-25$
$ \overrightarrow{CA}.\overrightarrow{BC} = \overrightarrow{CA}.(\overrightarrow{BA}+\overrightarrow{AC})$
$= \overrightarrow{CA}.\overrightarrow{BA} + \overrightarrow{CA}.\overrightarrow{AC} = 0 - AC^2 = -25$
Ta có $\overrightarrow{CA} = (-5;0)$ và $\overrightarrow{BC} = (-2; -5)$
$\Rightarrow \overrightarrow{CA}.\overrightarrow{BC} = -5.(-2) + 0.(-5) = 10$
$\overrightarrow{AC}.\overrightarrow{BC} = (5,0) \cdot (2,-5) = 10$
$\overrightarrow{CA}.\overrightarrow{BC} = -10$.
$\overrightarrow{CA}.\overrightarrow{BC}=AC.BC.cos(ACB)$
$\overrightarrow{CA}=(5;0)$
$\overrightarrow{BC}=(x_C-x_B;y_C-y_B)=(0-2;0-0)=(-2;0)$
$\overrightarrow{CA}.\overrightarrow{BC}=5(-2)+0.0=-10$
Do không có đáp án nào đúng, em xin phép chọn đáp án gần đúng nhất là C.