Cho tam giác $ABC$ biết $A(1;0;-2), B(-2;1;1), C(1;-3;-2). $ Gọi $D$ là điểm chia đoạn thẳng $AB$ theo tỉ số $-2$ và $E$ là điểm chia đoạn thẳng $BC$ theo tỉ số $2$
a) Tìm tọa độ các điểm $D, E$
b) Tìm Cosin của góc tạo bởi hai vectơ $\overrightarrow{AD}, \overrightarrow{AE}  $
a) Ta có $\overrightarrow{AD}=-2\overrightarrow{DB}  $. Suy ra:
$x_D=\frac{1+(-2)(-2)}{1-2}=5;  y_D=\frac{0+(-2).1}{1-2}=2; z_D=\frac{-2+(-2).1}{1-2}=4   $
Vậy $D(-5;2;4)$

Ta cũng có $\overrightarrow{BE}=2\overrightarrow{EC}  $ nên suy ra:
$x_E=\frac{-2+2.1}{1+2}=0; y_E=\frac{1+2.(-3)}{1+2}=-\frac{5}{3}; z_E=\frac{1+2.(-2)}{1+2}=-1    $
Vậy $E(0;\frac{5}{3};-1 )$

b) Ta tính được $\overrightarrow{AD}=(-6;2;6), \overrightarrow{AE}(-1;-\frac{5}{3};1 )  $
Vậy $\cos (\overrightarrow{AD}, \overrightarrow{AE}  )=\frac{(-6)(-1)+2(-\frac{5}{3})+6.1 }{\sqrt{(-6)^2+2^2+6^2}\sqrt{(-1)^2+(-\frac{5}{3} )^2+1^2}   }=\frac{13}{\sqrt{817} }\approx 0,4549  $

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