Cho tam giác $ABC$ có $A(1;-1;-3), B(2;1;-2), C(-5;2;-6)$
a) Các góc trong tam giác $ABC$ nhọn hay tù
b) Tính tọa độ chân đường phân giác trong của góc $A$
c) Tìm tọa độ giao điểm của đường thẳng $AB$ với mặt phẳng $Oxy$
a) Ta có:
$\begin{array}{l}
\overrightarrow {AB}  = (1;2;1),\,\,\overrightarrow {AC}  = ( - 6;3; - 3)\\
c{\rm{os}}\,\widehat A = c{\rm{os}}(\overrightarrow {AB} ,\overrightarrow {AC} ) = \frac{{\overrightarrow {AB} .\overrightarrow {AC} }}{{|\overrightarrow {AB} |.|\overrightarrow {AC} |}} = \frac{{ - 6 + 6 - 3}}{{\sqrt {1 + 4 + 1} .\sqrt {36 + 9 + 9} }} = \frac{{ - 3}}{{\sqrt 6 .\sqrt {54} }} = - \frac{1}{6}
\end{array}$
$\widehat{A}\Rightarrow  $ tù $\Rightarrow \widehat{B},\widehat{C} $ nhọn

b) Gọi $D$ là chân đường phân giác trong $\widehat{A}$ của $\Delta ABC$
$\begin{array}{l}
\frac{{DB}}{{DC}} = \frac{{AB}}{{AC}} = \frac{{\sqrt 6 }}{{3\sqrt 6 }} = \frac{1}{3} \Rightarrow 3DB = DC \Rightarrow 3\overrightarrow {DB}  =  - \overrightarrow {DC} \\
 \Rightarrow \left\{ \begin{array}{l}
3(2 - {x_D}) =  - ( - 5 - {x_D})\\
3(1 - {y_D}) =  - (2 - {y_D})\\
3( - 2 - {z_D}) =  - ( - 6 - {z_D})
\end{array} \right.\,\,\, \Rightarrow \left\{ \begin{array}{l}
{x_D} = \frac{1}{4}\\
{y_D} = \frac{5}{4}\\
{z_D} = - 3
\end{array} \right.\,\, \Rightarrow D\left( {\frac{1}{4};\frac{5}{4}; - 3} \right)
\end{array}$

c) Gọi $M(x;y;z)$ là tọa độ giao điểm của đường thẳng $AB$ với $(Oxy)$
Ta có:
$\begin{array}{l}
\overrightarrow {AB}  = (1;2;1),\,\,\overrightarrow {AM}  = (x - 1;y + 1;z + 3)\\
M = AB \cap ({\rm{Oxy}}) \Rightarrow \left\{ \begin{array}{l}
M \in ({\rm{Ox}}y)\\
M \in (AB)
\end{array} \right.
\end{array}$
Suy ra $M(x;y;0), \overrightarrow{AM} $ cùng phương $\overrightarrow{AB} $
$ \Rightarrow \frac{{x - 1}}{1} = \frac{{y + 1}}{2} = \frac{3}{1} = 3 \Rightarrow \left\{ \begin{array}{l}
x - 1 = 3\\
y + 1 = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = 5
\end{array} \right. \Rightarrow M(4;5;0)$

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