Đáp án:
a) $\left( \Delta \right):\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 - t\\
z = - 5
\end{array} \right.$
b) $\left( \Delta \right):\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 2}}{1}$
c) $\left( \Delta \right):\dfrac{{x + 1}}{5} = \dfrac{y}{4} = \dfrac{{z - 3}}{3}$
Giải thích các bước giải:
a) Ta có:
$(\Delta )//(d)\to $ $\overrightarrow {{u_\Delta }} = \overrightarrow {{u_d}} = \left( {2; - 1;0} \right)$
Mà $A \in \left( \Delta \right)$ nên $ \Rightarrow \left( \Delta \right):\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 - t\\
z = - 5
\end{array} \right.$
b) Ta có:
$(\Delta )//(d)\to $$\overrightarrow {{u_\Delta }} = \overrightarrow {{u_d}} = \left( {2;3;1} \right)$
Mà $B \in \left( \Delta \right)$ nên $\left( \Delta \right):\dfrac{{x - 1}}{2} = \dfrac{{y + 1}}{3} = \dfrac{{z - 2}}{1}$
c) Ta có:
$\overrightarrow {{n_{\left( \alpha \right)}}} = \left( {1; - 2;1} \right);\overrightarrow {{n_{\left( \beta \right)}}} = \left( {1;1; - 3} \right)$
Và:
$\begin{array}{l}
\left\{ \begin{array}{l}
\left( \Delta \right)//\left( \alpha \right)\\
\left( \Delta \right)//\left( \beta \right)
\end{array} \right.\\
\Rightarrow \overrightarrow {{u_\Delta }} = \left[ {\overrightarrow {{n_{\left( \alpha \right)}}} ,\overrightarrow {{n_{\left( \beta \right)}}} } \right] = \left( {5;4;3} \right)
\end{array}$
Mà $D \in \left( \Delta \right)$
$ \Rightarrow \left( \Delta \right):\dfrac{{x + 1}}{5} = \dfrac{y}{4} = \dfrac{{z - 3}}{3}$
