Trong không gian với hệ tọa độ Oxyz cho

C

- Hướng dẫn giải

Gọi \(\overrightarrow {{n_P}} = \left( {a;b;c} \right);\left( {{a^2} + {b^2} + {c^2} \ne 0} \right)\)

Ta có:

\(\begin{array}{l}A,B \in \left( P \right) \Rightarrow \overrightarrow {AB} \bot \overrightarrow {{n_P}} \Rightarrow 3a - 2b = 0\\ \Leftrightarrow 3a = 2b \Leftrightarrow 9{a^2} = 4{b^2}\left( 1 \right)\\cos\left( {\widehat {\left( P \right),\left( {Oyz} \right)}} \right) = \frac{2}{7} \Rightarrow \frac{{\left| {\overrightarrow {{n_P}} .\overrightarrow {{n_{Oyz}}} } \right|}}{{\left| {\overrightarrow {{n_P}} } \right|.\left| {\overrightarrow {{n_{Oyz}}} } \right|}} = \frac{{\left| a \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt 1 }} = \frac{2}{7}\\ \Leftrightarrow \frac{{\left| a \right|}}{{\sqrt {{a^2} + {{\left( {\frac{{3a}}{2}} \right)}^2} + {c^2}} }} = \frac{2}{7} \Leftrightarrow \frac{{\left| a \right|}}{{\sqrt {\frac{{13}}{4}{a^2} + {c^2}} }} = \frac{2}{7}\\ \Leftrightarrow {a^2} = \frac{4}{{49}}\left( {\frac{{13}}{4}{a^2} + {c^2}} \right) \Leftrightarrow 9{a^2} = {c^2}\left( 2 \right)\\\left( 1 \right),\left( 2 \right) \Rightarrow {c^2} = 4{b^2} \Leftrightarrow \left[ \begin{array}{l}c = 2b\\c = - 2b\end{array} \right.\end{array}\)

Chọn: \(\begin{array}{l}a = 2 \Rightarrow b = 3 \Rightarrow c = 6 \Rightarrow \left( P \right):2x + 3y + 6z - 12 = 0\\a = - 2 \Rightarrow b = - 3 \Rightarrow c = 6 \Rightarrow \left( P \right):2x + 3y - 6z = 0\end{array}\)