Trong không gian với hệ tọa độ Oxyz, cho đường thẳ...

C

- Hướng dẫn giải

Dễ thấy \(A\left( {1;0; - 1} \right);B\left( {3;1; - 2} \right) \in \left( \Delta \right)\)

Giả sử: \(\left( Q \right):a\left( {x - 1} \right) + by + c\left( {z + 1} \right) = 0\)

\(\begin{array}{l} \Rightarrow a\left( {3 - 1} \right) + b.1 + c\left( { - 2 + 1} \right) = 0\\ \Rightarrow b = c - 2a\\ \Rightarrow \left( Q \right):a\left( {x - 1} \right) + \left( {c - 2a} \right)y + c\left( {z + 1} \right) = 0\\\left( P \right):2x + y + 2z - 1 = 0\\ \Rightarrow cos\left( {\widehat {\left( P \right);\left( Q \right)}} \right) = \frac{{\left| {2a + \left( {c - 2a} \right) + 2c} \right|}}{{\sqrt {{a^2} + {{\left( {2a - c} \right)}^2} + {c^2}} \sqrt 9 }}\\ \Rightarrow cos\left( {\widehat {\left( P \right);\left( Q \right)}} \right) = \frac{{\left| {4c} \right|}}{{3\sqrt {5{a^2} - 4ac + 2{c^2}} }}\\ \Rightarrow cos\left( {\widehat {\left( P \right);\left( Q \right)}} \right) = \frac{{\left| c \right|}}{{\sqrt {5{{\left( {a - \frac{2}{5}c} \right)}^2} + \frac{6}{5}{c^2}} }} \le \frac{1}{{\sqrt {\frac{6}{5}} }}\end{array}\)

Dấu "=" xảy ra khi:

\(\begin{array}{l}a = \frac{2}{5}c \Leftrightarrow \left( Q \right):\frac{2}{5}\left( {x - 1} \right)\left( {1 - \frac{4}{5}} \right)y + \left( {z + 1} \right) = 0\\ \Leftrightarrow \left( Q \right):2x + y + 5z + 3 = 0\end{array}\)