Đáp án:
\({\left| {\overrightarrow u } \right|_{\min }} = 1\).
Giải thích các bước giải:
Gọi \(\overrightarrow u \left( {a;b} \right)\).
Lấy \(A\left( {1;2} \right) \in d\). Gọi \(A' = {T_{\overrightarrow u }}\left( A \right) \Rightarrow A'\left( {1 + a;2 + b} \right)\).
\(\begin{array}{l}A' \in d' \Rightarrow 3\left( {1 + a} \right) - 4\left( {2 + b} \right) = 0 \Leftrightarrow 3a - 4b - 5 = 0\\ \Rightarrow a = \frac{{4b + 5}}{3}\end{array}\)
Ta có:
\(\begin{array}{l}\overrightarrow u = \sqrt {{a^2} + {b^2}} = \sqrt {\frac{{{{\left( {4b + 5} \right)}^2}}}{9} + {b^2}} \\ = \frac{{\sqrt {16{b^2} + 40b + 25 + 9{b^2}} }}{3} = \frac{{\sqrt {25{b^2} + 40b + 25} }}{3}\\ = \frac{{\sqrt {{{\left( {5b} \right)}^2} + 2.5b.4 + 16 + 9} }}{3} = \frac{{\sqrt {{{\left( {5b + 4} \right)}^2} + 9} }}{3}\end{array}\)
Ta có: \({\left( {5b + 4} \right)^2} + 9 \ge 9\,\,\forall b \Rightarrow \sqrt {{{\left( {5b + 4} \right)}^2} + 9} \ge 3\,\,\forall b \Rightarrow \left| {\overrightarrow u } \right| \ge 1\,\,\forall b\)
Vậy \({\left| {\overrightarrow u } \right|_{\min }} = 1\). Dấu “=” xảy ra \( \Leftrightarrow b = - \frac{4}{5} \Rightarrow a = \frac{3}{5} \Rightarrow \overrightarrow u = \left( {\frac{3}{5}; - \frac{4}{5}} \right)\).
