Đáp án:
\(\left[ \begin{array}{l}
m = 0\\
m = \dfrac{3}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
vtpt:{\overrightarrow n _d} = \left( {3; - 1} \right) \to \left| {{{\overrightarrow n }_d}} \right| = \sqrt {10} \\
vtpt:{\overrightarrow n _{d'}} = \left( {m;1} \right) \to \left| {{{\overrightarrow n }_{d'}}} \right| = \sqrt {{m^2} + 1} \\
Do:\cos \left( {d;d'} \right) = \dfrac{1}{{\sqrt {10} }}\\
\to \dfrac{{\left| {{{\overrightarrow n }_d}.{{\overrightarrow n }_{d'}}} \right|}}{{\left| {{{\overrightarrow n }_d}} \right|.\left| {{{\overrightarrow n }_{d'}}} \right|}} = \dfrac{1}{{\sqrt {10} }}\\
\to \dfrac{{\left| {3m - 1} \right|}}{{\sqrt {10} .\sqrt {{m^2} + 1} }} = \dfrac{1}{{\sqrt {10} }}\\
\to \dfrac{{\left| {3m - 1} \right|}}{{\sqrt {{m^2} + 1} }} = 1\\
\to \sqrt {{m^2} + 1} = \left| {3m - 1} \right|\\
\to {m^2} + 1 = 9{m^2} - 6m + 1\\
\to 8{m^2} - 6m = 0\\
\to \left[ \begin{array}{l}
m = 0\\
m = \dfrac{3}{4}
\end{array} \right.
\end{array}\)
