Trong không gian Oxyz, cho đường thẳng d: $\left\{ \begin{aligned}...

Ta có điểm \)">A\left( 1;-3;5 \right)A\left( 1;-3;5 \right)\Delta \vec{v}\left( -3;0;-4 \right)\overrightarrow{{{u}_{1}}}=\dfrac{1}{\left| {\vec{u}} \right|}.\vec{u}=\dfrac{1}{3}\left( 1;2;-2 \right)=\left( \dfrac{1}{3};\dfrac{2}{3};-\dfrac{2}{3} \right)\overrightarrow{{{v}_{1}}}=\dfrac{1}{\left| {\vec{v}} \right|}.\vec{v}=\dfrac{1}{5}\left( -3;0;-4 \right)=\left( -\dfrac{3}{5};0;-\dfrac{4}{5} \right)\overrightarrow{{{u}_{1}}}.\overrightarrow{{{v}_{1}}}>0\overrightarrow{{{u}_{1}}}\overrightarrow{{{v}_{1}}}\Delta \overrightarrow{w}=\overrightarrow{{{u}_{1}}}+\overrightarrow{{{v}_{1}}}=\left( -\dfrac{4}{15};\dfrac{10}{15};-\dfrac{22}{15} \right)=-\dfrac{15}{2}\left( 2;-5;11 \right)\Delta \Delta \overrightarrow{{{w}_{1}}}=\left( 2;-5;11 \right)\left\{ \begin{aligned}
& x=-1+2t \\
& y=2-5t \\
& z=-6+11t \\
\end{aligned} \right.$