Giải thích các bước giải:
Ta có:
$\overrightarrow a = \left( {1; - 3;4} \right)$
a) Ta có:
$\overrightarrow b = \left( {2;y;z} \right)$
$ \Rightarrow {x_{\overrightarrow b }} = 2;{x_{\overrightarrow a }} = 1$
Để $\overrightarrow b ,\overrightarrow a $ cùng phương
$\begin{array}{l}
\Leftrightarrow \overrightarrow b = 2\overrightarrow a \\
\Leftrightarrow \left\{ \begin{array}{l}
y = 2{y_{\overrightarrow a }} = - 6\\
z = 2{z_{\overrightarrow a }} = 8
\end{array} \right.\\
\Rightarrow \overrightarrow b = \left( {2; - 6;8} \right)
\end{array}$
Vậy $\overrightarrow b = \left( {2; - 6;8} \right)$
b) Ta có:
$\overrightarrow c $ ngược hướng với $\overrightarrow b $
$ \Rightarrow \overrightarrow c = k\overrightarrow b \left( {k < 0} \right)$
$ \Rightarrow \overrightarrow c = \left( {2k; - 6k;8k} \right)$
Lại có:
$\begin{array}{l}
\overrightarrow a + \overrightarrow b = \left( {3; - 9;12} \right)\\
\Rightarrow \left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{3^2} + {{\left( { - 9} \right)}^2} + {{12}^2}} = 3\sqrt {26} \\
\Rightarrow \left| {\overrightarrow c } \right| = 3\left| {\overrightarrow a + \overrightarrow b } \right| = 9\sqrt {26} \\
\Rightarrow \sqrt {{{\left( {2k} \right)}^2} + {{\left( { - 6k} \right)}^2} + {{\left( {8k} \right)}^2}} = 9\sqrt {26} \\
\Leftrightarrow \left| k \right|.2\sqrt {26} = 9\sqrt {26} \\
\Leftrightarrow \left| k \right| = \dfrac{9}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
k = \dfrac{9}{2}\left( l \right)\\
k = \dfrac{{ - 9}}{2}\left( c \right)
\end{array} \right.\\
\Leftrightarrow k = \dfrac{{ - 9}}{2}\\
\Rightarrow \overrightarrow c = \left( { - 9;27;36} \right)
\end{array}$
Vậy $\overrightarrow c = \left( { - 9;27;36} \right)$
