$3$ つのベクトル $\overrightarrow{a_1} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$, $\overrightarrow{a_2} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $\overrightarrow{a_3} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ に対し
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|}$
$\overrightarrow{v_2} = \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|}$
$\overrightarrow{v_3} = \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|}$
と定める。
この時 $\overrightarrow{e_3}$ を以下の選択肢から選びなさい。
$\dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$
$\dfrac{1}{2\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$
$\dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$
$\dfrac{1}{2\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$
順に計算していくと
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|} = \dfrac{1}{\sqrt{3}} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_2} & = & \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_3} & = & \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}\\[1em] & = & \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} - \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\\[1em] & = & \dfrac{1}{2}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|} = \dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$
よって $\overrightarrow{e_3} = \dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ である。
$3$ つのベクトル $\overrightarrow{a_1} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$, $\overrightarrow{a_2} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$, $\overrightarrow{a_3} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ に対し
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|}$
$\overrightarrow{v_2} = \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|}$
$\overrightarrow{v_3} = \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|}$
と定める。
この時 $\overrightarrow{e_3}$ を以下の選択肢から選びなさい。
$\dfrac{1}{\sqrt{30}} \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}$
$\dfrac{1}{\sqrt{6}} \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$
$\dfrac{1}{\sqrt{43}} \begin{pmatrix} 3 \\ 5 \\ 3 \end{pmatrix}$
$\dfrac{1}{\sqrt{19}} \begin{pmatrix} 3 \\ 1 \\ 3 \end{pmatrix}$
順に計算していくと
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|} = \dfrac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_2} & = & \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|} = \dfrac{1}{\sqrt{5}} \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_3} & = & \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}\\[1em] & = & \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} - \dfrac{2}{\sqrt{5}}\cdot \dfrac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\\[1em] & = & \dfrac{1}{5}\begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|} = \dfrac{1}{\sqrt{30}} \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}$
よって $\overrightarrow{e_3} = \dfrac{1}{\sqrt{30}} \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}$ である。
$3$ つのベクトル $\overrightarrow{a_1} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $\overrightarrow{a_2} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}$, $\overrightarrow{a_3} = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}$ に対し
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|}$
$\overrightarrow{v_2} = \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|}$
$\overrightarrow{v_3} = \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|}$
と定める。
この時 $\overrightarrow{e_3}$ を以下の選択肢から選びなさい。
$\dfrac{1}{\sqrt{14}} \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$
$\dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$
$\dfrac{1}{\sqrt{14}} \begin{pmatrix} 3 \\ -1 \\ -2 \end{pmatrix}$
$\dfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$
順に計算していくと
$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|} = \dfrac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_2} & = & \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}\\[1em] & = & \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} - \left( -\dfrac{2}{\sqrt{3}} \right) \cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\\[1em] & = & \dfrac{1}{3} \begin{pmatrix} -4 \\ 5 \\ -1 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|} = \dfrac{1}{\sqrt{42}} \begin{pmatrix} -4 \\ 5 \\ -1 \end{pmatrix}$
$\begin{eqnarray*} \overrightarrow{v_3} & = & \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}\\[1em] & = & \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} - \left( - \dfrac{1}{\sqrt{3}} \right) \cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \dfrac{7}{\sqrt{42}}\cdot \dfrac{1}{\sqrt{42}}\begin{pmatrix} -4 \\ 5 \\ -1 \end{pmatrix}\\[1em] & = & \dfrac{1}{2}\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} \end{eqnarray*}$
$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|} = \dfrac{1}{\sqrt{14}} \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$
よって $\overrightarrow{e_3} = \dfrac{1}{\sqrt{14}} \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$ である。