A.\(A'\left( {1; - 1;5} \right).\)
B.\(A'\left( {1;1;5} \right).\)
C.\(A'\left( { - 1; - 1;5} \right).\)
D. \(A'\left( { - 1;1;5} \right).\)
D
Hình hộp \(ABCD.A'B'C'D' \Rightarrow \overrightarrow {AD} = \overrightarrow {BC} \) và \(\overrightarrow {AA'} = \overrightarrow {DD'} \)
* \(\overrightarrow {AD} = \overrightarrow {BC} \Leftrightarrow \left\{ \begin{array}{l}{x_D} - {x_A} = {x_C} - {x_B}\\{y_D} - {y_A} = {y_C} - {y_B}\\{z_D} - {z_A} = {z_C} - {z_B}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_D} - 0 = 1 - 1\\{y_D} - 0 = 2 - 0\\{z_D} - 0 = 0 - 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_D} = 0\\{y_D} = 2\\{z_D} = 0\end{array} \right.\)
* \(\overrightarrow {AA'} = \overrightarrow {DD'} \Leftrightarrow \Leftrightarrow \left\{ \begin{array}{l}{x_{A'}} - {x_A} = {x_{D'}} - {x_D}\\{y_{A'}} - {y_A} = {y_{D'}} - {y_D}\\{z_{A'}} - {z_A} = {z_{D'}} - {z_D}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_{A'}} - 0 = - 1 - 0\\{y_{A'}} - 0 = 3 - 2\\{z_{A'}} - 0 = 5 - 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_{A'}} = - 1\\{y_{A'}} = 1\\{z_{A'}} = 5\end{array} \right.\)
Vậy \(A'\left( { - 1;1;5} \right).\)
Đáp án D
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