Cho hình lập phương A(0;0;0),B(1;0;0),D(0;1;0),A′(0;0;1). Gọi M,N lần lượt là trung đ

* Hướng dẫn giải

Gọi C(x;y;z) ta có:

$\overrightarrow{AB}=\overrightarrow{DC}\iff \left\{\begin{array}{c}1-0=x-0\\ 0-0=y-1\\ 0-0=z-0\end{array}\right.\iff \left\{\begin{array}{c}x=1\\ y=1\\ z=0\end{array}\right.\Rightarrow C(1;1;0)$

Lại có

$\begin{array}{l}M\left(\frac{1}{2};0;0\right),N\left(\frac{1}{2};1;0\right)\Rightarrow \overrightarrow{MN}=\left(0;1;0\right),\overrightarrow{A^{\text{'}}C}=\left(1;1;-1\right),\overrightarrow{M{A}^{\text{'}}}=\left(-\frac{1}{2};0;1\right)\\ \Rightarrow \left[\overrightarrow{MN},\overrightarrow{A^{\text{'}}C}\right]=\left(\left|\begin{array}{cc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}0\\ -1\end{array}\end{array}\right|;\left|\begin{array}{cc}\begin{array}{l}0\\ -1\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right|;\left|\begin{array}{cc}\begin{array}{l}0\\ 1\end{array}& \begin{array}{l}1\\ 1\end{array}\end{array}\right|\right)=\left(-1;0;-1\right)\end{array}$

Vậy

$d\left(MN,A^{\text{'}}C\right)=\frac{\left|\left[\overrightarrow{MN},\overrightarrow{A^{\text{'}}C}\right].\overrightarrow{M{A}^{\text{'}}}\right|}{\left|\left[\overrightarrow{MN},\overrightarrow{A^{\text{'}}C}\right]\right|}=\frac{\left|\left(-1\right).\left(-\frac{1}{2}\right)+0.0+\left(-1\right).1\right|}{\sqrt{{\left(-1\right)}^2+0^2+{\left(-1\right)}^2}}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$

Đáp án cần chọn là: B