Cho hình lập phương ABCD A'B'C'D' có cạnh bằng a. Tính theo

* Hướng dẫn giải

Đáp án A.

Đặt $B\text{'}\left(0;0;0\right),A\text{'}\left(a;0;0\right),$$C\text{'}\left(0;a;0\right),B\left(0;0;a\right)$$\Rightarrow A\left(a;0;a\right)$

Ta có $\overrightarrow{B\text{'}A}=\left(a;0;a\right),\overrightarrow{BC\text{'}}=\left(0;a;-a\right),$$\overrightarrow{B\text{'}B}=\left(0;0;a\right)$

$\Rightarrow \left[\overrightarrow{B\text{'}A},\overrightarrow{BC\text{'}}\right]=\left(-a^2;a^2;a^2\right)$;$\left[\overrightarrow{B\text{'}A},\overrightarrow{BC\text{'}}\right].\overrightarrow{BB\text{'}}=a^3$

$\begin{array}{l}d\left(B\text{'}A,BC\text{'}\right)=\frac{\left|\left[\overrightarrow{B\text{'}A},\overrightarrow{BC\text{'}}\right].\overrightarrow{BB\text{'}}\right|}{\left|\left[\overrightarrow{B\text{'}A},\overrightarrow{BC\text{'}}\right]\right|}\\ =\frac{a^3}{\sqrt{3a^4}}=\frac{a^3}{a^2\sqrt{3}}=\frac{a\sqrt{3}}{3}\end{array}$