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Câu hỏi: Cho hình lập phương $ABCD.{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}$ có cạnh bằng 1. Hai điểm M, N lần lượt thay đổi trên các đoạn $A{{B}_{1}}$ và $B{{C}_{1}}$ sao cho $MN$ luôn tạo với mặt phẳng $\left( ABCD \right)$ một góc 60 (tham khảo hình vẽ). Giá trị bé nhất của đoạn $MN$ là
A. $\dfrac{\sqrt{3}}{3}$.
B. $2\left( \sqrt{2}-1 \right)$.
C. $2\left( \sqrt{3}-\sqrt{2} \right)$.
D. $\sqrt{3}-1$.
Chọn hệ trục Oxyz như hình vẽ ta có
$A\left( 0;0;0 \right),B\left( 0;1;0 \right),D\left( 1;0;0 \right),C\left( 1;1;0 \right),{{A}_{1}}\left( 0;0;1 \right),{{C}_{1}}\left( 1;1;1 \right)$
Ta có $\overrightarrow{AM}=m\overrightarrow{A{{B}_{1}}},\left( 0<m<1 \right)\Rightarrow M\left( 0;m;m \right)$ ;
$\overrightarrow{BN}=n\overrightarrow{B{{C}_{1}}},\left( 0<n<1 \right)\Rightarrow N\left( n;1;n \right)$
$\Rightarrow \overrightarrow{MN}\left( n;1-m;n-m \right)\Rightarrow M{{N}^{2}}={{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}$
MN tạo với mặt phẳng $\left( ABCD \right)\equiv \left( Oxy \right)$ góc 60
$\Rightarrow \sin 60{}^\circ =\dfrac{\left| \overrightarrow{MN}.\overrightarrow{k} \right|}{\left| \overrightarrow{MN} \right|.\left| \overrightarrow{k} \right|}\Leftrightarrow \dfrac{\left| n-m \right|}{\sqrt{{{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}}}=\dfrac{\sqrt{3}}{2}$
$\Leftrightarrow {{\left( n-m \right)}^{2}}=3\left[ {{n}^{2}}+{{\left( 1-m \right)}^{2}} \right]\ge 3.\dfrac{{{\left( n-m+1 \right)}^{2}}}{2}=\dfrac{3}{2}\left[ {{\left( n-m \right)}^{2}}+2\left( n-m \right)+1 \right]$
$\Leftrightarrow {{\left( n-m \right)}^{2}}+6\left( n-m \right)+3\le 0\Leftrightarrow -3-\sqrt{6}\le n-m\le -3+\sqrt{6}\Rightarrow 3-\sqrt{6}\le \left| n-m \right|\le 3+\sqrt{6}$
$\Rightarrow MN=\sqrt{{{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}}=\dfrac{2\sqrt{3}}{3}\left| n-m \right|\ge \dfrac{2\sqrt{3}}{3}\left( 3-\sqrt{6} \right)=2\left( \sqrt{3}-\sqrt{2} \right)$
$\Rightarrow \min MN=2\left( \sqrt{3}-\sqrt{2} \right)$ khi $m=\dfrac{4-\sqrt{6}}{2},n=\dfrac{\sqrt{6}-2}{2}$.
A. $\dfrac{\sqrt{3}}{3}$.
B. $2\left( \sqrt{2}-1 \right)$.
C. $2\left( \sqrt{3}-\sqrt{2} \right)$.
D. $\sqrt{3}-1$.
Chọn hệ trục Oxyz như hình vẽ ta có
$A\left( 0;0;0 \right),B\left( 0;1;0 \right),D\left( 1;0;0 \right),C\left( 1;1;0 \right),{{A}_{1}}\left( 0;0;1 \right),{{C}_{1}}\left( 1;1;1 \right)$
Ta có $\overrightarrow{AM}=m\overrightarrow{A{{B}_{1}}},\left( 0<m<1 \right)\Rightarrow M\left( 0;m;m \right)$ ;
$\overrightarrow{BN}=n\overrightarrow{B{{C}_{1}}},\left( 0<n<1 \right)\Rightarrow N\left( n;1;n \right)$
$\Rightarrow \overrightarrow{MN}\left( n;1-m;n-m \right)\Rightarrow M{{N}^{2}}={{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}$
MN tạo với mặt phẳng $\left( ABCD \right)\equiv \left( Oxy \right)$ góc 60
$\Rightarrow \sin 60{}^\circ =\dfrac{\left| \overrightarrow{MN}.\overrightarrow{k} \right|}{\left| \overrightarrow{MN} \right|.\left| \overrightarrow{k} \right|}\Leftrightarrow \dfrac{\left| n-m \right|}{\sqrt{{{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}}}=\dfrac{\sqrt{3}}{2}$
$\Leftrightarrow {{\left( n-m \right)}^{2}}=3\left[ {{n}^{2}}+{{\left( 1-m \right)}^{2}} \right]\ge 3.\dfrac{{{\left( n-m+1 \right)}^{2}}}{2}=\dfrac{3}{2}\left[ {{\left( n-m \right)}^{2}}+2\left( n-m \right)+1 \right]$
$\Leftrightarrow {{\left( n-m \right)}^{2}}+6\left( n-m \right)+3\le 0\Leftrightarrow -3-\sqrt{6}\le n-m\le -3+\sqrt{6}\Rightarrow 3-\sqrt{6}\le \left| n-m \right|\le 3+\sqrt{6}$
$\Rightarrow MN=\sqrt{{{n}^{2}}+{{\left( 1-m \right)}^{2}}+{{\left( n-m \right)}^{2}}}=\dfrac{2\sqrt{3}}{3}\left| n-m \right|\ge \dfrac{2\sqrt{3}}{3}\left( 3-\sqrt{6} \right)=2\left( \sqrt{3}-\sqrt{2} \right)$
$\Rightarrow \min MN=2\left( \sqrt{3}-\sqrt{2} \right)$ khi $m=\dfrac{4-\sqrt{6}}{2},n=\dfrac{\sqrt{6}-2}{2}$.
Đáp án C.