c.
\(K\left(2;2;z\right)\) với \(0\le x\le2\); \(M\left(0;0;1\right)\) ; \(N\left(1;2;0\right)\)
\(\overrightarrow{MK}=\left(2;2;z-1\right)\); \(\overrightarrow{NK}=\left(1;0;z\right)\)
\(KM+KN=\sqrt{8+\left(z-1\right)^2}+\sqrt{1+z^2}=\sqrt{8+\left(1-z\right)^2}+\sqrt{1+z^2}\)
\(\ge\sqrt{\left(2\sqrt{2}+1\right)^2+\left(1-z+z\right)^2}=\sqrt{10+4\sqrt{2}}\)
d.
\(\overrightarrow{MN}=\left(1;2;-1\right)\) ; \(\overrightarrow{BD'}=\left(-2;2;2\right)\)
\(\Rightarrow\overrightarrow{MN}.\overrightarrow{BD'}=-2+4-2=0\Rightarrow MN\perp BD'\)
Từ B kẻ BE vuông góc MN \(\Rightarrow MN\perp\left(BED'\right)\Rightarrow\left[B,MN,D'\right]=\widehat{BED'}\)
Do \(BM=BN=\sqrt{5}\Rightarrow\Delta BMN\) cân tại B
\(\Rightarrow E\) là trung điểm MN \(\Rightarrow E\left(\dfrac{1}{2};1;\dfrac{1}{2}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{EB}=\left(\dfrac{3}{2};-1;-\dfrac{1}{2}\right)\\\overrightarrow{ED'}=\left(-\dfrac{1}{2};1;\dfrac{3}{2}\right)\end{matrix}\right.\)
\(cos\left[B,MN,D'\right]=cos\left(\overrightarrow{EB};\overrightarrow{ED'}\right)=\dfrac{-\dfrac{3}{4}-1-\dfrac{3}{4}}{\sqrt{\dfrac{9}{4}+1+\dfrac{1}{4}}.\sqrt{\dfrac{1}{4}+1+\dfrac{9}{4}}}=-\dfrac{5}{7}\)
