Lời giải
Ta có: $\left\{ \begin{array}{l} OA^2+OC^2=AC^2\\ OB^2+OC^2=BC^2 \end{array} \right. $
$\Rightarrow 4OB^2-OA^2=4OA^2-OB^2\Rightarrow OA=OB$
Đặt $OA=a=OB\Rightarrow OC=a\sqrt{3} $
a) $\overrightarrow{AC}=(-a;0;a\sqrt{3} )=-a(1;0;-\sqrt{3} ) $
$\Rightarrow $ phương trình tham số $AC:\left\{ \begin{array}{l} x=a+t\\ y=0\\z=-\sqrt{3}t \end{array} \right. $
$\Rightarrow M(a+t;0;-\sqrt{3} t)$
$OM\bot AC\Rightarrow \overrightarrow{OM}.\overrightarrow{AC} =0\Leftrightarrow a+t+0+3t=0 $
$\Leftrightarrow t=-\frac{a}{4}\Rightarrow M(\frac{3a}{4};0;\frac{a\sqrt{3} }{4} ) $
$\overrightarrow{BC}=(0;-a;a\sqrt{3} )=-a(0;1;-\sqrt{3} ) $
Suy ra phương trình $BC:\left\{ \begin{array}{l} x=0\\ y=a+t\\z=-\sqrt{3} t \end{array} \right. \Rightarrow N(0;a+t;-\sqrt{3}t )$
$ON\bot BC\Rightarrow \overrightarrow{ON}.\overrightarrow{BC}=0\Leftrightarrow a+t+3t=0 $
$t=-\frac{a}{4}\Rightarrow N(0;\frac{3a}{4};\frac{a\sqrt{3} }{4} ) \Rightarrow \overrightarrow{MN}.\overrightarrow{OC}=(-\frac{3a}{4};\frac{3a}{4};0 ) (0;0;a\sqrt{3} ) =0$
Vậy $MN\bot OC$
b) $\widehat{MON}=\frac{\overrightarrow{OM}.\overrightarrow{ON} }{OM.ON} =\frac{\frac{3a^2}{16} }{\frac{12a^2}{16} } =\frac{1}{4} $
c) Đặt $\beta =\widehat{OCD}; \alpha =\widehat{OCA}$
$OC\bot (OAB)\Rightarrow OC\bot OD$
$OD=\frac{1}{2}AB=\frac{a\sqrt{2} }{2} $
$\Rightarrow tan \beta =\frac{OD}{OC}; tan \alpha =\frac{OA}{OC} \Rightarrow \frac{tan^4 \beta }{tan^4\alpha }=(\frac{OD}{OA})^2=\frac{1}{4} $
$\frac{MN}{AB}=\frac{\frac{3a\sqrt{2} }{4} }{a\sqrt{2} } \frac{3}{4} \Rightarrow \frac{tan^4\widehat{OCD}}{tan^4\widehat{OCA}}+\frac{MN}{AB}=1 $
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