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Chọn hệ trục tọa độ $Axyz$ sao cho: $A(0;0;0), B(a;0;0), D(0;a;0), A'(0;0;a), B'(a;0;a), C(a;a;0), D'(0;a;a), c'(a;a;a)$
$\Rightarrow M(0;\frac{a}{2};0 ), N(\frac{a}{2};a;0 ), P(a;0;\frac{3a}{4} )$
a) Ta có: $\overrightarrow{AC}=a(1;1;-1) , \overrightarrow{AB}=a(1;0;1), \overrightarrow{AD'}=a(0;1;1) \Rightarrow [\overrightarrow{AB'},\overrightarrow{AD'} ] =a^2(-1;-1;1)$
$[\overrightarrow{A'C}, \overrightarrow{n}_{(AB'D')} ]=[(a;a;-a),(-1;-1;1)]=\overrightarrow{0} $
$\Rightarrow \overrightarrow{A'C}// \overrightarrow{n}_{(AB'C')} \Rightarrow A'C\bot (AB'C')$
Ta có: $\overrightarrow{n}_1=\overrightarrow{n}_{(AB'D')}=(1;1;-1) $
$[\overrightarrow{BC'}, \overrightarrow{BD} ]=[(0;a;a),(-a;a;0)]=a^2(1;1;-1)$
$\overrightarrow{n}_2=\overrightarrow{n}_{(C'BD)}=(1;1;-1) $
Do $\overrightarrow{n}_1=\overrightarrow{n}_2 \Rightarrow (AB'D')//(C'BD)\Rightarrow d((AB'D'),(C'BD))=d(A,(C'BD))$
Phương trình $(C'BD):(x-a)+y-z=0\Leftrightarrow x+y-z-a=0$
$\Rightarrow d(A,(C'BD))=\frac{a}{\sqrt{3} } $
b) Ta có: $[\overrightarrow{DA'},\overrightarrow{DC'} ]=[(0;-a;a),(a;0;a)]=-a^2(1;-1;-1)$
$\Rightarrow \overrightarrow{n}_{(DA'C')}=(1;-1;-1); \overrightarrow{n}_{(ABB'A')}=(0;1;1) $
$\Rightarrow cos\varphi=\frac{|(1;-1;-1)(0;1;0)|}{\sqrt{3} }=\frac{\sqrt{3} }{3} $
c) Gọi $E, F$ là giao điểm của $(MNp)$ và $CC', AA'$
$\Rightarrow$ Thiết diện là ngũ giác $MNEPF$.
Ta có: $\overrightarrow{MN}=\frac{a}{2}(1;1;0), \overrightarrow{MP}=\frac{a}{4}=(4;-2;3) $
$\overrightarrow{n}_{(MNP)}=[(1;1;0),(4;-2;3)]=3(1;-1;-2) $
Phương trình $(MNP):x-(y-\frac{a}{2} )-2z=0\Leftrightarrow 2x-2y-4z+a=0$
$E\in CC'\Rightarrow E(a;a;z_E); F\in AA'\Rightarrow F(0;0;z_F)$
$E,F\in (MNP)\Rightarrow z_E=\frac{a}{4}, z_F =\frac{a}{4}\Rightarrow E(a;a;\frac{a}{4} ) , F(0;0;\frac{a}{4} )$
$S_{MNEPF}=S_{\Delta MFP}+S_{\Delta MPE}+S_{\Delta MEN}$
$=\frac{1}{2}(|[\overrightarrow{MF}, \overrightarrow{MP} ]|+|[\overrightarrow{MP}, \overrightarrow{ME} ]|+|[\overrightarrow{ME}, \overrightarrow{MN} ]|) $
$\begin{array}{l}
{\rm{[}}\overrightarrow {MF} {\rm{,}}\overrightarrow {MN} {\rm{] = }}\left[ {\left( {0; - \frac{a}{2};\frac{a}{4}} \right),\left( {a; - \frac{a}{2};\frac{{3a}}{4}} \right)} \right] = \left( { - \frac{{{a^2}}}{4};\frac{{{a^2}}}{4};\frac{{{a^2}}}{2}} \right)\\
{\rm{[}}\overrightarrow {MP} {\rm{,}}\overrightarrow {ME} {\rm{] = }}\left[ {\left( {a; - \frac{a}{2};\frac{{3a}}{4}} \right),\left( {a;\frac{a}{2};\frac{a}{4}} \right)} \right] = \left( { - \frac{{{a^2}}}{2};\frac{{{a^2}}}{2};{a^2}} \right)\\
{\rm{[}}\overrightarrow {ME} {\rm{,}}\overrightarrow {MN} {\rm{] = }}\left[ {\left( {a;\frac{a}{2};\frac{a}{4}} \right),\left( {\frac{a}{2};\frac{a}{2};0} \right)} \right] = \left( { - \frac{{{a^2}}}{8};\frac{{{a^2}}}{8};\frac{{{a^2}}}{4}} \right)\\
{S_{MNEPF}} = \frac{1}{2}\left[ {\frac{{{a^2}\sqrt 6 }}{4} + \frac{{{a^2}\sqrt 6 }}{2} + \frac{{{a^2}\sqrt 6 }}{8}} \right] = \frac{{7{a^2}\sqrt 6 }}{{16}}
\end{array}$
d) Gọi $V_1=V_{ABFMNCE}, V_2$ là phần còn lại
$I=BC\cap MN, J=AB\cap MN$
Dễ thấy $IB=\frac{3a}{2}, BJ=\frac{3a}{2} $. Ta có:
$\begin{array}{l}
{V_1} = {V_{P.BJ}} - ({V_{ICNE}} + {V_{J.{\rm{AMF}}}}) = \frac{1}{6}BP.BI.BJ - \frac{1}{6}(IC.CN.CE + {\rm{A}}J.AM.{\rm{A}}F)\\
= \frac{1}{6}.\frac{{3a}}{4}.\frac{{3a}}{2}.\frac{{3a}}{2} - \frac{1}{6}\left( {\frac{a}{2}.\frac{a}{2}.\frac{a}{4} + \frac{a}{2}.\frac{a}{2}.\frac{a}{4}} \right) = \frac{{25{a^3}}}{{96}}\\
{V_{ABCD.A'B'C'D'}} = {a^3} \Rightarrow {V_2} = {a^3} - \frac{{25{a^3}}}{{96}} = \frac{{71{a^3}}}{{96}}\\
\Rightarrow \frac{{{V_1}}}{{{V_2}}} = \frac{{25}}{{71}}
\end{array}$
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Đăng bài 07-06-12 04:55 PM
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