a.
+ \(SA\perp (ABCD)\Rightarrow AB\) là hình chiếu của SB lên (ABCD) ⇒ góc giữa SB và (ABCD) là \((SB; BA) =SBA=45^{\circ}\)
+ \(\triangle SAB\) vuông cân tại A ⇒ SA = a
+ \(S_{OABCD}=AB.AB=2a^{2}\)
+ \(V_{SABCD}=\frac{1}{2}S_{OABCD}.SA=\frac{1}{3}.2a^{2}.a=\frac{2a^{3}}{3}\)
+ \(S_{\triangle MCD}=\frac{1}{2}MC.CD=\frac{1}{2}a.a=\frac{a^{2}}{2}\)
\(d\left [ N;(MCD) \right ]=\frac{1}{2}SA=\frac{a}{2}\)
\(V_{N.MCD}=\frac{1}{3}S_{\triangle MCD}.d\left [ N;(MCD) \right ]\)
\(=\frac{1}{3}.\frac{a^{2}}{2}.\frac{a}{2}=\frac{a^{3}}{12}\)
b. Vẽ d qua C và \(d // BD;d\cap AB=E\Rightarrow BD/(SCE)\Rightarrow d(BD;SC)=d(B;(SCE))\)
+ BE = CD = AB suy ra B là trung điểm của AE
\(\Rightarrow d\left [ B;(SCE) \right ]=\frac{1}{2}d\left [ A;(SCE) \right ]\)
\(\left\{\begin{matrix} AH\perp CE\\AK\perp SH \end{matrix}\right.\Rightarrow AK\perp (SCE)\)
(Vì \(\left\{\begin{matrix} AK\perp SH\\AK\perp CE \end{matrix}\right.\; do\; CE\perp (SAH)\Rightarrow d\left [ A;(SCE) \right ]=AK\)
+ \(AH=\frac{4a}{\sqrt{5}}\)
+ \(\triangle SAH\) vuông tại A, AK là đường cao \(\Rightarrow \frac{1}{AK^{2}}=\frac{1}{SA^{2}}+\frac{1}{AH^{2}}\)
\(EC\cap AD=F\Rightarrow EF=2BD=2a\sqrt{5};AF=4A\)
\(\Rightarrow AH=\frac{AE.AF}{EF}=\frac{2a.4a}{2a\sqrt{5}}=\frac{4a}{\sqrt{5}}\Rightarrow AK=\frac{4a}{\sqrt{21}}\Rightarrow d(BD;SC)=\frac{2a}{\sqrt{21}}\)
Vẽ \(AP\perp SB\Rightarrow AP\perp (SBC)(do\; CB\perp (SAB))\)
Vẽ \(AQ\perp SD\Rightarrow AQ\perp (SCD)\)
⇒ Góc giữa hai mặt phẳng (SCB) và (SCD) bằng góc (AP; AQ)
+ \(AP=\frac{a\sqrt{2}}{2};AQ=\frac{2a}{\sqrt{5}}\)
+ \(SP=\frac{a\sqrt{2}}{2};SQ=\frac{a}{\sqrt{5}}\Rightarrow PQ^{2}=SP^{2}+SQ^{2}-2SP.SQ\cos BSD\)
Mà \(\cos BSD=\frac{SB^{2}+SD^{2}-BD^{2}}{2SB.SD}=\frac{2a^{2}}{2a\sqrt{2}.a\sqrt{5}}=\frac{1}{\sqrt{10}}\)
\(\Rightarrow PQ^{2}=\frac{2a^{2}}{4}+\frac{a^{2}}{5}-2\frac{a\sqrt{2}}{2}.\frac{a}{\sqrt{5}}.\frac{1}{\sqrt{10}}=\frac{a^{2}}{2}\)
+ \(\cos \left ( AP;AQ \right )=\left | \cos PAQ \right |=\left | \frac{AP^{2}+AQ^{2}-PQ^{2}}{2AP.AQ} \right |\)
\(=\left | \frac{\frac{2a^{2}}{4}+\frac{4a^{2}}{5}-\frac{2a^{2}}{4}}{2.\frac{a\sqrt{2}}{2}.\frac{2a}{\sqrt{5}}} \right |=\frac{4}{5}.\frac{\sqrt{10}}{4}=\frac{\sqrt{10}}{5}=\left | \frac{\frac{2a^{2}}{4}+\frac{4a^{2}}{5}-\frac{2a^{2}}{4}}{2.\frac{a\sqrt{2}}{2}.\frac{2a}{\sqrt{5}}} \right |=\frac{4}{5}.\frac{\sqrt{10}}{4}=\frac{\sqrt{10}}{5}\)
