Bài 4: Hai mặt phẳng vuông góc

\(\overrightarrow{BD'}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DD'}=-\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}\)

\(=-\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\)

\(AC=2OA=2a\sqrt{2}\) \(\Rightarrow AB=\dfrac{AC}{\sqrt{2}}=2a\)

Gọi M là trung điểm BC \(\Rightarrow\widehat{SMO}\) là góc giữa mặt bên và đáy

\(OM=\dfrac{1}{2}AB=a\)

\(\Rightarrow tan\widehat{SMO}=\dfrac{SO}{OM}=\sqrt{3}\Rightarrow\widehat{SMO}=60^0\)

C

\(A'C'||AC\Rightarrow\) góc cần tìm là góc \(\widehat{CAB'}\)

Mặt khác \(AB'=AC=B'C\) (các đường chéo của hình vuông bằng nhau)

\(\Rightarrow\Delta AB'C\) đều

\(\Rightarrow\widehat{CAB'}=60^0\)

A

\(S_{\Delta ACD}=\dfrac{1}{2}AC.AD.sin\widehat{CAD}=\dfrac{a^2\sqrt{3}}{4}\)

\(V=\dfrac{AB.AC.AD}{6}.\sqrt{1+2cos90^0.cos60^0.cos120^0-cos^290^0-cos^260^0-cos^2120^0}=\dfrac{a^3\sqrt{2}}{12}\)

\(\Rightarrow d\left(B;\left(ACD\right)\right)=\dfrac{3V}{S}=\dfrac{a\sqrt{6}}{3}\)

B

\(\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NC}+\overrightarrow{BM}+\overrightarrow{MN}+\overrightarrow{ND}\)

\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{BM}\right)+\left(\overrightarrow{NC}+\overrightarrow{ND}\right)\)

\(=2\overrightarrow{MN}\)

\(\Rightarrow k=\dfrac{1}{2}\)

A

A

A

D

Ta có: \(CD'||A'B\)

Mà \(A'B\perp AB'\) (hai đường chéo hv)

\(\Rightarrow AB'\perp CD'\)

\(V=\dfrac{a.a\sqrt{3}.a\sqrt{2}}{6}.\sqrt{1+2cos90^0.cos60^0.cos120^0-cos^290-cos^260-cos^2120}=\dfrac{a^3\sqrt{3}}{6}\)

A

A

\(\overrightarrow{x}=2\overrightarrow{a}+\overrightarrow{b}=2\left(\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}\right)-\left(-3\overrightarrow{b}-2\overrightarrow{c}\right)=2\overrightarrow{y}-\overrightarrow{z}\)

\(\Rightarrow\) 3 vecto đã cho đồng phẳng