19. In the quadrangular pyramid $Q - A B C D$ , the base $A B C D$ is a square with $A D = 2$ , $Q D = Q A = \sqrt { 5 }$ , $Q C = 3$ . [Figure] (1) Prove: plane $Q A D \perp$ plane $A B C D$ ; (2) Find the cosine of the dihedral angle $B - Q D - A$ . Answer: (1) See proof below; (2) $\frac { 2 } { 3 }$ .
[Solution]
[Analysis] (1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ . We can prove that $Q O \perp$ plane $A B C D$ , thus obtaining plane $Q A D \perp$ plane $A B C D$ . (2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ . Establish a coordinate system as shown in the figure. After finding the normal vectors of planes $Q A D$ and $B Q D$ , we can find the cosine of the dihedral angle.
[Detailed Solution]
[Figure] (1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ . Since $Q A = Q D$ and $O A = O D$ , we have $Q O \perp A D$ . Since $A D = 2$ and $Q A = \sqrt { 5 }$ , we have $Q O = \sqrt { 5 - 1 } = 2$ . In square $A B C D$ , since $A D = 2$ , we have $D O = 1$ , thus $C O = \sqrt { 1 + 4 } = \sqrt { 5 }$ . Since $Q C = 3$ , we have $Q C ^ { 2 } = 9 = 4 + 5 = Q O ^ { 2 } + O C ^ { 2 }$ , so $\triangle Q O C$ is a right triangle with $Q O \perp O C$ . Since $O C \cap A D = O$ , we have $Q O \perp$ plane $A B C D$ . Since $Q O \subset$ plane $Q A D$ , we have plane $Q A D \perp$ plane $A B C D$ . (2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ . Combined with $Q O \perp$ plane $A B C D$ from part (1), we can establish a coordinate system as shown in the figure. [Figure] Then $D ( 0,1,0 ) , Q ( 0,0,2 ) , B ( 2 , -1,0 )$ , so $\overrightarrow { B Q } = ( -2,1,2 ) , \overrightarrow { B D } = ( -2,2,0 )$ . Let the normal vector of plane $Q B D$ be $\vec { n } = ( x , y , z )$ . Then $\left\{ \begin{array} { l } \vec { n } \cdot \overrightarrow { B Q } = 0 \\ \vec { n } \cdot \overrightarrow { B D } = 0 \end{array} \right.$ , i.e., $\left\{ \begin{array} { l } - 2 x + y + 2 z = 0 \\ - 2 x + 2 y = 0 \end{array} \right.$ . Taking $x = 1$ , we get $y = 1 , z = \frac { 1 } { 2 }$ , Thus $\vec { n } = \left( 1,1 , \frac { 1 } { 2 } \right)$ . The normal vector of plane $Q A D$ is $\vec { m } = ( 1,0,0 )$ . Therefore $\cos \langle \vec { m } , \vec { n } \rangle = \frac { |\vec{m} \cdot \vec{n}| } { |\vec{m}| \cdot |\vec{n}| } = \frac
19. In the quadrangular pyramid $Q - A B C D$ , the base $A B C D$ is a square with $A D = 2$ , $Q D = Q A = \sqrt { 5 }$ , $Q C = 3$ .\\
\includegraphics[max width=\textwidth, alt={}, center]{3deb634f-22c0-49a5-8f06-4bdfee8ebbe0-13_383_399_1260_328}\\
(1) Prove: plane $Q A D \perp$ plane $A B C D$ ;\\
(2) Find the cosine of the dihedral angle $B - Q D - A$ .\\
\textbf{Answer:} (1) See proof below; (2) $\frac { 2 } { 3 }$ .
\section*{[Solution]}
[Analysis] (1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ . We can prove that $Q O \perp$ plane $A B C D$ , thus obtaining plane $Q A D \perp$ plane $A B C D$ .\\
(2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ . Establish a coordinate system as shown in the figure. After finding the normal vectors of planes $Q A D$ and $B Q D$ , we can find the cosine of the dihedral angle.
\section*{[Detailed Solution]}
\includegraphics[max width=\textwidth, alt={}, center]{3deb634f-22c0-49a5-8f06-4bdfee8ebbe0-14_700_737_264_508}\\
(1) Let $O$ be the midpoint of $A D$ , and connect $Q O$ and $C O$ .\\
Since $Q A = Q D$ and $O A = O D$ , we have $Q O \perp A D$ .\\
Since $A D = 2$ and $Q A = \sqrt { 5 }$ , we have $Q O = \sqrt { 5 - 1 } = 2$ .\\
In square $A B C D$ , since $A D = 2$ , we have $D O = 1$ , thus $C O = \sqrt { 1 + 4 } = \sqrt { 5 }$ .\\
Since $Q C = 3$ , we have $Q C ^ { 2 } = 9 = 4 + 5 = Q O ^ { 2 } + O C ^ { 2 }$ , so $\triangle Q O C$ is a right triangle with $Q O \perp O C$ .\\
Since $O C \cap A D = O$ , we have $Q O \perp$ plane $A B C D$ .\\
Since $Q O \subset$ plane $Q A D$ , we have plane $Q A D \perp$ plane $A B C D$ .\\
(2) In plane $A B C D$ , through $O$ draw $O T \parallel C D$ , intersecting $B C$ at $T$ . Then $O T \perp A D$ .\\
Combined with $Q O \perp$ plane $A B C D$ from part (1), we can establish a coordinate system as shown in the figure.\\
\includegraphics[max width=\textwidth, alt={}, center]{3deb634f-22c0-49a5-8f06-4bdfee8ebbe0-15_758_784_300_339}
Then $D ( 0,1,0 ) , Q ( 0,0,2 ) , B ( 2 , -1,0 )$ , so $\overrightarrow { B Q } = ( -2,1,2 ) , \overrightarrow { B D } = ( -2,2,0 )$ .\\
Let the normal vector of plane $Q B D$ be $\vec { n } = ( x , y , z )$ .\\
Then $\left\{ \begin{array} { l } \vec { n } \cdot \overrightarrow { B Q } = 0 \\ \vec { n } \cdot \overrightarrow { B D } = 0 \end{array} \right.$ , i.e., $\left\{ \begin{array} { l } - 2 x + y + 2 z = 0 \\ - 2 x + 2 y = 0 \end{array} \right.$ . Taking $x = 1$ , we get $y = 1 , z = \frac { 1 } { 2 }$ ,\\
Thus $\vec { n } = \left( 1,1 , \frac { 1 } { 2 } \right)$ .\\
The normal vector of plane $Q A D$ is $\vec { m } = ( 1,0,0 )$ . Therefore $\cos \langle \vec { m } , \vec { n } \rangle = \frac { |\vec{m} \cdot \vec{n}| } { |\vec{m}| \cdot |\vec{n}| } = \frac