gaokao 2015 Q17
11 marks
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17. (11 points)
A student uses the ``five-point method'' to sketch the graph of $f(x) = A\sin(\omega x + \varphi)$ (where $\omega > 0$, $|\varphi| < \frac{\pi}{2}$) over one period and creates a table with partial data filled in as follows:
\begin{tabular}{ | c | c | c | c | c | c | } \hline $\omega x + \varphi$ & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3\pi}{2}$ & $2\pi$ \hline $x$ & & $\frac Thus $\overrightarrow { P B } \cdot \overrightarrow { D E } = 0$, that is, $P B \perp D E$. Since $E F \perp P B$ and $D E \cap E F = E$, we have $P B \perp$ plane $D E F$. Because $\overrightarrow { P C } = ( 0,1 , - 1 ) , \overrightarrow { D E } \cdot \overrightarrow { P C } = 0$, then $D E \perp P C$, so $D E \perp$ plane $P B C$. From $D E \perp$ plane $P B C$ and $P B \perp$ plane $D E F$, we know that all four faces of tetrahedron $B D E F$ are right triangles, that is, tetrahedron $B D E F$ is an orthocentric tetrahedron, with right angles on its four faces being $\angle D E B , \angle D E F , \angle E F B , \angle D F B$ respectively.
[Figure]
Solution diagram 1 for Question 19
[Figure]
Solution diagram 2 for Question 19
(II) Since $P D \perp$ plane $A B C D$, we have $\overrightarrow { D P } = ( 0,0,1 )$ is a normal vector to plane $A B C D$; From (I), $P B \perp$ plane $D E F$, so $\overrightarrow { B P } = ( - \lambda , - 1,1 )$ is a normal vector to plane $D E F$. If the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, then $\cos \frac { \pi } { 3 } = \left| \frac { \overrightarrow { B P } \cdot \overrightarrow { D P } } { | \overrightarrow { B P } | \cdot | \overrightarrow { D P } | } \right| = \left| \frac { 1 } { \sqrt { \lambda ^ { 2 } + 2 } } \right| = \frac { 1 } { 2 }$, solving gives $\lambda = \sqrt { 2 }$. Therefore $\frac { D C } { B C } = \frac { 1 } { \lambda } = \frac { \sqrt { 2 } } { 2 }$. Thus when the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, we have $\frac { D C } { B C } = \frac { \sqrt { 2 } } { 2 }$.