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2023年高考数学新高考Ⅱ-20

  2023-07-08 11:20:35  

(12分)如图,三棱锥$A-BCD$中,$DA=DB=DC$,$BD\bot CD$,$\angle ADB=\angle ADC=60^\circ$,$E$为$BC$中点.
(1)证明$BC\bot DA$;
(2)点$F$满足$\overrightarrow{EF}=\overrightarrow{DA}$,求二面角$D-AB-F$的正弦值.

分析:(1)根据已知条件,推得$DE\bot BC$,$AE\bot BC$,再结合线面垂直的判定定理,即可求证.
(2)根据已知条件,推得$AE\bot$平面$BCD$,依次求出两个平面的法向量,再结合向量的夹角公式,即可求解.
证明:(1)连接$AE$,$DE$,
$\because DB=DC$,$E$为$BC$中点.
$\therefore DE\bot BC$,
又$\because DA=DB=DC$,$\angle ADB=\angle ADC=60^\circ$,
$\therefore \Delta ACD$与$\Delta ABD$ 均为等边三角形,
$\therefore AC=AB$,
$\therefore AE\bot BC$,$AE\bigcap DE=E$,
$\therefore BC\bot$平面$ADE$,
$\because AD\subset$平面$ADE$,
$\therefore BC\bot DA$.
(2)解:设$DA=DB=DC=2$,
$\therefore$$BC=2\sqrt{2}$,
$\because$$DE=AE=\sqrt{2}$,$AD=2$,
$\therefore AE^{2}+DE^{2}=4=AD^{2}$,
$\therefore AE\bot DE$,
又$\because AE\bot BC$,$DE\bigcap BC=E$,
$\therefore AE\bot$平面$BCD$,
以$E$为原点,建立如图所示空间直角坐标系,

$D(\sqrt{2},0,0)$,$A(0,0,\sqrt{2})$,$B(0,\sqrt{2},0)$,$E(0$,0,$0)$,
$\because$$\overrightarrow{EF}=\overrightarrow{DA}$,
$\therefore$$F(-\sqrt{2},0,\sqrt{2})$,
$\therefore$$\overrightarrow{DA}=(-\sqrt{2},0,\sqrt{2})$,$\overrightarrow{AB}=(0,\sqrt{2},-\sqrt{2})$,$\overrightarrow{AF}=(-\sqrt{2},0,0)$,
设平面$DAB$与平面$ABF$的一个法向量分别为$\overrightarrow{{n}_{1}}=({x}_{1},{y}_{1},{z}_{1})$,$\overrightarrow{n_2}=(x_2,y_2,z_2)$,
则$\left\{\begin{array}{l}{-\sqrt{2}{x}_{1}+\sqrt{2}{z}_{1}=0}\\ {\sqrt{2}{y}_{1}-\sqrt{2}{z}_{1}=0}\end{array}\right.$,令$x_{1}=1$,解得$y_{1}=z_{1}=1$,
$\left\{\begin{array}{l}{\sqrt{2}{y}_{2}-\sqrt{2}{z}_{2}=0}\\ {-\sqrt{2}{x}_{2}=0}\end{array}\right.$,令$y_{2}=1$,解得$x_{2}=0$,$z_{2}=1$,
故$\overrightarrow{{n}_{1}}=(1$,1,$1)$,$\overrightarrow{{n}_{2}}=(0$,1,$1)$,
设二面角$D-AB-F$的平面角为$\theta$,
则$\vert \cos \theta \vert =\dfrac{\vert \overrightarrow{{n}_{1}}\cdot \overrightarrow{{n}_{2}}\vert }{\vert \overrightarrow{{n}_{1}}\vert \;\vert \overrightarrow{{n}_{2}}\vert }=\dfrac{2}{\sqrt{3}\times \sqrt{2}}=\dfrac{\sqrt{6}}{3}$,
故$\sin \theta =\dfrac{\sqrt{3}}{3}$,
所以二面角$D-AB-F$的正弦值为$\dfrac{\sqrt{3}}{3}$.
点评:本题主要考查二面角的平面角及其求法,考查转化能力,属于中档题.

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