2023年高考数学新高考Ⅰ-18 |
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2023-07-08 11:03:38 |
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(12分)如图,在正四棱柱$ABCD-A_{1}B_{1}C_{1}D_{1}$中,$AB=2$,$AA_{1}=4$.点$A_{2}$,$B_{2}$,$C_{2}$,$D_{2}$分别在棱$AA_{1}$,$BB_{1}$,$CC_{1}$,$DD_{1}$上,$AA_{2}=1$,$BB_{2}=DD_{2}=2$,$CC_{2}=3$. (1)证明:$B_{2}C_{2}//A_{2}D_{2}$; (2)点$P$在棱$BB_{1}$上,当二面角$P-A_{2}C_{2}-D_{2}$为$150^\circ$时,求$B_{2}P$.
分析:(1)建系,根据坐标法及向量共线定理,即可证明; (2)建系,根据向量法,向量夹角公式,方程思想,即可求解. 解:
(1)证明:根据题意建系如图,则有: $B_{2}(0$,2,$2)$,$C_{2}(0$,0,$3)$,$A_{2}(2$,2,$1)$,$D_{2}(2$,0,$2)$, $\therefore$$\overrightarrow{{B}_{2}{C}_{2}}=(0,-2,1)$,$\overrightarrow{{A}_{2}{D}_{2}}=(0,-2,1)$, $\therefore$$\overrightarrow{{B}_{2}{C}_{2}}=\overrightarrow{{A}_{2}{D}_{2}}$,又$B_{2}$,$C_{2}$,$A_{2}$,$D_{2}$四点不共线, $\therefore B_{2}C_{2}//A_{2}D_{2}$; (2)在(1)的坐标系下,可设$P(0$,2,$t)$,$t\in [0$,$4]$, 又由(1)知$C_{2}(0$,0,$3)$,$A_{2}(2$,2,$1)$,$D_{2}(2$,0,$2)$, $\therefore$$\overrightarrow{{C}_{2}{A}_{2}}=(2,2,-2)$,$\overrightarrow{{C}_{2}P}=(0,2,t-3)$,$\overrightarrow{{A}_{2}{D}_{2}}=(0,-2,1)$, 设平面$PA_{2}C_{2}$的法向量为$\overrightarrow{m}=(x,y,z)$, 则$\left\{\begin{array}{l}{\overrightarrow{m}\cdot \overrightarrow{{C}_{2}{A}_{2}}=2x+2y-2z=0}\\ {\overrightarrow{m}\cdot \overrightarrow{{C}_{2}P}=2y+(t-3)z=0}\end{array}\right.$,取$\overrightarrow{m}=(t-1,3-t,2)$, 设平面$A_{2}C_{2}D_{2}$的法向量为$\overrightarrow{n}=(a,b,c)$, 则$\left\{\begin{array}{l}{\overrightarrow{n}\cdot \overrightarrow{{C}_{2}{A}_{2}}=2a+2b-2c=0}\\ {\overrightarrow{n}\cdot \overrightarrow{{A}_{2}{D}_{2}}=-2b+c=0}\end{array}\right.$,取$\overrightarrow{n}=(1,1,2)$, $\therefore$根据题意可得$\vert \cos 150^\circ \vert =\vert \cos < \overrightarrow{m}$,$\overrightarrow{n} > \vert =\dfrac{\vert \overrightarrow{m}\cdot \overrightarrow{n}\vert }{\vert \overrightarrow{m}\vert \vert \overrightarrow{n}\vert }$, $\therefore$$\dfrac{\sqrt{3}}{2}=\dfrac{6}{\sqrt{(t-1)^{2}+(3-t)^{2}+4}\times \sqrt{6}}$, $\therefore t^{2}-4t+3=0$,又$t\in [0$,$4]$, $\therefore$解得$t=1$或$t=3$, $\therefore P$为$B_{1}B_{2}$的中点或$B_{2}B$的中点, $\therefore B_{2}P=1$. 点评:本题考查利用向量法证明线线平行,利用向量法求解二面角问题,向量共线定理及向量夹角公式的应用,方程思想,属中档题.
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