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点面距的求解

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前言

求解策略

例题暂缺;

【2022届高三文科用题】已知直三棱柱\(A_{1}B_{1}C_{1}-ABC\)中,\(AB=AC=AA_{1}=1\),\(\angle BAC=90^{\circ}\).

(1).求异面直线\(A_{1}B\)与\(B_{1}C_{1}\)所成角;

解: 在直三棱柱\(A_{1}B_{1}C_{1}-ABC\)中,\(AA_{1}\perp AB\),\(AA_{1}\perp AC\),\(AB=AC=AA_{1}=1\),\(\angle BAC=90^{\circ}\)

所以,\(A_{1}B=A_{1}C=BC=\sqrt{2}\)

因为,\(BC//B_{1}C_{1}\),所以\(\angle A_{1}BC\)为异面直线 \(A_{1}B\) 与 \(B_{1}C_{1}\) 所成的角或补角.

在\(\triangle A_{1}BC\)中,因为\(A_{1}B=A_{1}C=BC=\sqrt{2}\),

所以,异面直线\(A_{1}B\)与\(B_{1}C_{1}\)所成角为\(\cfrac{\pi}{3}\).

(2).求点\(B_{1}\)到平面\(A_{1}BC\)的距离.

解:设点\(B_{1}\)到平面\(A_{1}BC\)的距离为\(h\),

由(1)得\(S_{\Delta ABC}=\cfrac{1}{2}\times\sqrt{2}\times\sqrt{2}\cdot\sin\cfrac{\pi}{3}\)

\(=\cfrac{\sqrt{3}}{2}\)\(S_{\triangle AB_{1}B}=\cfrac{1}{2}\times1\times1=\cfrac{1}{2}\),

因为,\(V_{B_{1}-ABC}=V_{C-A_{1}B_1B}\),

所以,\(\cfrac{1}{3}S_{\Delta ABC}\cdot h=\cfrac{1}{3}S_{\Delta A_{1}B_{1}B}\cdot CA\),解得,\(h=\cfrac{\sqrt{3}}{3}\).

所以,点\(B_{1}\)到平面\(A_{1}BC\)的距离为\(\cfrac{\sqrt{3}}{3}\).

标签:AA,ABC,BC,点面,sqrt,cfrac,AB,求解
来源: https://www.cnblogs.com/wanghai0666/p/15415936.html