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    18．解法一: 依题设知.． (Ⅰ)连结交于点.则． 由三垂线定理知.．····················································································· 3分 在平面内.连结交于点. 由于. 故.. 与互余． 于是． 与平面内两条相交直线都垂直. 所以平面．······························································································· 6分 (Ⅱ)作.垂足为.连结．由三垂线定理知. 故是二面角的平面角．································································· 8分 . .． .． 又.． ． 所以二面角的大小为．·························································· 12分 解法二: 以为坐标原点.射线为轴的正半轴. 建立如图所示直角坐标系． 依题设.． . ．················································································ 3分 (Ⅰ)因为.. 故.． 又. 所以平面．································································································ 6分 (Ⅱ)设向量是平面的法向量.则 .． 故.． 令.则..．······························································ 9分 等于二面角的平面角. ． 所以二面角的大小为．························································· 12分 【查看更多】

     

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