高二数学必修5,与选修2-1总测试题

BNANMCANN有此时能使点坐标为时可知当?ANBMCBNMCANMCBNMCAN???????所以得由.,0,0为所求二面角的平面角30 30 4| | ,| | , .5 5 52cos( , ) .3| | | |os( ).3AN BN AN BNAN BNAN BNAN BN? ? ??? ????????? ?????????????????????????????????????故所求的二面角为3如图,在四棱锥P ABCD?中,底面ABCD为矩形,侧棱PA?底面ABCD,3AB?,1BC?,2PA?,E为PD的中点(Ⅰ)求直线AC与PB所成角的余弦值;(Ⅱ)在侧面PAB内找一点N,使NE?面PAC,并求出点N到AB和AP的距离解:(Ⅰ)建立如图所示的空间直角坐标系,则, , , , ,A B C D P E的坐标为(0, 0, 0)A、( 3, 0, 0)B、( 3,1, 0)C、(0,1, 0)D、(0, 0, 2)P、1(0, ,1)2E,从而).2,0,3(),0,1,3(???PBAC设PBAC与的夹角为?,则,1473723||||cos?????PBACPBAC?∴AC与PB所成角的余弦值为1473(Ⅱ)由于N点在侧面PAB内,故可设N点坐标为( , 0, )x z,则)1,21,(zxNE???,由NE?面PAC可得,??????????????????????????????????.0213,01.0)0,1,3()1,21,(,0)2,0,0()1,21,(.0,EAPNE化简得即∴???????163zx即N点的坐标为)1,0,63(,从而N点到AB和AP的距离分别为31,6邯郸市荀子中学高二数学期末测试题胡明明9