共50条信息
如图所示,在正方体\(ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}\)中,过定点\(A\)在空间作直线\(l\),使\(l\)与直线\(AC\)和\(B{{C}_{1}}\)所成角都等于\(60{}^\circ \),这样的直线\(l\)可以作( )条
已知\(P\)是四边形\(ABCD\)所在平面外一点,\(PA\)\(=\)\(PB\)\(=\)\(PD\),在四边形\(ABCD\)中\(BA\)\(=\)\(AD\),\(BA\)\(⊥\)\(AD\),\(O\)是\(BD\)的中点,\(OC\)\(=\dfrac{1}{2}OA=\dfrac{1}{3}OP\).
\((1)\)求证:\(PD\)\(⊥\)\(AC\);
\((2)\)若\(E\)是\(PD\)的中点,求平面\(EAC\)将四棱锥\(P−ABCD\)分成两部分的体积之比.
\(21.\)四面体\(ABCD\)及其三视图如图所示,过棱\(AB\)的中点\(E\)作平行于\(AD\),\(BC\)的平面分别交四面体的棱\(BD\),\(DC\),\(CA\)于点\(F\),\(G\),\(H\).
\((1)\)证明:四边形\(EFGH\)是矩形;
\((2)\)求直线\(AB\)与平面\(EFGH\)夹角\(θ\)的正弦值.
如图,四棱柱\(ABCD-A′B′C′D′\)中,侧棱\(AA′⊥ABCD\),\(AB/\!/DC\),\(AB⊥AD\),\(AD=CD=1\),\(AA′=AB=2\),\(E\)为棱\(AA′\)的中点.
\((1)\)求证:\(B′C′⊥CE\);
\((2)\)求二面角\(B′-CE-C′\)的余弦值;\((3)\)设点\(M\)在线段\(C′E\)上,且直线\(AM\)与平面\(ADD′A′\)所成角的正弦值为,求线段\(AM\)的长.
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