共50条信息
如图,在四棱锥\(P—ABCD\)中,四边形\(ABCD\)是直角梯形,\(AB⊥AD\),\(AB/\!/CD\),\(PC⊥\)底面\(ABCD\),\(AB=2AD=2CD=4\),\(PC=2a\),\(E\)是\(PB\)的中点.
\((1)\)求证:\(AC⊥\)平面\(PBC\);
\((2)\)若二面角\(P—AC—E\)的余弦值为\(\dfrac{\sqrt{6}}{3}\),求直线\(PA\)与平面\(EAC\)所成角的正弦值.
如图,正四棱柱\((\)底面为正方形,侧棱垂直于底面\()ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(AA_{1}=2AB\),则异面直线\(A_{1}B\)与\(AD_{1}\)所成角的余弦值为\((\) \()\)
如下图,直四棱柱\(ABCD—A_{1}B_{1}C_{1}D_{1}\)中,底面\(ABCD\)为等腰梯形,\(AD=2BC=2CD=4\),\(A{{A}_{1}}=2\sqrt{3}\).
\((\)Ⅰ\()\)证明:\(AD_{1}⊥B_{1}D\);
\((\)Ⅱ\()\)设\(E\)是线段\(A_{1}B_{1}\)上的动点,是否存在这样的点\(E\),使得二面角\(E—BD_{1}—A\)的余弦值为\(\dfrac{\sqrt{7}}{7}\),如果存在,求出\(B_{1}E\)的长;如果不存在,请说明理由.
如图,在四棱锥\(P-ABCD\)中,\(PA⊥\)平面\(ABCD\),\(PA=AB=AD=2\),四边形\(ABCD\)满足\(AB⊥AD\),\(BC/\!/AD\)且\(BC =4\),点\(M\)为\(PC\)的中点,点\(E\)为\(BC\)边上的动点,且\(\dfrac{BE}{EC}=\lambda \).
\((\)Ⅱ\()\)是否存在实数\(\lambda \),使得二面角\(P-DE-B\)的余弦值为\(\dfrac{2}{3}\)?若存在,试求出实数\(\lambda \)的值;若不存在,说明理由.
如图,在三棱柱\(ABC-{A}_{1}{B}_{1}{C}_{1} \)中,\(CA=AB=A{A}_{1} \),\(∠BA{A}_{1}=∠BAC={60}^{^{\circ}} \),点 \(O\)是线段 \(AB\) 的中点.
\((1)\)证明:\(B{C}_{1} /\!/\)平面\(O{A}_{1}C \);
\((2)\)若 \(AB=2\),\({A}_{1}C= \sqrt{6} \),求二面角\(A-BC-{A}_{1} \) 的余弦值.
如图,正方形\(ABCD\)与等边三角形\(ABE\)所在的平面互相垂直,\(M,N\)分别是\(DE,AB\)的中点.
\((1)\)证明:\(MN/\!/\)平面\(BCE\);
\((2)\)求锐二面角\(M-AB-E\)的余弦值.
已知正四棱锥\(P-ABCD\)中,\(PA=AB=2,E,F\)分别是\(PB,PC\)的中点,则异面直线\(AE\)与\(BF\)所成角的余弦值为\((\) \()\)
如图,四棱柱\(ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}\)的底面\(ABCD\)平面是平行四边形,且\(AB=1\),\(BC=2,∠ABC=60^{\circ} \),点\(E\)为\(BC\)的中点,\(AA_{1}⊥ \)平面\(ABCD\).
如图,矩形\(ABCD\)所在平面与三角形\(ECD\)所在平面相交于\(CD,AE\bot \)平面\(ECD\), \(2AE=AB=DE=2\),点\(M\)在线段\(AE\)上,\(N\)为线段\(CD\)的中点.
\((1)\)证明\(AB\bot \)平面\(ADE\);
\((2)\)若\(M\)为\(AE\)中点,求平面\(BDM\)与平面\(BNE\)所成锐二面角的余弦值;
\((3)\)若\(EN/\!/\)平面\(BDM\),求\(MN\)的长.
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