优优班--学霸训练营 > 知识点挑题
全部资源
          排序:
          最新 浏览

          50条信息

            • 1. 直三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(∠BCA=90^{\circ}\),\(M\),\(N\)分别是\(A_{1}B_{1}\),\(A_{1}C_{1}\)的中点,\(BC=CA=CC_{1}\),则\(BM\)与\(AN\)所成角的余弦值为\((\)  \()\)
              A.\( \dfrac {1}{10}\)
              B.\( \dfrac {2}{5}\)
              C.\( \dfrac { \sqrt {30}}{10}\)
              D.\( \dfrac { \sqrt {2}}{2}\)
            • 2.
              如图,在直四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,底面\(ABCD\)是边长为\(2\)的正方形,\(E\),\(F\)分别为线段\(DD_{1}\),\(BD\)的中点.
              \((1)\)求证:\(EF/\!/\)平面\(ABC_{1}D_{1}\);
              \((2)\)四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)的外接球的表面积为\(16π\),求异面直线\(EF\)与\(BC\)所成的角的大小.
            • 3.
              如图,在棱长为\(3\)的正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(E\),\(F\)分别在棱\(AB\),\(CD\)上,且\(AE=CF=1\).
              \((1)\)求异面直线\(A_{1}E\)与\(C_{1}F\)所成角的余弦值.
              \((2)\)求四面体\(EFC_{1}A_{1}\)的体积.
            • 4.
              已知四棱锥\(S-ABCD\)的底面是正方形,侧棱长均相等,\(E\)是线段\(AB\)上的点\((\)不含端点\().\)设\(SE\)与\(BC\)所成的角为\(θ_{1}\),\(SE\)与平面\(ABCD\)所成的角为\(θ_{2}\),二面角\(S-AB-C\)的平面角为\(θ_{3}\),则\((\)  \()\)
              A.\(θ_{1}\leqslant θ_{2}\leqslant θ_{3}\)
              B.\(θ_{3}\leqslant θ_{2}\leqslant θ_{1}\)
              C.\(θ_{1}\leqslant θ_{3}\leqslant θ_{2}\)
              D.\(θ_{2}\leqslant θ_{3}\leqslant θ_{1}\)
            • 5.
              设正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)的棱长为\(2\),动点\(E\),\(F\)在棱\(A_{1}B_{2}\)上,动点\(P\)、\(Q\)分别在棱\(AD\)、\(CD\)上,若\(EF=1\),\(A_{1}E=x\),\(DQ=y\),\(DP=Z\),\((x,y,z > 0)\),则下列结论中正确的是 ______ .
              \(①EF/\!/\)平面\(DPQ\);
              \(②\)三菱锥\(P\)---\(EFQ\)的体积与\(Y\)的变化有关,与\(x\)、\(z\)的变化无关;
              \(③\)异面直线\(EQ\)和\(AD_{1}\)所成角的大小与\(x\)、\(y\)、\(z\)的变化无关.
            • 6.
              如图所示,在正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,已知\(M\),\(N\)分别是\(BD\)和\(AD\)的中点,则\(B_{1}M\)与\(D_{1}N\)所成角的余弦值为\((\)  \()\)
              A.\( \dfrac { \sqrt {30}}{10}\)
              B.\( \dfrac { \sqrt {30}}{15}\)
              C.\( \dfrac { \sqrt {30}}{30}\)
              D.\( \dfrac { \sqrt {15}}{15}\)
            • 7.
              如图,在长方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(AB=2\),\(BC=BB_{1}=1\),\(P\)为\(A_{1}C\)的中点,则异面直线\(BP\)与\(AD_{1}\)所成角的余弦值为\((\)  \()\)
              A.\( \dfrac {1}{3}\)
              B.\( \dfrac { \sqrt {6}}{4}\)
              C.\( \dfrac { \sqrt {2}}{3}\)
              D.\( \dfrac { \sqrt {3}}{3}\)
            • 8.
              正四棱锥的侧棱长为\( \sqrt {2}\),底面的边长为\( \sqrt {3}\),\(E\)是\(PA\)的中点,则异面直线\(BE\)与\(PC\)所成的角为\((\)  \()\)
              A.\( \dfrac {π}{6}\)
              B.\( \dfrac {π}{4}\)
              C.\( \dfrac {π}{3}\)
              D.\( \dfrac {π}{2}\)
            • 9.
              在直角梯形\(ABCD\)中,\(AD/\!/BC\),\(AB⊥AD\),\(E\),\(F\)分别是\(AB\),\(AD\)的中点,\(PF⊥\)平面\(ABCD\),且\(AB=BC=PF= \dfrac {1}{2}AD=2\),则异面直线\(PE\),\(CD\)所成的角为\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(45^{\circ}\)
              C.\(60^{\circ}\)
              D.\(90^{\circ}\)
            • 10.
              如图\(1\),在\(\triangle ABC\)中,\(D\),\(E\)分别为\(AB\),\(AC\)的中点,\(O\)为\(DE\)的中点,\(AB=AC=2 \sqrt {5}\),\(BC=4.\)将\(\triangle ADE\)沿\(DE\)折起到\(\triangle A_{1}DE\)的位置,使得平面\(A_{1}DE⊥\)平面\(BCED\),如图\(2\).
              \((\)Ⅰ\()\)求证:\(A_{1}O⊥BD\);
              \((\)Ⅱ\()\)求直线\(A_{1}C\)和平面\(A_{1}BD\)所成角的正弦值;
              \((\)Ⅲ\()\)线段\(A_{1}C\)上是否存在点\(F\),使得直线\(DF\)和\(BC\)所成角的余弦值为\( \dfrac { \sqrt {5}}{3}\)?若存在,求出\( \dfrac {A_{1}F}{A_{1}C}\)的值;若不存在,说明理由.
            0/40

            进入组卷