优优班--学霸训练营 > 知识点挑题
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            • 1.

              在正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(E\)是\(AD\)的中点,则异面直线\(C_{1}E\)与\(BC\)所成的角的余弦值是\((\)  \()\)
              A.\( \dfrac { \sqrt {10}}{5}\)
              B.\( \dfrac { \sqrt {10}}{10}\)
              C.\( \dfrac {1}{3}\)
              D.\( \dfrac {2 \sqrt {2}}{3}\)
            • 2.
              在棱长为\(3\)的正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,点\(E\),\(F\)分别在棱\(A_{1}B_{1}\),\(C_{1}D_{1}\)上且\(A_{1}E=1\),\(C_{1}F=1\),则异面直线\(AE\),\(B_{1}F\)所成角的余弦值为\((\)  \()\)
              A.\( \dfrac {3}{10}\)
              B.\( \dfrac {1}{9}\)
              C.\( \dfrac {1}{11}\)
              D.\(0\)
            • 3.
              如图,在直角梯形\(ABCD\)中,\(AD/\!/BC\),\(AD=AB\),\(∠A=90^{\circ}\),\(BD⊥DC\),将\(\triangle ABD\)沿\(BD\)折起到\(\triangle EBD\)的位置,使平面\(EBD⊥\)平面\(BDC\).
              \((1)\)求证:平面\(EBD⊥\)平面\(EDC\);
              \((2)\)求\(ED\)与\(BC\)所成的角.
            • 4.
              在正三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,若\(AB= \sqrt {2}BB_{1}\),则\(AB_{1}\)与\(C_{1}B\)所成的角的大小为\((\)  \()\)
              A.\(60^{\circ}\)
              B.\(90^{\circ}\)
              C.\(75^{\circ}\)
              D.\(105^{\circ}\)
            • 5.
              如图,在正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(E\)、\(F\)分别是\(BB_{1}\)、\(CD\)的中点,
              \((1)\)证明:\(AD⊥D_{1}F\);
              \((2)\)求异面直线\(AE\)与\(D_{1}F\)所成的角;
              \((3)\)证明:平面\(AED⊥\)平面\(A_{1}FD_{1}\).
            • 6.
              正四棱锥\(P-ABCD\)的底面积为\(3\),体积为\( \dfrac { \sqrt {2}}{2}\),\(E\)为侧棱\(PC\)的中点,则\(PA\)与\(BE\)所成的角为\((\)  \()\)
              A.\( \dfrac {π}{6}\)
              B.\( \dfrac {π}{3}\)
              C.\( \dfrac {π}{4}\)
              D.\( \dfrac {π}{2}\)
            • 7.
              如图,在三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(AA_{1}⊥\)面\(ABC\),\(AB=BC=2BB_{1}\),\(∠ABC=90^{\circ}\),\(D\)为\(BC\)的中点.
              \((1)\)求证:\(A_{1}B/\!/\)平面\(ADC_{1}\);
              \((2)\)求二面角\(C-AD-C_{1}\)的余弦值;
              \((3)\)若\(E\)为\(A_{1}B_{1}\)的中点,求\(AE\)与\(DC_{1}\)所成的角.
            • 8.
              如图,在四棱锥\(O-ABCD\)中,底面\(ABCD\)是边长为\(1\)的正方形,\(OA⊥\)底面\(ABCD\),\(OA=2\),\(M\)为\(OA\)的中点,\(N\)为\(BC\)的中点,建立适当的空间坐标系,利用空间向量解答以下问题:
              \((1)\)证明:直线\(MN/\!/\)平面\(OCD\);
              \((2)\)求异面直线\(AC\)与\(MD\)所成角的大小;
              \((3)\)求直线\(AC\)与平面\(OCD\)所成角的余弦值.
            • 9. 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.\( \dfrac { \sqrt {2}}{2}\)
            • 10.
              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.
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