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            • 1.
              如图,在长方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(AB=2\),\(AD=1\),\(A_{1}A=1\).
              \((1)\)求异面直线\(BC_{1}\)与\(CD_{1}\)所成的角;
              \((2)\)求三棱锥\(B-D_{1}AC\)的体积.
            • 2.
              正四棱锥\(S-ABCD\)的侧棱长为\( \sqrt {2}\),底面边长为\( \sqrt {3}\),\(E\)为\(SA\)的中点,则异面直线\(BE\)和\(SC\)所成的角为\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(45^{\circ}\)
              C.\(60^{\circ}\)
              D.\(90^{\circ}\)
            • 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.
              已知\(\triangle ABC\)与\(\triangle BCD\)均为正三角形,且\(AB=4.\)若平面\(ABC\)与平面\(BCD\)垂直,且异面直线\(AB\)和\(CD\)所成角为\(θ\),则\(\cos θ=(\)  \()\)
              A.\(- \dfrac { \sqrt {15}}{4}\)
              B.\( \dfrac { \sqrt {15}}{4}\)
              C.\(- \dfrac {1}{4}\)
              D.\( \dfrac {1}{4}\)
            • 7.
              如图所示,在矩形\(ABCD\)中,\(AB=4\),\(AD=2\),\(P\)为边\(AB\)的中点,现将\(\triangle DAP\)绕直线\(DP\)翻转至\(\triangle DA{{"}}P\)处,若\(M\)为线段\(A{{"}}C\)的中点,则异面直线\(BM\)与\(PA{{"}}\)所成角的正切值为\((\)  \()\)
              A.\( \dfrac {1}{2}\)
              B.\(2\)
              C.\( \dfrac {1}{4}\)
              D.\(4\)
            • 8.
              如图,在各棱长均为\(2\)的正三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(D\),\(E\)分别为棱\(A_{1}B_{1}\)与\(BB_{1}\)的中点,\(M\),\(N\)为线段\(C_{1}D\)上的动点,其中,\(M\)更靠近\(D\),且\(MN=C_{1}N.\)
              \((1)\)证明:\(A_{1}E⊥\)平面\(AC_{1}D\);
              \((2)\)若\(NE\)与平面\(BCC_{1}B_{1}\)所成角的正弦值为\( \dfrac { \sqrt {10}}{20}\),求异面直线\(BM\)与\(NE\)所成角的余弦值.
            • 9.
              将正方形\(ABCD\)沿对角线\(BD\)折叠成一个四面体\(ABCD\),当该四面体的体积最大时,直线\(AB\)与\(CD\)所成的角为\((\)  \()\)
              A.\(90^{\circ}\)
              B.\(60^{\circ}\)
              C.\(45^{\circ}\)
              D.\(30^{\circ}\)
            • 10.
              三棱锥\(A-BCD\)的所有棱长都相等,\(M\),\(N\)别是棱\(AD\),\(BC\)的中点,则异面直线\(BM\)与\(AN\)所成角的余弦值为\((\)  \()\)
              A.\( \dfrac {1}{3}\)
              B.\( \dfrac { \sqrt {2}}{4}\)
              C.\( \dfrac { \sqrt {3}}{3}\)
              D.\( \dfrac {2}{3}\)
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