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            • 1.

              \(19.\)如图,在直角梯形\(A{{A}_{1}}{{B}_{1}}B\)中,\(\angle {{A}_{1}}AB=90{}^\circ \),\({{A}_{1}}{{B}_{1}}/\!/AB\),\({{A}_{1}}{{B}_{1}}=1\),\(AB=A{{A}_{1}}=2.\)直角梯形\(A{{A}_{1}}{{C}_{1}}C\)通过直角梯形\(A{{A}_{1}}{{B}_{1}}B\)以直线\(A{{A}_{1}}\)为轴旋转得到,且使得平面\(A{{A}_{1}}{{C}_{1}}C\bot \)平面\(A{{A}_{1}}{{B}_{1}}B\).


              \((1)\)求证:平面\(CA{{B}_{1}}\bot \)平面\(A{{A}_{1}}{{B}_{1}}B\);

              \((2)\)延长\({{B}_{1}}{{A}_{1}}\)至点\({{D}_{1}}\),使\({{B}_{1}}{{A}_{1}}={{A}_{1}}{{D}_{1}}\),\(E\)为平面\(ABC\)内的动点,若直线\({{D}_{1}}E\)与平面\(CA{{B}_{1}}\)所成的角为\(\alpha \),且\(\sin \alpha =\dfrac{2\sqrt{5}}{5}\),求点\(E\)到点\(B\)的距离的最小值.

            • 2.

              如图,四棱柱\(ABCD-A_{1}B_{1}C_{1}D_{1}\)的底面\(ABCD\)是菱形,\(AC\cap BD=0\),\(A_{1}O⊥\)底面\(ABCD\),\(AB=2\),\(AA_{1}=3\).

              \((1)\)证明:平面\(A_{1}CO⊥\)平面\(BB_{1}D_{1}D\);

              \((2)\)若\(∠BAD=60^{\circ}\),求二面角\(B-OB_{1}-C\)的余弦值.

            • 3.

              已知\(A(1,0,0)\),\(B(0,1,0)\),\(C(0,0,1)\),则平面\(ABC\)的一个单位法向量是(    )

              A.\(\left( \left. \dfrac{ \sqrt{3}}{3}, \dfrac{ \sqrt{3}}{3},- \dfrac{ \sqrt{3}}{3} \right. \right)\)      
              B.\(\left( \left. \dfrac{ \sqrt{3}}{3},- \dfrac{ \sqrt{3}}{3}, \dfrac{ \sqrt{3}}{3} \right. \right)\)

              C.\(\left( \left. - \dfrac{ \sqrt{3}}{3}, \dfrac{ \sqrt{3}}{3}, \dfrac{ \sqrt{3}}{3} \right. \right)\)
              D.\(\left( \left. - \dfrac{ \sqrt{3}}{3},- \dfrac{ \sqrt{3}}{3},- \dfrac{ \sqrt{3}}{3} \right. \right)\)
            • 4.

              若\(a=(1,2,3)\)是平面\(γ\)的一个法向量,则下列向量中能作为平面\(γ\)的法向量的是 (    )

              A.\((0,1,2)\)
              B.\((3,6,9)\)
              C.\((-1,-2,3)\)
              D.\((3,6,8)\)
            • 5.

              在正方体\(ABCD - A_{1}B_{1}C_{1}D_{1}\)中,\(O\)是\(BD\)的中点,点\(P\)在线段\(B_{1}D_{1}\)上,直线\(OP\)与平面\(A_{1}BD\)所成的角为\(α\),则\(\sin α\)的取值范围是 \((\)  \()\)

              A.\([\dfrac{\sqrt{2}}{3},\dfrac{\sqrt{3}}{3}]\) 
              B.\([\dfrac{1}{3},\dfrac{1}{2}]\)
              C.\([\dfrac{\sqrt{3}}{4},\dfrac{\sqrt{3}}{3}]\) 
              D.\([\dfrac{1}{4},\dfrac{1}{3}]\)
            • 6.

              如图,四棱锥\(P-ABCD\)中,底面\(ABCD\)为平行四边形,\(PA⊥\)底面\(ABCD\),\(M\)是棱\(PD\)的中点,且\(PA=AB=AC=2\),\(BC=2\sqrt{2}\).



              \((\)Ⅰ\()\)求证:\(CD⊥\)平面\(PAC\);

              \((\)Ⅱ\()\)如果\(N\)是棱\(AB\)上一点,且直线\(CN\)与平面\(MAB\)所成角的正弦值为\(\dfrac{\sqrt{10}}{5}\),求\(\dfrac{AN}{NB}\)的值.

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