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.

              四面体\(ABCD\)中,\(AB=AC=AD=BC=CD=2 \) ,\(BD=2 \sqrt{2} \), \(E\)为\(AC\)中点,则异面直线\(BE\)与\(CD\)所成角的余弦值为______________

            • 2.

              已知棱长为\(a\)的正方体\(ABCD-A′B′C′D′\)中,\(M\)、\(N\)分别为\(C′D′\)、\(A′D′\)的中点,则\(MN\)与\(BC′\)的所成的角为___________.

            • 3.

              矩形\(ABCD\)中,\(AB=\sqrt{3}\),\(BC=1\),将\(\triangle ABC\)与\(\triangle ADC\)沿\(AC\)所在的直线进行随意翻折,在翻折过程中直线\(AD\)与直线\(BC\)成的角范围\((\)包含初始状态\()\)为\((\)    \()\)

              A.\([0,\dfrac{{ }\!\!\pi\!\!{ }}{6}]\)
              B.\([0,\dfrac{{ }\!\!\pi\!\!{ }}{3}]\)
              C.\([0,\dfrac{{ }\!\!\pi\!\!{ }}{2}]\)
              D.\([0,\dfrac{2{ }\!\!\pi\!\!{ }}{3}]\)
            • 4. 如图,在直三棱柱\(ABC-A_{1}B_{l}C_{1}\)中,\(AC=BC= \sqrt {2}\),\(∠ACB=90^{\circ}.AA_{1}=2\),\(D\)为\(AB\)的中点.
              \((\)Ⅰ\()\)求证:\(AC⊥BC_{1}\);
              \((\)Ⅱ\()\)求证:\(AC_{1}/\!/\)平面\(B_{1}CD\):
              \((\)Ⅲ\()\)求异面直线\(AC_{1}\)与\(B_{1}C\)所成角的余弦值.
            • 5.

              三棱锥\(A\)\(—\)\(BCD\)的棱长全相等, \(E\)\(AD\)中点,则直线\(CE\)与直线\(BD\)所成角的余弦值为\((\)   \()\)

              A.\( \dfrac{ \sqrt{3}}{6} \)
              B.\( \dfrac{ \sqrt{3}}{2} \)
              C.\( \dfrac{ \sqrt{33}}{6} \)
              D.\( \dfrac{1}{2} \)
            • 6. 如图,多面体\(ABCDS\)中,面\(ABCD\)为矩形,\(SD⊥AD\),且\(SD⊥AB\),\(AD=1\),\(AB=2\),\(SD= \sqrt {3}\).
              \((1)\)求证:\(CD⊥\)平面\(ADS\);
              \((2)\)求\(AD\)与\(SB\)所成角的余弦值;
              \((3)\)求二面角\(A-SB-D\)的余弦值.
            • 7.

              如图,在直四棱柱\(ABCD\)\(-\)\(A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\)\(C\)\({\,\!}_{1}\)\(D\)\({\,\!}_{1}\)中,底面四边形\(ABCD\)为菱形,\(A\)\({\,\!}_{1}\)\(A\)\(=\)\(AB\)\(=2\),\(∠ABC=\dfrac{\pi }{3}\),\(E\)\(F\)分别是\(BC\)\(A\)\({\,\!}_{1}\)\(C\)的中点.

              \((1)\)求异面直线\(EF\)\(AD\)所成角的余弦值;

              \((2)\)点\(M\)在线段\(A\)\({\,\!}_{1}\)\(D\)上,\(\dfrac{{{A}_{1}}M}{{{A}_{1}}D}=\lambda .\)若\(CM\)\(/\!/\)平面\(AEF\),求实数\(λ\)的值.

            • 8. 如图:已知矩形\(B{B}_{1}{C}_{1}C \)所在平面与底面\(ABB_{1}N\)垂直,直角梯形\(ABB\)\(1\)\(N\)中\(AN{/\!/}BB_{1}\),\(AB⊥AN\),\(CB=BA=AN=2\),\(BB_{1}=4\).
                
               \((\)Ⅰ\()\)求证:\(BN\)拓展平面\(C_{1}B_{1}N\);
               \((\)Ⅱ\()\)求二面角\(C-C_{1}N-B_{1}\)的正弦值;
               \((\)Ⅲ\()\)在\(BC\)边上找一点\(P\),使\(B_{1}P\)与\(CN\)所成角的余弦值为\( \dfrac{5 \sqrt{51}}{51} \),并求线段\(B\)\(1\)\(P\)的长.
            • 9.

              如图正方形\(BCDE\)的边长为\(a\),已知\(AB=\sqrt{3}BC\),将\(\Delta ABE\)沿\(BE\)边折起,折起后\(A\)点在平面\(BCDE\)上的射影为\(D\)点,则翻折后的几何体中\(AB\)与\(DE\)所成角的正切值是           

              \(\sin C \)

            • 10.

              如图所示,正四棱锥\(P-ABCD\)的底面面积为\(3\),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.\(30{}^\circ \)
              B.\(45{}^\circ \)
              C.\(60{}^\circ \)
              D.\(90{}^\circ \)
            0/40

            进入组卷