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.

              正方体\(ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}\)中,异面直线\({{A}_{1}}{{C}_{1}}\)与\({{B}_{1}}C\)所成角的大小是             

            • 2.
              如图,在四棱锥\(P-ABCD\)中,底面正方形\(ABCD\)为边长为\(2\),\(PA⊥\)底面\(ABCD\),\(E\)为\(BC\)的中点,\(PC\)与平面\(PAD\)所成的角为\(\arctan \dfrac { \sqrt {2}}{2}\).
              \((1)\)求异面直线\(AE\)与\(PD\)所成角的大小\((\)结果用反三角函数表示\()\);
              \((2)\)求点\(B\)到平面\(PCD\)的距离.
            • 3.

              正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,异面直线\(A_{1}B\)与\(B_{1}C\)所成角的大小为         


            • 4.

              如下图,在直三棱柱\(ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,\(∠BCA={{90}^{\circ }}\),\(M\)、\(N\)分别是\(AB\)、\(AC\)的中点,\(BC=AC=C{{C}_{1}}\),则直线\({{B}_{1}}M\)与\({{A}_{1}}N\)所成角的余弦值为

              A.\( \dfrac{1}{10} \)                   
              B.\(\dfrac{2}{5}\)                     
              C.\(\dfrac{\sqrt{30}}{10}\)                
              D.\(\dfrac{\sqrt{2}}{2}\)
            • 5.

              如图,直三棱柱\(ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,侧棱\(A{{A}_{1}}\bot \)平面\(ABC\),若\(AB=AC=A{{A}_{1}}=1\),\(BC=\sqrt{2}\),则异面直线\({{A}_{1}}C\)与\({{B}_{1}}{{C}_{1}}\)所成的角为(    )

              A.\(30{}^\circ \)
              B.\(45{}^\circ \)
              C.\(60{}^\circ \)
              D.\(90{}^\circ \) 
            • 6.

              如下图,正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,\(M\)、\(N\)分别为\(A_{1}B_{1}\)、\(CC_{1}\)的中点,\(P\)为\(AD\)上一动点,记\(a\)为异面直线\(PM\)与\(D\)\({\,\!}_{1}\)\(N\)所成的角,则\(a\)的集合是\((\)     \()\)

              A.\(\{ \dfrac{π}{2} \}\)
              B.\(\{a| \dfrac{π}{6} \leqslant a\leqslant \dfrac{π}{2} \}\)  

              C.\(\{a| \dfrac{π}{4} \leqslant a\leqslant \dfrac{π}{2} \}\)
              D.\(\{a| \dfrac{π}{3} \leqslant a\leqslant \dfrac{π}{2} \}\)
            • 7. 如图,平面\(PAD⊥\)平面\(ABCD\),四边形\(ABCD\)为正方形,\(∠PAD=90^{\circ}\),且\(PA=AD=2\),\(E\),\(F\)分别是线段\(PA\),\(CD\)的中点,则异面直线\(EF\)与\(BD\)所成角的余弦值为 ______ .
            • 8.

              直线\(l_{1}\),\(l\)的方向向量分别是\(v_{1}\),\(v\),若\(v_{1}\)与\(v_{2}\)所成的角为\(θ\),直线\(l_{1}\),\(l_{2}\)所成的角为\(α\),则\((\)   \()\)

              A.\(α=θ\)
              B.\(α=π-θ\)
              C.\(\cos θ=|\cos α|\)
              D.\(\cos α=|\cos θ|\)
            • 9.

              如图所示,直三棱柱\(ABC—A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\)\(C\)\({\,\!}_{1}\),底面\(\triangle ABC\)中,\(CA=CB=1\),\(∠BCA=90^{\circ}\),棱\(AA\)\({\,\!}_{1}\)\(=2\),\(M\),\(N\)分别是\(A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\),\(A\)\({\,\!}_{1}\)\(A\)的中点.



              \((\)Ⅰ\()\)求\(\overrightarrow{BN}\) 的模;

              \((\)Ⅱ\()\)求\(\cos 〈\overrightarrow{BA_{1}}\),\(\overrightarrow{CB_{1}}〉\)的值;

              \((\)Ⅲ\()\)求证:\(A_{1}B⊥C_{1}M.\)

            • 10.

              长方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中\(AB=AA_{1}=2\),\(AD=1\),\(E\)为\(CC_{1}\)的中点且\(AE= \sqrt{6} \),则异面直线\(BC_{1}\)与\(AE\)所成角的余弦值为\((\)  \()\)


              A.\( \dfrac{ \sqrt{10}}{10} \)
              B.\( \dfrac{ \sqrt{30}}{10} \)
              C.\( \dfrac{2 \sqrt{15}}{10} \)
              D.\( \dfrac{3 \sqrt{10}}{10} \)
            0/40

            进入组卷