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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\(A(0,2,\dfrac{19}{8})\),\(B(1,-1,\dfrac{5}{8})\),\(C(-2,1,\dfrac{5}{8})\)是平面\(\alpha \)内的三点,设平面\(\alpha \)的法向量\(\vec{a}=(x,y,z)\),则\(x:y:z=\)________________。

            • 5.
              若平面\(α\)、\(β\)的法向量分别为\( \overrightarrow{n_{1}}=(2,3,5)\),\( \overrightarrow{n_{2}}=(-3,1,-4)\),则\((\)  \()\)
              A.\(α/\!/β\)
              B.\(α⊥β\)
              C.\(α\),\(β\)相交但不垂直
              D.以上均有可能
            • 6.

              已知梯形\(CEPD\)如如图所示,其中\(PD=8\),\(CE=6\),\(A\)为线段\(PD\)的中点,四边形\(ABCD\)为正方形,现沿\(AB\)进行折叠,使得平面\(PABE⊥\)平面\(ABCD\),得到如图所示的几何体,已知当点\(F\)满足\(\overrightarrow{AF}=\lambda \overrightarrow{AB}(0 < \lambda < 1)\)时,平面\(DEF⊥\)平面\(PCE\),则\(λ\)的值为

              A.\(\dfrac{1}{2}\)
              B.\(\dfrac{2}{3}\)
              C.\(\dfrac{3}{5}\)
              D.\(\dfrac{4}{5}\)
            • 7.

              若直线 \(l\) 的方向向量为\( \overrightarrow{b} \),平面\(α \) 的法向量为\( \overrightarrow{n} \),则可能使\(l/\!/α \) 的是

              A.\( \overrightarrow{b}=\left(1,0,0\right) \),\( \overrightarrow{n}=\left(-2,0,0\right) \)
              B.\( \overrightarrow{b}=\left(1,3,5\right), \overrightarrow{n}=\left(1,0,1\right) \)

              C.\( \overrightarrow{b}=\left(0,2,1\right) \),\( \overrightarrow{n}=\left(-1,0,-1\right) \)
              D.\( \overrightarrow{b}=\left(1,-1,3\right) \),\( \overrightarrow{n}=\left(0,3,1\right) \)
            • 8.
              已知向量\( \overrightarrow{AB}=(0,2,1)\),\( \overrightarrow{AC}=(-1,1,-2)\),则平面\(ABC\)的一个法向量可以是\((\)  \()\)
              A.\((3,-1,-2)\)
              B.\((-4,2,2)\)
              C.\((5,1,-2)\)
              D.\((5,-2,1)\)
            • 9.

              在三棱锥\(P-ABC\)中,\(CP\),\(CA\),\(CB\)两两垂直,\(AC=CB=1\),\(PC=2\),如图,建立空间直角坐标系,则下列向量中是平面\(PAB\)的法向量的是(    )

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

              已知直线\(l\)过点\(P(1,0,-1)\),平行于向量\(a=(2,1,1)\),平面\(α\)过直线\(l\)与点\(M(1,2,3)\),则平面\(α\)的法向量不可能是(    )

              A.\((1,-4,2)\)
              B.\((\dfrac{1}{4},-1,\dfrac{1}{2})\)
              C.\((-\dfrac{1}{4},1,-\dfrac{1}{2})\)
              D.\((0,-1,1)\)
            0/40

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