优优班--学霸训练营 > 知识点挑题
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            • 1.

              给出以下命题,其中真命题的个数是

              \(①\)若“\(¬p \)或\(q\)”是假命题,则“\(p\)且\(¬q \)”是真命题

              \(②\)命题“若\(a+b\neq 5 \),则\(a\neq 2 \)或\(b\neq 3 \)”为真命题

              \(③\)已知空间任意一点\(O\)和不共线的三点\(A\),\(B\),\(C\),若\( \overrightarrow{OP}= \dfrac{1}{6} \overrightarrow{OA}+ \dfrac{1}{3} \overrightarrow{OB}+ \dfrac{1}{2} \overrightarrow{OC} \),则\(P\),\(A\),\(B\),\(C\)四点共面;

              \(④\)直线\(y=k\left(x-3\right) \)与双曲线\( \dfrac{{x}^{2}}{4}- \dfrac{{y}^{2}}{5}=1 \)交于\(A\),\(B\)两点,若\(\left|AB\right|=5 \),则这样的直线有\(3\)条;

              A.\(1\)                   
              B.\(2\)                 
              C.  \(3\)                
              D.\(4\)
            • 2.
              在平行六面体\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中,设\( \overrightarrow{AC_{1}}=x \overrightarrow{AB}+2y \overrightarrow{BC}+3z \overrightarrow{CC_{1}}\),则\(x+y+z\)等于\((\)  \()\)
              A.\(1\)
              B.\( \dfrac {2}{3}\)
              C.\( \dfrac {5}{6}\)
              D.\( \dfrac {11}{6}\)
            • 3.
              已知\(\{ \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\}\)是空间向量的一个基底,则可以与向量\( \overrightarrow{p}= \overrightarrow{a}+ \overrightarrow{b}\),\( \overrightarrow{q}= \overrightarrow{a}- \overrightarrow{b}\)构成基底的向量是\((\)  \()\)
              A.\( \overrightarrow{a}\)
              B.\( \overrightarrow{b}\)
              C.\( \overrightarrow{a}+2 \overrightarrow{b}\)
              D.\( \overrightarrow{a}+2 \overrightarrow{c}\)
            • 4.
              如图,在三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(M\)为\(A_{1}C_{1}\)的中点,若\( \overrightarrow{AB}= \overrightarrow{a}\),\( \overrightarrow{BC}= \overrightarrow{b}\),\( \overrightarrow{AA_{1}}= \overrightarrow{c}\),则\( \overrightarrow{BM}\)可表示为\((\)  \()\)
              A.\(- \dfrac {1}{2} \overrightarrow{a}+ \dfrac {1}{2} \overrightarrow{b}+ \overrightarrow{c}\)
              B.\( \dfrac {1}{2} \overrightarrow{a}+ \dfrac {1}{2} \overrightarrow{b}+ \overrightarrow{c}\)
              C.\(- \dfrac {1}{2} \overrightarrow{a}- \dfrac {1}{2} \overrightarrow{b}+ \overrightarrow{c}\)
              D.\( \dfrac {1}{2} \overrightarrow{a}- \dfrac {1}{2} \overrightarrow{b}+ \overrightarrow{c}\)
            • 5.
              如图,已知空间四边形\(OABC\),其对角线为\(OB\)、\(AC\),\(M\)、\(N\)分别是对边\(OA\)、\(BC\)的中点,点\(G\)在线段\(MN\)上,且\( \overrightarrow{MG}=2 \overrightarrow{GN}\),现用基向量\( \overrightarrow{OA}\),\( \overrightarrow{OB}\),\( \overrightarrow{OC}\)表示向量,设\( \overrightarrow{OG}=x \overrightarrow{OA}+y \overrightarrow{OB}+z \overrightarrow{OC}\),则\(x\)、\(y\)、\(z\)的值分别是\((\)  \()\)
              A.\(x= \dfrac {1}{3}\),\(y= \dfrac {1}{3}\),\(z= \dfrac {1}{3}\)
              B.\(x= \dfrac {1}{3}\),\(y= \dfrac {1}{3}\),\(z= \dfrac {1}{6}\)
              C.\(x= \dfrac {1}{3}\),\(y= \dfrac {1}{6}\),\(z= \dfrac {1}{3}\)
              D.\(x= \dfrac {1}{6}\),\(y= \dfrac {1}{3}\),\(z= \dfrac {1}{3}\)
            • 6.

              圆\(O\)上两点\(C\),\(D\)在直径\(AB\)的两侧\((\)如图甲\()\),沿直径\(AB\)将圆\(O\)折起形成一个二面角\((\)如图乙\()\),若\(∠DOB\)的平分线交弧\(\overline {BD} \)于点\(G\),交弦\(BD\)于点\(E\),\(F\)为线段\(BC\)的中点.

              \((\)Ⅰ\()\)证明:平面\(OGF/\!/\)平面\(CAD\);\((\)Ⅱ\()\)若二面角\(C-AB-D\)为直二面角,且\(AB=2\),\(∠CAB=45^{\circ}\),\(∠DAB=60^{\circ}\),求直线\(FG\)与平面\(BCD\)所成角的正弦值.

            • 7.

              如图,在三棱柱\(ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}\)中,\(M\)为\({{A}_{1}}{{C}_{1}}\)的中点,若\(\overrightarrow{AB}=\vec{a}\),\(\overrightarrow{BC}=\vec{b}\),\(\overrightarrow{A{{A}_{1}}}=\vec{c}\),则\(\overrightarrow{BM}\)可表示为\((\)  \()\)

              A.\(-\dfrac{1}{2}\vec{a}+\dfrac{1}{2}\vec{b}+\vec{c}\)
              B.\(\dfrac{1}{2}\vec{a}+\dfrac{1}{2}\vec{b}+\vec{c}\)
              C.\(-\dfrac{1}{2}\vec{a}-\dfrac{1}{2}\vec{b}+\vec{c}\)
              D.\(\dfrac{1}{2}\vec{a}-\dfrac{1}{2}\vec{b}+\vec{c}\)
            • 8.

              在平行六面体 \(ABCD\)\(−\) \(EFGH\)中,若\({\,\!} \overset{→}{AG}=2x \overset{→}{AB}+3y \overset{→}{BC}+3z \overset{→}{HD} \),则 \(x\)\(+\) \(y\)\(+\) \(z\)等于\((\) \()\)

              A.\( \dfrac{7}{6}\)
              B.\( \dfrac{2}{3}\)
              C.\( \dfrac{5}{6}\)
              D.\( \dfrac{1}{2}\)
            • 9.

              如图\((1)\),在直角梯形\(ABCD\)中,\(O\)为\(BD\)的中点,\(AD\)\(/\!/\)\(BC\),把沿翻折如图\((2)\),使得平面

              \((1)\)求证:

              \((2)\)在线段上是否存在点\(N\),使得与平面所成角为\({{30}^{\circ }}\)?若存在,求出\( \dfrac{BN}{BC} \)的值;若不存在,说明理由.

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