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“\((\neg p)\)或\(q\)”是假命题,则“\(p\)且\((\neg q)\)”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\(\overrightarrow{OP}=\dfrac{1}{6}\overrightarrow{PA}+\dfrac{1}{3}\overrightarrow{OB}+\dfrac{1}{2}\overrightarrow{OC}\),则\(P\),\(A\),\(B\),\(C\)四点共面;

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

              A.\(1\)
              B.\(2\)
              C.\(3\)
              D.\(4\)
            • 2. 若\(|\) \(a|\)\(=1\), \(|b|\)\(=2\), \(c\)\(=\) \(a\)\(+\) \(b\)\(c\)\(⊥\) \(a\),则向量 \(a\)\(b\)的夹角是\((\)  \()\)
              A.\(30^{\circ}\)
              B.\(60^{\circ}\)
              C.\(120^{\circ}\)
              D.\(150^{\circ}\)
            • 3.
              设\(O-ABC\)是正三棱锥,\(G_{1}\)是\(\triangle ABC\)的重心,\(G\)是\(OG_{1}\)上的一点,且\(OG=3GG_{1}\),若,则 \( \overrightarrow{OG}=x \overrightarrow{OA}+y \overrightarrow{OB}+z \overrightarrow{OC}\),则\((x,y,z)\)为\((\)  \()\)
              A.\(( \dfrac {1}{4}, \dfrac {1}{4}, \dfrac {1}{4})\)
              B.\(( \dfrac {3}{4}, \dfrac {3}{4}, \dfrac {3}{4})\)
              C.\(( \dfrac {1}{3}, \dfrac {1}{3}, \dfrac {1}{3})\)
              D.\(( \dfrac {2}{3}, \dfrac {2}{3}, \dfrac {2}{3})\)
            • 4.
              空间四边形\(OABC\)中,\(M\),\(N\)分别是对边\(OA\),\(BC\)的中点,点\(G\)为\(MN\)中点,设\( \overrightarrow{OA}= \overrightarrow{a}\),\( \overrightarrow{OB}= \overrightarrow{b}\),\( \overrightarrow{OC}= \overrightarrow{c}\),则\( \overrightarrow{OG}\)可以用基底\(\{ \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\}\)表示为\((\)  \()\)
              A.\( \dfrac {1}{4} \overrightarrow{a}+ \dfrac {1}{4} \overrightarrow{b}+ \dfrac {1}{4} \overrightarrow{c}\)
              B.\( \dfrac {1}{4} \overrightarrow{a}+ \dfrac {1}{4} \overrightarrow{b}+ \dfrac {1}{3} \overrightarrow{c}\)
              C.\( \dfrac {1}{4} \overrightarrow{a}+ \dfrac {1}{4} \overrightarrow{b}+ \dfrac {1}{6} \overrightarrow{c}\)
              D.\( \dfrac {1}{4} \overrightarrow{a}+ \dfrac {1}{4} \overrightarrow{b}+ \dfrac {1}{4} \overrightarrow{c}\)
            • 5. 如图所示,在空间直角坐标系中\(BC=2\),原点\(O\)是\(BC\)的中点,点\(A\)的坐标是\(( \dfrac { \sqrt {3}}{2}, \dfrac {1}{2},0)\),点\(D\)在平面\(yOz\)上,且\(∠BDC=90^{\circ}\),\(∠DCB=30^{\circ}\),则向量\( \overrightarrow{AD}\)的坐标为\((\)  \()\)
              A.\((- \dfrac { \sqrt {3}}{2},- \dfrac {1}{2}, \dfrac { \sqrt {3}}{2})\)
              B.\((- \dfrac { \sqrt {3}}{2},-1, \dfrac { \sqrt {3}}{2})\)
              C.\((- \dfrac {1}{2},- \dfrac { \sqrt {3}}{2}, \dfrac { \sqrt {3}}{2})\)
              D.\(( \dfrac { \sqrt {3}}{2},1, \dfrac { \sqrt {3}}{2})\)
            • 6.

              空间四边形\(ANCD\)中,若向量\( \overrightarrow{AB}=(-3,5,2) \),\( \overrightarrow{CD}=(-7,-1,-4) \),点\(E\),\(F\)分别为线段\(BC\),\(AD\)的中点,则\( \overrightarrow{EF} \)的坐标为(    ) 

                

              A.\((2,3,3)\)       
              B.\((-2,-3,-3)\)
              C.\((5,-2,1)\)     
              D.\((-5,2,-1)\)
            • 7.

              在空间中,已知\(\overrightarrow{{AB}}{=}(2{,}4{,}0){,}\overrightarrow{{BC}}{=}({-}1{,}3{,}0)\),则\({∠}ABC\)的大小为\(({  })\)

              A.\(45^{{∘}}\)
              B.\(90^{{∘}}\)
              C.\(120^{{∘}}\)
              D.\(135^{{∘}}\)
            • 8. 三棱柱ABC-A1B1C1中,底面边长和侧棱长都相等,∠BAA1=∠CAA1=60°,则异面直线AB1与BC1所成角的余弦值为(  )
              A.
              3
              3
              B.
              6
              6
              C.
              3
              4
              D.
              3
              6
            • 9. 已知点P为三棱锥O-ABC的底面ABC所在平面内的一点,且
              OP
              =
              1
              2
              OA
              +k
              OB
              -
              OC
              ,则实数k的值为(  )
              A.-
              1
              2
              B.
              1
              2
              C.1
              D.
              3
              2
            • 10. 已知正方体ABCD-A′B′C′D′中,点F是侧面CDD′C′的中心,若
              AF
              =
              AD
              +x
              AB
              +y
              AA′
              ,则x-y等于(  )
              B.1
              C.
              1
              2
              D.-
              1
              2
            0/40

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