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            • 1.

              \(O\)为\(\triangle ABC\)的外心,\(AB+BC=\dfrac{2\sqrt{3}}{3}AC,\sin C(\cos A-\sqrt{3})+\cos C\sin A=0.\) 若\(\overrightarrow{{AO}}=x\overrightarrow{{AB}}+y\overrightarrow{{AC}}\) \((x,y∈R)\)则\(\dfrac{{x}}{{y}}=\)(    )

              A.\(1\)            
              B.\(-1\)          
              C.\(\sqrt{3}\)
              D.\(-\sqrt{3}\)
            • 2.

              在边长为\(1\)的等边\(\triangle ABC\)中,\(E\)为\(AC\)上一点,且\(\overrightarrow{AC}=4\overrightarrow{AE}\),\(P\)为\(BE\)上一点,且满足\(\overrightarrow{AP}=m\overrightarrow{AB}+n\overrightarrow{AC}(m > 0,n > 0)\),则\(\dfrac{1}{m}+\dfrac{1}{n}\)取最小值时,\(\left| \overrightarrow{AP} \right|=\)__________.

            • 3.

              已知\(\Delta ABC,\angle BAC={{60}^{\circ }},AB=2,AC=1,E,F\)为边\(BC\)的两个三等分点,则\(\overset{\to }{{AE}}\,\cdot \overset{\to }{{AF}}\,=(\)   \()\)

              A.\(\dfrac{5}{4}\)
              B.\(\dfrac{10}{9}\)
              C.\(\dfrac{15}{8}\)
              D.\(\dfrac{5}{3}\)
            • 4.

              如图,在\(\Delta ABC\)中,已知点\(D,E\)分别在边\(AB,BC\)上,且\(AB=3AD,BC=2BE\)


              \((1)\)用向量\(\overrightarrow{AB},\overrightarrow{AC}\)表示\(\overrightarrow{DE}\);

              \((2)\)设\(AB=6,AC=4,A={{60}^{\circ }}\),求线段\(DE\)的长.

            • 5.

              已知\( \overrightarrow{AB}⊥ \overrightarrow{AC},\left| \overrightarrow{AB}\right|= \dfrac{1}{t},\left| \overrightarrow{AC}\right|=t \),若\(P \)点是\(∆ABC \)所在平面内一点,且\( \overrightarrow{AP}= \dfrac{ \overrightarrow{AB}}{\left| \overrightarrow{AB}\right|}+ \dfrac{4 \overrightarrow{AC}}{\left| \overrightarrow{AC}\right|} \) ,则\( \overrightarrow{PB}· \overrightarrow{PC} \) 的最大值等于\((\)    \()\)

              A.\(13\)
              B.\(15\)
              C.\(19\)
              D.\(21\)
            • 6.

              将圆的六个等分点分成相同的两组,它们每组三个点构成的两个正三角形除去内部的六条线段后可以形成一个正六角星\(.\)如图所示的正六角星的中心为点\(O\),其中\(x,y\)分别为点\(O\)到两个顶点的向量\(.\)若将点\(O\)到正六角星\(12\)个顶点的向量都写成\(a\overrightarrow{x}+b\overrightarrow{y}\)的形式,则\(a+b\)的最大值为         

            • 7.

              设\(D\)为\(\triangle ABC\)所在平面内一点\( \overset{→}{BC}=3 \overset{→}{CD} \),则\((\) \()\)

              A.\( \overset{→}{AD}=- \dfrac{1}{3} \overset{→}{AB}+ \dfrac{4}{3} \overset{→}{AC} \) 
              B.\( \overset{→}{AD}= \dfrac{1}{3} \overset{→}{AB}- \dfrac{4}{3} \overset{→}{AC} \) 
              C.\( \overset{→}{AD}= \dfrac{4}{3} \overset{→}{AB}+ \dfrac{1}{3} \overset{→}{AC} \)  
              D.\( \overset{→}{AD}= \dfrac{4}{3} \overset{→}{AB}- \dfrac{1}{3} \overset{→}{AC} \)
            • 8.

              已知\(a\)\(b\)\(c\)分别是\(\triangle ABC\)中角\(A\),\(B\),\(C\)的对边,\(G\)是\(\triangle ABC\)的三条边上中线的交点,若\( \overrightarrow{GA}+\left(a+b\right) \overrightarrow{GB}+2c \overrightarrow{GC}= \overrightarrow{0} = \overrightarrow{0} \),且\( \dfrac{1}{a}+ \dfrac{2}{b} \geqslant \)\(\cos \)\(2\)\(x\)\(-\)\(m\sin x\)\((\)\(x\)\(∈R)\)恒成立,则实数\(m\)的取值范围为______.

            • 9.

              如图,直三棱柱\(ABC\)\(-\)\(A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\)\(C\)\({\,\!}_{1}\)中,\(AB\)\(=\)\(AC\)\(=\)\(AA\)\({\,\!}_{1}=4\),\(BC\)\(=2 \sqrt{2}\).\(BD\)\(⊥\)\(AC\),垂足为\(D\)\(E\)为棱\(BB\)\({\,\!}_{1}\)上一点,\(BD\)\(/\!/\)平面\(AC\)\({\,\!}_{1}\)\(E\)



              \((I)\)求线段\(B\)\({\,\!}_{1}\)\(E\)的长;

              \((II)\)求二面角\(C\)\({\,\!}_{1}-\)\(AC\)\(-\)\(E\)的余弦值.

            • 10.

              \((\)本小题满分\(12\)分\()\)

              平面内给定三个向量\( \overrightarrow{a}=(3,2)\),\( \overrightarrow{b}=(-1,2)\),\( \overrightarrow{c}=(4,1)\)
              \((1)\)求\(3 \overrightarrow{a}+ \overrightarrow{b}-2 \overrightarrow{c}\);
              \((2)\)求满足\( \overrightarrow{a}=m \overrightarrow{b}+n \overrightarrow{c}\)的实数\(m\)、\(n\).

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