平面向量數(shù)學(xué)試題及答案

單項(xiàng)選擇題（每題2分，共10題）1.已知向量\(\vec{a}=(1,2)\)，\(\vec=(3,4)\)，則\(\vec{a}+\vec\)等于（）A.\((4,6)\)B.\((2,2)\)C.\((-2,-2)\)D.\((-4,-6)\)2.若向量\(\vec{a}=(x,1)\)，\(\vec=(-1,2)\)，且\(\vec{a}\perp\vec\)，則\(x\)的值為（）A.\(\frac{1}{2}\)B.\(-\frac{1}{2}\)C.\(2\)D.\(-2\)3.向量\(\vec{a}=(3,-4)\)的模\(\vert\vec{a}\vert\)是（）A.\(5\)B.\(6\)C.\(7\)D.\(8\)4.已知\(\vec{a}=(1,1)\)，\(\vec=(2,m)\)，若\(\vec{a}\parallel\vec\)，則\(m\)的值為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)5.設(shè)向量\(\vec{a}=(-1,2)\)，\(\vec=(m,1)\)，若\(\vec{a}+2\vec\)與\(2\vec{a}-\vec\)平行，則\(m\)的值為（）A.\(-\frac{1}{2}\)B.\(\frac{1}{2}\)C.\(1\)D.\(-1\)6.向量\(\vec{a}=(2,3)\)在向量\(\vec=(1,0)\)方向上的投影為（）A.\(2\)B.\(3\)C.\(\sqrt{13}\)D.\(0\)7.已知\(\vec{a}\cdot\vec=-12\sqrt{2}\)，\(\vert\vec{a}\vert=4\)，\(\vert\vec\vert=6\)，則\(\vec{a}\)與\(\vec\)的夾角\(\theta\)為（）A.\(120^{\circ}\)B.\(150^{\circ}\)C.\(30^{\circ}\)D.\(60^{\circ}\)8.若\(\vec{a}=(x_1,y_1)\)，\(\vec=(x_2,y_2)\)，則\(\vec{a}-\vec\)的坐標(biāo)為（）A.\((x_1-x_2,y_1-y_2)\)B.\((x_2-x_1,y_2-y_1)\)C.\((x_1+x_2,y_1+y_2)\)D.\((x_1x_2,y_1y_2)\)9.已知\(\vec{a}=(2,-1)\)，\(\vec=(-3,4)\)，則\(3\vec{a}+4\vec\)的坐標(biāo)為（）A.\((-6,13)\)B.\((6,-13)\)C.\((-12,19)\)D.\((12,-19)\)10.與向量\(\vec{a}=(1,-1)\)平行的單位向量是（）A.\((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)B.\((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)C.\((\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)或\((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)D.\((\frac{1}{2},-\frac{1}{2})\)多項(xiàng)選擇題（每題2分，共10題）1.下列關(guān)于平面向量的說法正確的是（）A.零向量與任意向量平行B.若\(\vec{a}\parallel\vec\)，\(\vec\parallel\vec{c}\)，則\(\vec{a}\parallel\vec{c}\)C.向量的模一定是非負(fù)的D.單位向量都相等2.已知向量\(\vec{a}=(1,2)\)，\(\vec=(-2,m)\)，若\(\vec{a}\)與\(\vec\)的夾角為鈍角，則\(m\)的取值范圍是（）A.\(m\lt1\)B.\(m\gt1\)C.\(m\lt1\)且\(m\neq-4\)D.\(m\gt-4\)3.以下向量運(yùn)算正確的是（）A.\(\vec{a}+\vec=\vec+\vec{a}\)B.\((\vec{a}+\vec)+\vec{c}=\vec{a}+(\vec+\vec{c})\)C.\(\lambda(\vec{a}+\vec)=\lambda\vec{a}+\lambda\vec\)D.\(\lambda(\mu\vec{a})=(\lambda\mu)\vec{a}\)4.若向量\(\vec{a}=(x_1,y_1)\)，\(\vec=(x_2,y_2)\)，則下列能使\(\vec{a}\perp\vec\)的條件是（）A.\(\vec{a}\cdot\vec=0\)B.\(x_1x_2+y_1y_2=0\)C.\(\vec{a}\cdot\vec=\vert\vec{a}\vert\vert\vec\vert\)D.\(\vec{a}\cdot\vec=-\vert\vec{a}\vert\vert\vec\vert\)5.已知\(\vec{a}=(3,-1)\)，\(\vec=(1,2)\)，向量\(\vec{c}\)滿足\(\vec{c}\cdot\vec{a}=9\)，\(\vec{c}\cdot\vec=-4\)，則向量\(\vec{c}\)的坐標(biāo)可能是（）A.\((2,-3)\)B.\((-2,3)\)C.\((1,-4)\)D.\((-1,4)\)6.下列向量中，與向量\(\vec{a}=(1,\sqrt{3})\)共線的向量有（）A.\((\sqrt{3},3)\)B.\((-1,-\sqrt{3})\)C.\((\sqrt{3},-3)\)D.\((-\sqrt{3},-3)\)7.向量\(\vec{a}\)，\(\vec\)滿足\(\vert\vec{a}\vert=1\)，\(\vert\vec\vert=2\)，\(\vec{a}\cdot\vec=1\)，則\(\vec{a}\)與\(\vec\)的夾角的余弦值為（）A.\(\frac{1}{2}\)B.\(\frac{\sqrt{2}}{2}\)C.\(\frac{\sqrt{3}}{2}\)D.\(1\)8.已知\(\vec{a}=(x,y)\)，\(\vec=(-1,2)\)，且\(\vec{a}+2\vec=(0,6)\)，則（）A.\(x=2\)B.\(y=2\)C.\(x=-2\)D.\(y=2\)9.對于向量\(\vec{a}\)，\(\vec\)，\(\vec{c}\)，下列命題正確的是（）A.\((\vec{a}\cdot\vec)\vec{c}=\vec{a}(\vec\cdot\vec{c})\)B.若\(\vec{a}\cdot\vec=\vec{a}\cdot\vec{c}\)，則\(\vec=\vec{c}\)C.\(\vec{a}\cdot\vec=\vec\cdot\vec{a}\)D.\(\vert\vec{a}\cdot\vec\vert\leqslant\vert\vec{a}\vert\vert\vec\vert\)10.已知向量\(\vec{a}=(1,-2)\)，\(\vec=(3,4)\)，則（）A.\(\vec{a}\cdot\vec=-5\)B.\(\vert\vec{a}\vert=\sqrt{5}\)C.\(\vec{a}\)在\(\vec\)方向上的投影為\(-\frac{\sqrt{5}}{5}\)D.\(\vec{a}\)與\(\vec\)夾角的余弦值為\(-\frac{\sqrt{5}}{5}\)判斷題（每題2分，共10題）1.向量\(\vec{a}=(1,2)\)與向量\(\vec=(2,4)\)是相等向量。（）2.若\(\vec{a}\cdot\vec=0\)，則\(\vec{a}=\vec{0}\)或\(\vec=\vec{0}\)。（）3.向量的減法滿足結(jié)合律。（）4.單位向量的模都為\(1\)。（）5.若\(\vec{a}\)與\(\vec\)是共線向量，則\(\vec{a}\)與\(\vec\)所在直線平行。（）6.對于任意向量\(\vec{a}\)，\(\vec\)，都有\(zhòng)((\vec{a}+\vec)^2=\vec{a}^2+2\vec{a}\cdot\vec+\vec^2\)。（）7.若\(\vec{a}=(x_1,y_1)\)，\(\vec=(x_2,y_2)\)，則\(\vec{a}\times\vec=x_1y_2-x_2y_1\)（這里的\(\times\)表示向量叉乘）。（）8.向量\(\vec{a}\)在向量\(\vec\)方向上的投影是一個向量。（）9.若\(\vert\vec{a}\vert=\vert\vec\vert\)，則\(\vec{a}=\vec\)。（）10.零向量沒有方向。（）簡答題（每題5分，共4題）1.已知向量\(\vec{a}=(2,-3)\)，\(\vec=(1,4)\)，求\(2\vec{a}-3\vec\)的坐標(biāo)。答案：先計算\(2\vec{a}=(4,-6)\)，\(3\vec=(3,12)\)，則\(2\vec{a}-3\vec=(4-3,-6-12)=(1,-18)\)。2.已知\(\vec{a}=(3,-2)\)，\(\vec=(-1,4)\)，求\(\vec{a}\cdot\vec\)以及\(\vec{a}\)與\(\vec\)夾角的余弦值。答案：\(\vec{a}\cdot\vec=3\times(-1)+(-2)\times4=-3-8=-11\)。\(\vert\vec{a}\vert=\sqrt{3^2+(-2)^2}=\sqrt{13}\)，\(\vert\vec\vert=\sqrt{(-1)^2+4^2}=\sqrt{17}\)，夾角余弦值\(\cos\theta=\frac{\vec{a}\cdot\vec}{\vert\vec{a}\vert\vert\vec\vert}=\frac{-11}{\sqrt{13\times17}}=-\frac{11}{\sqrt{221}}\)。3.已知向量\(\vec{a}\)與\(\vec\)的夾角為\(60^{\circ}\)，\(\vert\vec{a}\vert=2\)，\(\vert\vec\vert=3\)，求\((\vec{a}+2\vec)\cdot\vec{a}\)的值。答案：\((\vec{a}+2\vec)\cdot\vec{a}=\vec{a}^2+2\vec\cdot\vec{a}=\vert\vec{a}\vert^2+2\vert\vec{a}\vert\vert\vec\vert\cos60^{\circ}=4+2\times2\times3\times\frac{1}{2}=4+6=10\)。4.已知\(\vec{a}=(1,1)\)，\(\vec=(-1,2)\)，求\(\vec{a}\)在\(\vec\)方向上的投影。答案：\(\vec{a}\cdot\vec=1\times(-1)+1\times2=1\)，\(\vert\vec\vert=\sqrt{(-1)^2+2^2}=\sqrt{5}\)，投影為\(\frac{\vec{a}\cdot\vec}{\vert\vec\vert}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)。討論題（每題5分，共4題）1.討論向量平行與垂直的判定方法在實(shí)際解題中的應(yīng)用及注意事項(xiàng)。答案：平行判定用坐標(biāo)關(guān)系\(x_1y_2-x_2y_1=0\)或\(\vec{a}=\lambda\vec\)；垂直用\(\vec{a}\cdot\vec=0\)即\(x_1x_2+y_1y_2=0\)。應(yīng)用時要準(zhǔn)確找坐標(biāo)，注意零向量特殊性，如零