平面向量測試題含答案一

一、單項(xiàng)選擇題(每題2分,共10題)1.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.已知向量\(\vec{a}=(2,m)\),\(\vec=(1,-2)\),若\(\vec{a}\parallel\vec\),則\(m\)的值為()A.1B.-1C.4D.-43.向量\(\vec{a}\)的模\(\vert\vec{a}\vert=3\),向量\(\vec\)的模\(\vert\vec\vert=4\),且\(\vec{a}\)與\(\vec\)夾角為\(60^{\circ}\),則\(\vec{a}\cdot\vec\)等于()A.\(6\sqrt{3}\)B.6C.\(12\sqrt{3}\)D.124.若向量\(\vec{a}=(x,1)\),\(\vec=(-1,2)\),且\(\vec{a}\perp\vec\),則\(x\)的值為()A.2B.-2C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)5.已知\(\overrightarrow{AB}=(2,3)\),\(A(-1,2)\),則點(diǎn)\(B\)的坐標(biāo)為()A.\((1,5)\)B.\((-3,-1)\)C.\((3,1)\)D.\((-1,-3)\)6.向量\(\vec{a}=(3,-4)\),則與\(\vec{a}\)同向的單位向量\(\vec{e}\)為()A.\((\frac{3}{5},-\frac{4}{5})\)B.\((-\frac{3}{5},\frac{4}{5})\)C.\((\frac{3}{5},\frac{4}{5})\)D.\((-\frac{3}{5},-\frac{4}{5})\)7.已知\(\vec{a}=(1,1)\),\(\vec=(2,-1)\),則\(3\vec{a}-2\vec\)等于()A.\((-1,5)\)B.\((-1,-5)\)C.\((1,5)\)D.\((1,-5)\)8.若\(\vec{a}\),\(\vec\)是兩個(gè)非零向量,且\(\vert\vec{a}+\vec\vert=\vert\vec{a}\vert+\vert\vec\vert\),則()A.\(\vec{a}\perp\vec\)B.\(\vec{a}\)與\(\vec\)反向C.\(\vec{a}\)與\(\vec\)同向D.\(\vec{a}\)與\(\vec\)夾角為\(90^{\circ}\)9.向量\(\vec{a}=(-1,2)\),\(\vec=(m,1)\),若\(\vec{a}+\vec\)與\(\vec{a}\)垂直,則\(m\)的值為()A.7B.-7C.3D.-310.已知\(\vec{a}=(2,1)\),\(\vec=(x,-2)\),且\(\vert\vec{a}+\vec\vert^2=\vert\vec{a}\vert^2+\vert\vec\vert^2\),則\(x\)的值為()A.3B.-3C.1D.-1二、多項(xiàng)選擇題(每題2分,共10題)1.下列關(guān)于向量的說法正確的是(）A.零向量與任意向量平行B.向量的模一定是非負(fù)實(shí)數(shù)C.相等向量一定是平行向量D.平行向量一定是相等向量2.已知向量\(\vec{a}=(1,2)\),\(\vec=(-2,m)\),若\(\vec{a}\)與\(\vec\)平行,則\(m\)的值可能為()A.4B.-4C.1D.-13.設(shè)向量\(\vec{a}=(x_1,y_1)\),\(\vec=(x_2,y_2)\),則下列運(yùn)算正確的是()A.\(\vec{a}+\vec=(x_1+x_2,y_1+y_2)\)B.\(\vec{a}-\vec=(x_1-x_2,y_1-y_2)\)C.\(\lambda\vec{a}=(\lambdax_1,\lambday_1)\)(\(\lambda\)為實(shí)數(shù))D.\(\vec{a}\cdot\vec=x_1x_2+y_1y_2\)4.已知向量\(\vec{a}=(3,-4)\),\(\vec=(-3,4)\),則（）A.\(\vec{a}\)與\(\vec\)平行B.\(\vec{a}\)與\(\vec\)反向C.\(\vert\vec{a}\vert=\vert\vec\vert\)D.\(\vec{a}\cdot\vec=-25\)5.若向量\(\vec{a}\),\(\vec\)滿足\(\vert\vec{a}\vert=1\),\(\vert\vec\vert=2\),\(\vec{a}\cdot\vec=1\),則()A.向量\(\vec{a}\)與\(\vec\)夾角為\(60^{\circ}\)B.\(\vert\vec{a}+\vec\vert=\sqrt{7}\)C.\(\vert\vec{a}-\vec\vert=\sqrt{3}\)D.向量\(\vec{a}\)在\(\vec\)方向上的投影為\(\frac{1}{2}\)6.以下向量中,與向量\(\vec{a}=(1,\sqrt{3})\)垂直的向量有()A.\((\sqrt{3},-1)\)B.\((-\sqrt{3},1)\)C.\((\frac{\sqrt{3}}{2},-\frac{1}{2})\)D.\((-\frac{\sqrt{3}}{2},\frac{1}{2})\)7.已知向量\(\vec{m}=(2,1)\),\(\vec{n}=(x,y)\),若\(\vec{m}\)與\(\vec{n}\)共線,則()A.\(x=2y\)B.\(y=2x\)C.\(x-2y=0\)D.\(2x-y=0\)8.向量\(\vec{a}=(1,-1)\),\(\vec=(-1,2)\),則()A.\(\vec{a}+\vec=(0,1)\)B.\(\vec{a}-\vec=(2,-3)\)C.\(\vec{a}\cdot\vec=-3\)D.\(\vert\vec{a}\vert=\sqrt{2}\)9.對于向量\(\vec{a}\),\(\vec\),\(\vec{c}\),下列說法正確的是()A.\(\vec{a}\cdot(\vec+\vec{c})=\vec{a}\cdot\vec+\vec{a}\cdot\vec{c}\)B.\((\vec{a}\cdot\vec)\vec{c}=\vec{a}(\vec\cdot\vec{c})\)C.若\(\vec{a}\cdot\vec=\vec{a}\cdot\vec{c}\),則\(\vec=\vec{c}\)D.若\(\vec{a}\perp\vec\),則\(\vec{a}\cdot\vec=0\)10.已知向量\(\vec{a}=(x,1)\),\(\vec=(1,-2)\),若\(\vec{a}\)與\(\vec\)夾角為銳角,則\(x\)的取值范圍是（）A.\(x\gt2\)B.\(x\lt2\)C.\(x\neq-\frac{1}{2}\)D.\(x\lt-\frac{1}{2}\)三、判斷題(每題2分,共10題)1.向量\(\vec{a}=(1,0)\)與向量\(\vec=(0,1)\)的夾角為\(90^{\circ}\)。()2.若\(\vec{a}\cdot\vec=0\),則\(\vec{a}=\vec{0}\)或\(\vec=\vec{0}\)。（)3.向量\(\vec{a}\)的模\(\vert\vec{a}\vert\)等于\(\sqrt{\vec{a}^2}\)。()4.兩個(gè)向量相加的結(jié)果還是向量。()5.若向量\(\vec{a}\)與\(\vec\)平行,則存在唯一實(shí)數(shù)\(\lambda\),使得\(\vec{a}=\lambda\vec\)。（)6.零向量的方向是任意的。()7.向量\(\vec{a}=(2,3)\)與向量\(\vec=(4,6)\)是相等向量。()8.若\(\vert\vec{a}\vert=\vert\vec\vert\),則\(\vec{a}=\vec\)。（)9.向量\(\vec{a}\)在向量\(\vec\)方向上的投影是一個(gè)向量。()10.對于非零向量\(\vec{a}\),\(\vec\),若\(\vec{a}\cdot\vec\gt0\),則它們的夾角是銳角。（）四、簡答題(每題5分,共4題)1.已知向量\(\vec{a}=(1,-2)\),\(\vec=(3,4)\),求\(\vec{a}\cdot\vec\)的值。答案:根據(jù)向量點(diǎn)積公式\(\vec{a}\cdot\vec=x_1x_2+y_1y_2\),這里\(x_1=1\),\(y_1=-2\),\(x_2=3\),\(y_2=4\),則\(\vec{a}\cdot\vec=1×3+(-2)×4=3-8=-5\)。2.已知\(\vert\vec{a}\vert=3\),\(\vert\vec\vert=4\),\(\vec{a}\)與\(\vec\)夾角為\(120^{\circ}\),求\(\vert\vec{a}+\vec\vert\)。答案:先求\((\vec{a}+\vec)^2=\vec{a}^2+2\vec{a}\cdot\vec+\vec^2\)。\(\vec{a}^2=\vert\vec{a}\vert^2=9\),\(\vec^2=\vert\vec\vert^2=16\),\(\vec{a}\cdot\vec=\vert\vec{a}\vert\vert\vec\vert\cos120^{\circ}=3×4×(-\frac{1}{2})=-6\),所以\((\vec{a}+\vec)^2=9-12+16=13\),則\(\vert\vec{a}+\vec\vert=\sqrt{13}\)。3.已知向量\(\vec{a}=(2,-3)\),\(\vec=(x,6)\),且\(\vec{a}\parallel\vec\),求\(x\)的值。答案:兩向量平行,對應(yīng)坐標(biāo)成比例。因?yàn)閈(\vec{a}\parallel\vec\),則\(\frac{2}{x}=\frac{-3}{6}\),交叉相乘得\(-3x=12\),解得\(x=-4\)。4.已知向量\(\vec{a}=(1,1)\),求與\(\vec{a}\)垂直的單位向量坐標(biāo)。答案：設(shè)所求單位向量為\(\vec{e}=(x,y)\),因?yàn)閈(\vec{a}\perp\vec{e}\),則\(\vec{a}\cdot\vec{e}=x+y=0\),又\(x^2+y^2=1\),由\(y=-x\)代入\(x^2+y^2=1\)得\(2x^2=1\),解得\(x=\pm\frac{\sqrt{2}}{2}\),\(y=\mp\frac{\sqrt{2}}{2}\),所以單位向量為\((\frac{