2025年高一數(shù)學(xué)向量真題及答案

一、單項選擇題1.已知向量\(\vec{a}=(1,2)\)，\(\vec=(x,-1)\)，若\(\vec{a}\parallel\vec\)，則\(x\)的值為（）A.\(2\)B.\(-2\)C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)【答案】D2.向量\(\vec{a}=(3,4)\)，則\(\vert\vec{a}\vert\)等于（）A.\(25\)B.\(7\)C.\(5\)D.\(\sqrt{7}\)【答案】C3.已知\(\vec{AB}=(2,3)\)，\(A(-1,2)\)，則點\(B\)的坐標(biāo)為（）A.\((1,5)\)B.\((-5,-1)\)C.\((5,1)\)D.\((-1,-5)\)【答案】A4.若向量\(\vec{a}=(1,1)\)，\(\vec=(1,-1)\)，\(\vec{c}=(-1,2)\)，則\(\vec{c}\)等于（）A.\(-\frac{1}{2}\vec{a}+\frac{3}{2}\vec\)B.\(\frac{1}{2}\vec{a}-\frac{3}{2}\vec\)C.\(\frac{3}{2}\vec{a}-\frac{1}{2}\vec\)D.\(-\frac{3}{2}\vec{a}+\frac{1}{2}\vec\)【答案】B5.已知向量\(\vec{a}\)，\(\vec\)滿足\(\vert\vec{a}\vert=2\)，\(\vert\vec\vert=3\)，\(\vec{a}\cdot\vec=-3\)，則\(\vec{a}\)與\(\vec\)的夾角為（）A.\(\frac{\pi}{6}\)B.\(\frac{\pi}{3}\)C.\(\frac{2\pi}{3}\)D.\(\frac{5\pi}{6}\)【答案】C6.已知\(\vec{a}=(x,2)\)，\(\vec=(-3,5)\)，且\(\vec{a}\)與\(\vec\)的夾角為鈍角，則\(x\)的取值范圍是（）A.\(x\gt\frac{10}{3}\)B.\(x\geq\frac{10}{3}\)C.\(x\lt\frac{10}{3}\)D.\(x\leq\frac{10}{3}\)【答案】A7.已知向量\(\vec{a}=(2,-1)\)，\(\vec=(-1,m)\)，\(\vec{c}=(-1,2)\)，若\((\vec{a}+\vec)\parallel\vec{c}\)，則\(m\)的值為（）A.\(-1\)B.\(1\)C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)【答案】A8.已知\(\vec{a}\)，\(\vec\)是單位向量，\(\vec{a}\cdot\vec=0\)，若向量\(\vec{c}\)滿足\(\vert\vec{c}-\vec{a}-\vec\vert=1\)，則\(\vert\vec{c}\vert\)的最大值為（）A.\(\sqrt{2}-1\)B.\(\sqrt{2}\)C.\(\sqrt{2}+1\)D.\(\sqrt{2}+2\)【答案】C9.設(shè)\(D\)為\(\triangleABC\)所在平面內(nèi)一點，\(\overrightarrow{BC}=3\overrightarrow{CD}\)，則（）A.\(\overrightarrow{AD}=-\frac{1}{3}\overrightarrow{AB}+\frac{4}{3}\overrightarrow{AC}\)B.\(\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AB}-\frac{4}{3}\overrightarrow{AC}\)C.\(\overrightarrow{AD}=\frac{4}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)D.\(\overrightarrow{AD}=\frac{4}{3}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AC}\)【答案】A10.已知\(\vec{a}=(1,\sin\theta)\)，\(\vec=(1,\cos\theta)\)，則\(\vert\vec{a}-\vec\vert\)的最大值為（）A.\(1\)B.\(\sqrt{2}\)C.\(\sqrt{3}\)D.\(2\)【答案】B二、多項選擇題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.若\(\vec{a}\parallel\vec\)，\(\vec\parallel\vec{c}\)，則\(\vec{a}\parallel\vec{c}\)C.單位向量都相等D.向量的模是非負(fù)實數(shù)【答案】AD2.已知向量\(\vec{a}=(1,2)\)，\(\vec=(-2,1)\)，則（）A.\(\vec{a}\perp\vec\)B.\(\vert\vec{a}\vert=\vert\vec\vert\)C.\(\vec{a}\)與\(\vec\)的夾角為\(90^{\circ}\)D.\(\vec{a}+\vec=(-1,3)\)【答案】BD3.設(shè)向量\(\vec{a}=(x_1,y_1)\)，\(\vec=(x_2,y_2)\)，則下列為\(\vec{a}\)與\(\vec\)共線的充要條件的有（）A.存在實數(shù)\(\lambda\)，使得\(\vec{a}=\lambda\vec\)B.\(x_1y_2-x_2y_1=0\)C.\(\frac{x_1}{x_2}=\frac{y_1}{y_2}\)D.\(\vec{a}\)與\(\vec\)是兩個非零向量時，\(\vert\vec{a}\cdot\vec\vert=\vert\vec{a}\vert\vert\vec\vert\)【答案】ABD4.已知向量\(\vec{a}=(-1,2)\)，\(\vec=(m,1)\)，若向量\(\vec{a}+\vec\)與\(\vec{a}\)垂直，則\(m\)的值可以是（）A.\(-3\)B.\(-2\)C.\(2\)D.\(3\)【答案】A5.下列向量組中，能作為平面內(nèi)所有向量的一組基底的是（）A.\(\vec{e_1}=(0,0)\)，\(\vec{e_2}=(1,-2)\)B.\(\vec{e_1}=(-1,2)\)，\(\vec{e_2}=(5,7)\)C.\(\vec{e_1}=(3,5)\)，\(\vec{e_2}=(6,10)\)D.\(\vec{e_1}=(2,-3)\)，\(\vec{e_2}=(\frac{1}{2},-\frac{3}{4})\)【答案】B6.已知向量\(\vec{a}=(3,-1)\)，\(\vec=(1,-2)\)，則（）A.\(\vec{a}\cdot\vec=5\)B.向量\(\vec{a}\)在向量\(\vec\)上的投影向量為\((\frac{1}{5},-\frac{2}{5})\)C.\(\vert\vec{a}-\vec\vert=\sqrt{5}\)D.與\(\vec{a}\)同向的單位向量為\((\frac{3\sqrt{10}}{10},-\frac{\sqrt{10}}{10})\)【答案】AD7.已知\(\vec{a}\)，\(\vec\)，\(\vec{c}\)是三個非零向量，則下列命題中真命題的是（）A.\(\vert\vec{a}\cdot\vec\vert=\vert\vec{a}\vert\vert\vec\vert\)，則\(\vec{a}\parallel\vec\)B.若\(\vec{a}\cdot\vec{c}=\vec\cdot\vec{c}\)，則\(\vec{a}=\vec\)C.若\(\vec{a}\)與\(\vec\)反向，則\(\vec{a}\cdot\vec=-\vert\vec{a}\vert\vert\vec\vert\)D.若\(\vert\vec{a}\vert=\vert\vec\vert\)，則\(\vert\vec{a}\cdot\vec{c}\vert=\vert\vec\cdot\vec{c}\vert\)【答案】AC8.已知向量\(\vec{a}=(2,1)\)，\(\vec=(-1,k)\)，\(\vec{a}\cdot(2\vec{a}-\vec)=0\)，則\(k\)的值可以是（）A.\(-12\)B.\(-6\)C.\(6\)D.\(12\)【答案】D9.設(shè)\(\vec{a}\)，\(\vec\)是兩個非零向量，下列說法正確的是（）A.若\(\vert\vec{a}+\vec\vert=\vert\vec{a}\vert-\vert\vec\vert\)，則\(\vec{a}\perp\vec\)B.若\(\vec{a}\perp\vec\)，則\(\vert\vec{a}+\vec\vert=\vert\vec{a}\vert-\vert\vec\vert\)C.若\(\vert\vec{a}+\vec\vert=\vert\vec{a}\vert-\vert\vec\vert\)，則存在實數(shù)\(\lambda\)，使得\(\vec=\lambda\vec{a}\)D.若存在實數(shù)\(\lambda\)，使得\(\vec=\lambda\vec{a}\)，則\(\vert\vec{a}+\vec\vert=\vert\vec{a}\vert-\vert\vec\vert\)【答案】C10.已知向量\(\vec{a}=(1,\cos\theta)\)，\(\vec=(\sin\theta,2)\)，且\(\vec{a}\perp\vec\)，則下列說法正確的是（）A.\(\sin2\theta=-2\)B.\(\sin2\theta=2\)C.\(\cos2\theta=-1\)D.\(\cos2\theta=1\)【答案】C三、判斷題1.向量\(\vec{a}=(1,2)\)與向量\(\vec=(2,4)\)是共線向量。（√）2.若\(\vec{a}\cdot\vec=0\)，則\(\vec{a}=\vec{0}\)或\(\vec=\vec{0}\)。（×）3.單位向量都平行。（×）4.向量\(\vec{a}=(x_1,y_1)\)，\(\vec=(x_2,y_2)\)，若\(\vec{a}\parallel\vec\)，則\(x_1y_2-x_2y_1=0\)。（√）5.若\(\vert\vec{a}\vert=\vert\vec\vert\)，則\(\vec{a}=\vec\)。（×）6.向量\(\vec{a}\)在向量\(\vec\)上的投影是一個向量。（×）7.已知\(\vec{AB}=(2,3)\)，\(A(1,2)\)，則點\(B\)的坐標(biāo)為\((3,5)\)。（√）8.若\(\vec{a}\)，\(\vec\)，\(\vec{c}\)是三個向量，\((\vec{a}\cdot\vec)\vec{c}=\vec{a}(\vec\cdot\vec{c})\)一定成立。（×）9.向量\(\vec{a}=(-1,2)\)的模為\(\sqrt{5}\)。（√）10.若\(\vec{a}\)與\(\vec\)的夾角為\(60^{\circ}\)，\(\vert\vec{a}\vert=2\)，\(\vert\vec\vert=3\)，則\(\vec{a}\cdot\vec=3\)。（√）四、簡答題1.已知向量\(\vec{a}=(1,-2)\)，\(\vec=(m,4)\)，且\(\vec{a}\parallel\vec\)，求\(m\)的值，并求\(\vec{a}\cdot\vec\)?！敬鸢浮恳驗閈(\vec{a}\parallel\vec\)，根據(jù)兩向量平行坐標(biāo)關(guān)系有\(zhòng)(1\times4-(-2)\timesm=0\)，即\(4+2m=0\)，解得\(m=-2\)。此時\(\vec=(-2,4)\)，那么\(\vec{a}\cdot\vec=1\times(-2)+(-2)\times4=-2-8=-10\)。2.已知\(\vec{a}=(2,1)\)，\(\vec=(-1,3)\)，求\(\vert\vec{a}+\vec\vert\)。【答案】先求\(\vec{a}+\vec\)，\(\vec{a}+\vec=(2-1,1+3)=(1,4)\)。再求其模，根據(jù)向量模的計算公式\(\vert\vec{c}\vert=\sqrt{x^{2}+y^{2}}\)（\(\vec{c}=(x,y)\)），則\(\vert\vec{a}+\vec\vert=\sqrt{1^{2}+4^{2}}=\sqrt{1+16}=\sqrt{17}\)。3.已知向量\(\vec{a}\)，\(\vec\)滿足\(\vert\vec{a}\vert=2\)，\(\vert\vec\vert=3\)，\(\vec{a}\cdot\vec=3\)，求\(\vec{a}\)與\(\vec\)的夾角\(\theta\)?！敬鸢浮扛鶕?jù)向量點積公式\(\vec{a}\cdot\vec=\vert\vec{a}\vert\vert\vec\vert\cos\theta\)，已知\(\vert\vec{a}\vert=2\)，\(\vert\vec\vert=3\)，\(\vec{a}\cdot\vec=3\)