數(shù)學(xué)幾何矢量題目及答案一、選擇題1.若向量\(\vec{a}=(2,3)\)，向量\(\vec=(-1,2)\)，則向量\(\vec{a}+\vec\)的坐標(biāo)為：A.\((1,5)\)B.\((1,1)\)C.\((3,5)\)D.\((3,1)\)答案：A2.已知向量\(\vec{a}=(3,-2)\)，向量\(\vec=(1,4)\)，求向量\(\vec{a}\)與向量\(\vec\)的數(shù)量積：A.10B.-2C.2D.-10答案：A3.若向量\(\vec{a}=(4,-3)\)，向量\(\vec=(2,1)\)，則向量\(\vec{a}\)與向量\(\vec\)的夾角的余弦值為：A.\(\frac{1}{5}\)B.\(\frac{2}{5}\)C.\(-\frac{2}{5}\)D.\(-\frac{1}{5}\)答案：B4.已知向量\(\vec{a}=(x,y)\)，向量\(\vec=(2,3)\)，若\(\vec{a}\)與\(\vec\)垂直，則\(x\)和\(y\)的關(guān)系為：A.\(2x+3y=0\)B.\(2x-3y=0\)C.\(3x+2y=0\)D.\(3x-2y=0\)答案：A5.已知向量\(\vec{a}=(1,2)\)，向量\(\vec=(3,-1)\)，求向量\(\vec{a}\)與向量\(\vec\)的叉積：A.5B.-5C.1D.-1答案：A二、填空題6.若向量\(\vec{a}=(3,4)\)，向量\(\vec=(2,-1)\)，則向量\(\vec{a}\)與向量\(\vec\)的模之比為\(\frac{|\vec{a}|}{|\vec|}=\)________。答案：\(\frac{5}{\sqrt{5}}\)7.已知向量\(\vec{a}=(1,0)\)，向量\(\vec=(0,1)\)，求向量\(\vec{a}\)與向量\(\vec\)的點(diǎn)積為\(\vec{a}\cdot\vec=\)________。答案：08.若向量\(\vec{a}=(x,y)\)，向量\(\vec=(2,3)\)，且\(\vec{a}\)與\(\vec\)平行，則\(x\)和\(y\)的關(guān)系為\(\frac{x}{2}=\frac{y}{3}=\)________。答案：常數(shù)9.已知向量\(\vec{a}=(2,1)\)，向量\(\vec=(1,-2)\)，求向量\(\vec{a}\)與向量\(\vec\)的單位向量分別為\(\vec{a}_{\text{unit}}\)和\(\vec_{\text{unit}}\)，則\(\vec{a}_{\text{unit}}\cdot\vec_{\text{unit}}=\)________。答案：\(-\frac{3}{\sqrt{10}}\)10.若向量\(\vec{a}=(1,-1)\)，向量\(\vec=(2,2)\)，則向量\(\vec{a}\)與向量\(\vec\)的叉積的模為\(|\vec{a}\times\vec|=\)________。答案：0三、解答題11.已知向量\(\vec{a}=(3,-1)\)，向量\(\vec=(2,4)\)，求向量\(\vec{a}\)與向量\(\vec\)的夾角。解：首先計(jì)算向量\(\vec{a}\)與向量\(\vec\)的數(shù)量積：\[\vec{a}\cdot\vec=3\times2+(-1)\times4=6-4=2\]然后計(jì)算向量\(\vec{a}\)與向量\(\vec\)的模：\[|\vec{a}|=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}\]\[|\vec|=\sqrt{2^2+4^2}=\sqrt{4+16}=\sqrt{20}\]最后計(jì)算夾角的余弦值：\[\cos\theta=\frac{\vec{a}\cdot\vec}{|\vec{a}|\cdot|\vec|}=\frac{2}{\sqrt{10}\cdot\sqrt{20}}=\frac{2}{2\sqrt{50}}=\frac{1}{5}\]所以，向量\(\vec{a}\)與向量\(\vec\)的夾角為\(\arccos\left(\frac{1}{5}\right)\)。12.已知三角形ABC的頂點(diǎn)坐標(biāo)分別為A(1,2)，B(4,6)，C(5,1)，求三角形ABC的面積。解：首先計(jì)算向量\(\vec{AB}\)和向量\(\vec{AC}\)：\[\vec{AB}=(4-1,6-2)=(3,4)\]\[\vec{AC}=(5-1,1-2)=(4,-1)\]然后計(jì)算向量\(\vec{AB}\)與向量\(\vec{AC}\)的叉積：\[\vec{AB}\times\vec{AC}=3\times(-1)-4\times4=-3-16=-19\]叉積的模為：\[|\vec{AB}\times\vec{AC}|=|-19|=19\]三角形ABC的面積為叉積模的一半：\[\text{面積}=\frac{1}{2}\times19=9.5\]13.已知向量\(\vec{a}=(2,1)\)，向量\(\vec=(1,-1)\)，求向量\(\vec{a}\)在向量\(\vec\)上的投影。解：首先計(jì)算向量\(\vec{a}\)與向量\(\vec\)的數(shù)量積：\[\vec{a}\cdot\vec=2\times1+1\times(-1)=2-1=1\]然后計(jì)算向量\(\vec\)的模：\[|\vec|=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\]向量\(\vec{a}\)在向量\(\vec\)上的投影為：\[\text{投影}=\frac{\vec{a}\cdot\vec}{|\vec|}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]14.已知向量\(\vec{a}=(3,-2)\)，向量\(\vec=(1,4)\)，求向量\(\vec{a}\)與向量\(\vec\)的單位向量。解：首先計(jì)算向量\(\vec{a}\)與向量\(\vec\)的模：\[|\vec{a}|=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\]\[|\vec|=\sqrt{1^2+4^2}=\sqrt{1+16}=\sqrt{17}\]然后計(jì)算單位向量：\[\vec{a}_{\text{unit}}=\left(\frac{3}{\sqrt{13}},\frac{-2}{\sqrt{13}}\right)\]\[\vec_{\text{unit}}=\left(\frac{1}{\sqrt{17}},\frac{4}{\sqrt{17}}\right)\]15.已知向量\(\vec{a}=(4,3)\)，向量\(\vec=(2,-1)\)，求向量\(\vec{a}\)與向量\(\vec\)的法向量。解：向量\(\vec{a}\)與向量\(\vec\)的法向量可以通過(guò)計(jì)算它們的叉積得到：\[\vec{a}\tim