2024年高中數(shù)學(xué)新高二暑期培優(yōu)講義第05講 空間向量及其運(yùn)算的坐標(biāo)表示（教師版）

第第頁第05講空間向量及其運(yùn)算的坐標(biāo)表示【題型歸納目錄】題型一：空間向量的坐標(biāo)表示題型二：空間向量的直角坐標(biāo)運(yùn)算題型三：空間向量的共線與共面題型四：空間向量模長坐標(biāo)表示題型五：空間向量平行坐標(biāo)表示題型六：空間向量垂直坐標(biāo)表示題型七：空間向量夾角坐標(biāo)表示【知識(shí)點(diǎn)梳理】知識(shí)點(diǎn)一、空間直角坐標(biāo)系1、空間直角坐標(biāo)系從空間某一定點(diǎn)O引三條互相垂直且有相同單位長度的數(shù)軸，這樣就建立了空間直角坐標(biāo)系SKIPIF1<0，點(diǎn)O叫做坐標(biāo)原點(diǎn)，x軸、y軸、z軸叫做坐標(biāo)軸，這三條坐標(biāo)軸中每兩條確定一個(gè)坐標(biāo)平面，分別是SKIPIF1<0平面、yOz平面、zOx平面.2、右手直角坐標(biāo)系在空間直角坐標(biāo)系中，讓右手拇指指向x軸的正方向，食指指向y軸的正方向，如果中指指向z軸的正方向，則稱這個(gè)坐標(biāo)系為右手直角坐標(biāo)系.3、空間點(diǎn)的坐標(biāo)空間一點(diǎn)A的坐標(biāo)可以用有序數(shù)組(x，y，z)來表示，有序數(shù)組(x，y，z)叫做點(diǎn)A的坐標(biāo)，記作A(x，y，z)，其中x叫做點(diǎn)A的橫坐標(biāo)，y叫做點(diǎn)A的縱坐標(biāo)，z叫做點(diǎn)A的豎坐標(biāo).知識(shí)點(diǎn)二、空間直角坐標(biāo)系中點(diǎn)的坐標(biāo)1、空間直角坐標(biāo)系中點(diǎn)的坐標(biāo)的求法通過該點(diǎn)，作兩條軸所確定平面的平行平面，此平面交另一軸于一點(diǎn)，交點(diǎn)在這條軸上的坐標(biāo)就是已知點(diǎn)相應(yīng)的一個(gè)坐標(biāo).特殊點(diǎn)的坐標(biāo)：原點(diǎn)SKIPIF1<0；SKIPIF1<0軸上的點(diǎn)的坐標(biāo)分別為SKIPIF1<0；坐標(biāo)平面SKIPIF1<0上的點(diǎn)的坐標(biāo)分別為SKIPIF1<0.2、空間直角坐標(biāo)系中對(duì)稱點(diǎn)的坐標(biāo)在空間直角坐標(biāo)系中，點(diǎn)SKIPIF1<0，則有點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于橫軸(x軸)的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于縱軸(y軸)的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于豎軸(z軸)的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于坐標(biāo)平面SKIPIF1<0的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于坐標(biāo)平面SKIPIF1<0的對(duì)稱點(diǎn)是SKIPIF1<0；點(diǎn)SKIPIF1<0關(guān)于坐標(biāo)平面SKIPIF1<0的對(duì)稱點(diǎn)是SKIPIF1<0.知識(shí)點(diǎn)三、空間向量的坐標(biāo)運(yùn)算(1)空間兩點(diǎn)的距離公式若SKIPIF1<0，則①SKIPIF1<0即:一個(gè)向量在直角坐標(biāo)系中的坐標(biāo)等于表示這個(gè)向量的有向線段的終點(diǎn)的坐標(biāo)減去起點(diǎn)的坐標(biāo)。②SKIPIF1<0，或SKIPIF1<0.知識(shí)點(diǎn)詮釋：兩點(diǎn)間距離公式是模長公式的推廣，首先根據(jù)向量的減法推出向量SKIPIF1<0的坐標(biāo)表示，然后再用模長公式推出。(2)空間線段中點(diǎn)坐標(biāo)空間中有兩點(diǎn)SKIPIF1<0，則線段AB的中點(diǎn)C的坐標(biāo)為SKIPIF1<0.(3)向量加減法、數(shù)乘的坐標(biāo)運(yùn)算若SKIPIF1<0，則①SKIPIF1<0;②SKIPIF1<0;③SKIPIF1<0;(4)向量數(shù)量積的坐標(biāo)運(yùn)算若SKIPIF1<0，則SKIPIF1<0即：空間兩個(gè)向量的數(shù)量積等于他們的對(duì)應(yīng)坐標(biāo)的乘積之和。(5)空間向量長度及兩向量夾角的坐標(biāo)計(jì)算公式若SKIPIF1<0，則(1)SKIPIF1<0.(2)SKIPIF1<0.知識(shí)點(diǎn)詮釋：①夾角公式可以根據(jù)數(shù)量積的定義推出：SKIPIF1<0，其中SKIPIF1<0的范圍是SKIPIF1<0②SKIPIF1<0.③用此公式求異面直線所成角等角度時(shí)，要注意所求角度與θ的關(guān)系(相等，互余，互補(bǔ))。(6)空間向量平行和垂直的條件若SKIPIF1<0，則①SKIPIF1<0②SKIPIF1<0規(guī)定：SKIPIF1<0與任意空間向量平行或垂直作用：證明線線平行、線線垂直.【典例例題】題型一：空間向量的坐標(biāo)表示例1．已知平行四邊形SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則頂點(diǎn)SKIPIF1<0的坐標(biāo)為(

)A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由題意得：SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，故頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0.故選：D例2．已知點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)為______．【答案】SKIPIF1<0【解析】點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0由SKIPIF1<0，則SKIPIF1<0，即x=0y=12z=1，所以點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0故答案為：SKIPIF1<0例3．已知點(diǎn)SKIPIF1<0，SKIPIF1<0，若點(diǎn)SKIPIF1<0為線段AB上靠近SKIPIF1<0的三等分點(diǎn)，則點(diǎn)SKIPIF1<0的坐標(biāo)為___________.【答案】SKIPIF1<0【解析】由題設(shè)，SKIPIF1<0，而SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0題型二：空間向量的直角坐標(biāo)運(yùn)算例4．若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0(

)A．-11 B．3 C．4 D．15【答案】C【解析】由已知，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0.故選：C．例5．已知向量SKIPIF1<0,則SKIPIF1<0(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵向量SKIPIF1<0,∴SKIPIF1<0.故選:B.題型三：空間向量的共線與共面例6．已知SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0與SKIPIF1<0共線，則實(shí)數(shù)SKIPIF1<0(

)A．-2 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【解析】∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0.∵SKIPIF1<0與SKIPIF1<0共線，∴SKIPIF1<0，即SKIPIF1<0.故選：B.例7．已知向量SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0共面，則實(shí)數(shù)SKIPIF1<0的值為(

)A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0共面，則設(shè)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故選：D.題型四：空間向量模長坐標(biāo)表示例8．設(shè)空間向量SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0＝______．【答案】3【解析】SKIPIF1<0，則顯然SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故答案為：3.題型五：空間向量平行坐標(biāo)表示例9．在空間直角坐標(biāo)系中，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0__________.【答案】4【解析】由題意得，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：4題型六：空間向量垂直坐標(biāo)表示例10．已知空間向量SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0【解析】空間向量SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0題型七：空間向量夾角坐標(biāo)表示例11．已知向量SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0與SKIPIF1<0的夾角為鈍角，則實(shí)數(shù)SKIPIF1<0的取值范圍為______.【答案】SKIPIF1<0【解析】因?yàn)橄蛄縎KIPIF1<0，SKIPIF1<0，且SKIPIF1<0與SKIPIF1<0的夾角為鈍角，所以SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0，故答案為：SKIPIF1<0例12．已知向量SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若向量SKIPIF1<0與SKIPIF1<0所成角為鈍角，則實(shí)數(shù)SKIPIF1<0的范圍是______.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)橄蛄縎KIPIF1<0與SKIPIF1<0所成角為鈍角，所以SKIPIF1<0，解得SKIPIF1<0，若向量SKIPIF1<0與SKIPIF1<0共線，則SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0與SKIPIF1<0共線同向，綜上可得SKIPIF1<0.故答案為：SKIPIF1<0例13．已知向量SKIPIF1<0，若向量SKIPIF1<0與SKIPIF1<0的夾角為銳角，則實(shí)數(shù)SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0且SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)橄蛄縎KIPIF1<0與SKIPIF1<0的夾角為銳角，所以SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0且SKIPIF1<0.故答案為：SKIPIF1<0且SKIPIF1<0.【過關(guān)測(cè)試】一、單選題1．若向量SKIPIF1<0，且SKIPIF1<0與SKIPIF1<0夾角的余弦值為SKIPIF1<0，則SKIPIF1<0等于(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．2【答案】A【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0與SKIPIF1<0夾角的余弦值為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，注意到SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：A.2．已知向量SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的值為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故選：D.3．已知向量SKIPIF1<0,若三個(gè)向量SKIPIF1<0共面,則實(shí)數(shù)m等于(

)A．4 B．6 C．8 D．10【答案】A【解析】由SKIPIF1<0共面可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故選：A．4．長方體SKIPIF1<0中，SKIPIF1<0，E為SKIPIF1<0與SKIPIF1<0的交點(diǎn)，F(xiàn)為SKIPIF1<0與SKIPIF1<0的交點(diǎn)，又SKIPIF1<0，則長方體的高SKIPIF1<0等于(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)長方體的長為SKIPIF1<0，由長方體的性質(zhì)建立如圖所示的空間直角坐標(biāo)系，則SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0.故選：C.5．設(shè)SKIPIF1<0，則AB的中點(diǎn)M到點(diǎn)C的距離SKIPIF1<0(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)镾KIPIF1<0的中點(diǎn)SKIPIF1<0，所以SKIPIF1<0故選：C．6．已知空間直角坐標(biāo)系SKIPIF1<0中，SKIPIF1<0，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上運(yùn)動(dòng)，則當(dāng)SKIPIF1<0取得最小值時(shí)，點(diǎn)SKIPIF1<0的坐標(biāo)為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因點(diǎn)Q在直線SKIPIF1<0上運(yùn)動(dòng)，則SKIPIF1<0，設(shè)SKIPIF1<0，于是有SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因此SKIPIF1<0，SKIPIF1<0，于是得SKIPIF1<0SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)點(diǎn)QSKIPIF1<0，所以當(dāng)SKIPIF1<0取得最小值時(shí)，點(diǎn)Q的坐標(biāo)為SKIPIF1<0.故選：C二、填空題7．已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0為______.【答案】SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0又SKIPIF1<0SKIPIF1<0故答案為：SKIPIF1<08．已知空間向量SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0共面，則SKIPIF1<0______．【答案】SKIPIF1<0【解析】若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0共面，則存在實(shí)數(shù)SKIPIF1<0，使SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0．故答案為：SKIPIF1<09．已知向量SKIPIF1<0，若向量SKIPIF1<0與SKIPIF1<0的夾角為銳角，求實(shí)數(shù)SKIPIF1<0的取值范圍______.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因?yàn)橄蛄縎KIPIF1<0與SKIPIF1<0的夾角為銳角，所以SKIPIF1<0，解得SKIPIF1<0，而當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0三、解答題10．已知空間三點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0．(1)設(shè)SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0；(2)求SKIPIF1<0與SKIPIF1<0的夾角；(3)若SKIPIF1<0與SKIPIF1<0互相垂直，求k．【解析】(1)由題可知，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKI