5.3平面向量的應(yīng)用(精練)(基礎(chǔ)版)題組一題組一證線段垂直1.(2022·全國·高一課前預(yù)習(xí))在平行四邊形ABCD中,M、N分別在BC、CD上,且滿足BC=3MC,DC=4NC,若AB=4,AD=3,則△AMN的形狀是(
)A.銳角三角形 B.鈍角三角形C.直角三角形 D.等腰三角形2.(2022·新疆)在△ABC中,若SKIPIF1<0,則△ABC的形狀是(
)A.等腰三角形 B.直角三角形 C.等邊三角形 D.等腰直角三角形3.(2021·浙江)在SKIPIF1<0中,若SKIPIF1<0,則SKIPIF1<0的形狀為(
)A.等邊三角形 B.等腰三角形C.直角三角形 D.等腰直角三角形4.(2022·黑龍江)如圖,正方形ABCD的邊長為a,E是AB的中點(diǎn),F(xiàn)是BC的中點(diǎn),求證:DE⊥AF.5.(2022·湖南)如圖所示,在等腰直角三角形ACB中,SKIPIF1<0,SKIPIF1<0,D為BC的中點(diǎn),E是AB上的一點(diǎn),且SKIPIF1<0,求證:SKIPIF1<0.6.(2022·浙江)如圖所示,若D是△ABC內(nèi)的一點(diǎn),且AB2-AC2=DB2-DC2,求證:AD⊥BC.7.(2022·浙江)如圖,在平行四邊形ABCD中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,BD,AC相交于點(diǎn)O,M為BO中點(diǎn).設(shè)向量SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0的值;(2)用SKIPIF1<0,SKIPIF1<0表示SKIPIF1<0和SKIPIF1<0;(3)證明:SKIPIF1<0.題組二題組二夾角問題1.(2022·云南)SKIPIF1<0中,若SKIPIF1<0,SKIPIF1<0,點(diǎn)SKIPIF1<0滿足SKIPIF1<0,直線SKIPIF1<0與直線SKIPIF1<0相交于點(diǎn)SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·江西)已知菱形SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,點(diǎn)SKIPIF1<0為SKIPIF1<0上一點(diǎn),且SKIPIF1<0,則SKIPIF1<0的余弦值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·江蘇)(多選)已知向量SKIPIF1<0,記向量SKIPIF1<0的夾角為SKIPIF1<0,則(
)A.SKIPIF1<0時SKIPIF1<0為銳角 B.SKIPIF1<0時SKIPIF1<0為鈍角C.SKIPIF1<0時SKIPIF1<0為直角 D.SKIPIF1<0時SKIPIF1<0為平角4.(2023·全國·高三專題練習(xí))已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,若向量SKIPIF1<0與SKIPIF1<0的夾角是銳角,則實數(shù)入的取值范圍是:______.5.(2022·四川省平昌中學(xué))已知SKIPIF1<0,且SKIPIF1<0的夾角為鈍角,則實數(shù)SKIPIF1<0的范圍_______6.(2022·全國·期末)一扇中式實木仿古正方形花窗如圖1所示,該窗有兩個正方形,將這兩個正方形(它們有共同的對稱中心與對稱軸)單獨(dú)拿出來放置于同一平面,如圖2所示.已知SKIPIF1<0分米,SKIPIF1<0分米,點(diǎn)SKIPIF1<0在正方形SKIPIF1<0的四條邊上運(yùn)動,當(dāng)SKIPIF1<0取得最大值時,SKIPIF1<0與SKIPIF1<0夾角的余弦值為___________.7(2022·福建·廈門一中模擬預(yù)測)已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0均為單位向量,且SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0夾角的余弦值為______.8.(2022·安徽·池州市第一中學(xué))如圖,在SKIPIF1<0中,已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,線段AM,BN相交于點(diǎn)P,則SKIPIF1<0的余弦值為___________.9.(2021·湖南)已知平面四邊形SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0_______.10.(2022·湖北)已知SKIPIF1<0=(1,2),SKIPIF1<0=(1,SKIPIF1<0),分別確定實數(shù)SKIPIF1<0的取值范圍,使得:(1)SKIPIF1<0與SKIPIF1<0的夾角為直角;(2)SKIPIF1<0與SKIPIF1<0的夾角為鈍角;(3)SKIPIF1<0與SKIPIF1<0的夾角為銳角.11.(2022·全國·高三專題練習(xí))已知△ABC的面積為S滿足SKIPIF1<0,且SKIPIF1<0·SKIPIF1<0=3,SKIPIF1<0與SKIPIF1<0的夾角為θ.求SKIPIF1<0與SKIPIF1<0夾角的取值范圍.題組三題組三線段長度1.(2022·全國·高三專題練習(xí))在平行四邊形SKIPIF1<0中,點(diǎn)SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.SKIPIF1<02.(2022·湖南)(多選)已知SKIPIF1<0分別是三棱錐SKIPIF1<0的棱SKIPIF1<0,SKIPIF1<0的中點(diǎn),SKIPIF1<0.若異面直線SKIPIF1<0與SKIPIF1<0所成角的大小為60°,則線段SKIPIF1<0的長為(
)A.3 B.6 C.SKIPIF1<0 D.SKIPIF1<03.(2022·全國·信陽高中)已知四邊形SKIPIF1<0是矩形,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·山東濟(jì)寧)已知兩點(diǎn)SKIPIF1<0分別是四邊形SKIPIF1<0的邊SKIPIF1<0的中點(diǎn),且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則線段SKIPIF1<0的長為是___________5(2022·全國·高三專題練習(xí))如圖,SKIPIF1<0,SKIPIF1<0分別是四邊形SKIPIF1<0的邊SKIPIF1<0,SKIPIF1<0的中點(diǎn),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則線段SKIPIF1<0的長是___________.6.(2021·上海市市西中學(xué))空間四邊形SKIPIF1<0中,SKIPIF1<0分別是SKIPIF1<0邊的中點(diǎn),且SKIPIF1<0,則SKIPIF1<0____________.7.(2022·上海理工大學(xué)附屬中學(xué))如圖,定圓SKIPIF1<0的半徑為3,A,B為圓SKIPIF1<0上的兩點(diǎn),且SKIPIF1<0的最小值為2,則SKIPIF1<0______.題組四題組四幾何中的最值1.(2022·河南南陽·高一期末)已知SKIPIF1<0是SKIPIF1<0的邊SKIPIF1<0上一點(diǎn),且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·湖南張家界)如圖,在梯形ABCD中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若M,N是線段BC上的動點(diǎn),且SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·湖南)線段SKIPIF1<0是圓SKIPIF1<0的一條直徑,直線SKIPIF1<0上有一動點(diǎn)SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·廣東廣州·)平面四邊形SKIPIF1<0中,SKIPIF1<0,則SKIPIF1<0最小值(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<05.(2022·浙江·鎮(zhèn)海中學(xué))已知平面向量SKIPIF1<0、SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0所成夾角的最大值是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06.(2022·湖南·周南中學(xué))已知邊長為2的菱形ABCD中,點(diǎn)F為BD上一動點(diǎn),點(diǎn)E滿足SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.0 B.SKIPIF1<0 C.SKIPIF1<0 D.27.(2022·浙江麗水)已知平面向量SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最小值是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<08.(2022·河南)已知點(diǎn)SKIPIF1<0是圓:SKIPIF1<0上的動點(diǎn),點(diǎn)SKIPIF1<0是以坐標(biāo)原點(diǎn)為圓心的單位圓上的動點(diǎn),且SKIPIF1<0,則SKIPIF1<0的最大值為(
)A.5 B.6 C.7 D.8題組五題組五三角的四心1.(2022·湖北武漢)在三棱錐SKIPIF1<0中.作SKIPIF1<0平面SKIPIF1<0,垂足為SKIPIF1<0.①若三條側(cè)棱SKIPIF1<0與底面SKIPIF1<0所成的角相等,則SKIPIF1<0是SKIPIF1<0的(
)心;②若三個側(cè)面SKIPIF1<0與底面SKIPIF1<0所成的二面角相等,則SKIPIF1<0是SKIPIF1<0的(
)心:③若三組對棱SKIPIF1<0與SKIPIF1<0與SKIPIF1<0與SKIPIF1<0中有兩組互相垂直,則SKIPIF1<0是SKIPIF1<0的(
)心以上三個空依次填(
)A.外,垂,內(nèi) B.內(nèi),外,垂 C.垂,內(nèi),外 D.外,內(nèi),垂2.(2022·全國·專題練習(xí))若O在△ABC所在的平面內(nèi),a,b,c是△ABC的三邊,滿足以下條件SKIPIF1<0,則O是△ABC的(
)A.垂心 B.重心 C.內(nèi)心 D.外心3.(2022·重慶市長壽中學(xué)校)奔馳定理:已知SKIPIF1<0是SKIPIF1<0內(nèi)的一點(diǎn),若SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的面積分別記為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0.“奔馳定理”是平面向量中一個非常優(yōu)美的結(jié)論,這個定理對應(yīng)的圖形與“奔馳”轎車的SKIPIF1<0很相似,故形象地稱其為“奔馳定理”.如圖,已知SKIPIF1<0是SKIPIF1<0的垂心,且SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·重慶市實驗中學(xué))在平面上有SKIPIF1<0及內(nèi)一點(diǎn)O滿足關(guān)系式:SKIPIF1<0即稱為經(jīng)典的“奔馳定理”,若SKIPIF1<0的三邊為a,b,c,現(xiàn)有SKIPIF1<0則O為SKIPIF1<0的(
)A.外心 B.內(nèi)心 C.重心 D.垂心5.(2022·浙江省杭州第二中學(xué))在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0為SKIPIF1<0的重心,若SKIPIF1<0,則SKIPIF1<0外接圓的半徑為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<06(2022·四川達(dá)州)在SKIPIF1<0中,SKIPIF1<0為重心,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0___________.題組六題組六三角形的面積1.(2022·河南·新密市第一高級中學(xué))若點(diǎn)M是△ABC所在平面內(nèi)的一點(diǎn),且滿足3SKIPIF1<0-SKIPIF1<0-SKIPIF1<0=SKIPIF1<0,則△ABM與△ABC的面積之比為()A.1∶2 B.1∶3 C.1∶4 D.2∶52.(2022·江西宜春)已知SKIPIF1<0,點(diǎn)M是△ABC內(nèi)一點(diǎn)且SKIPIF1<0,則△MBC的面積為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03(2022·廣東·東莞市東華高級中學(xué))已知SKIPIF1<0是SKIPIF1<0內(nèi)部(不含邊界)一點(diǎn),若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0
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