版權(quán)說(shuō)明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡(jiǎn)介
第03講平面向量的數(shù)量積(精講)目錄第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶第二部分:課前自我評(píng)估測(cè)試第三部分:典型例題剖析高頻考點(diǎn)一:平面向量數(shù)量積的定義角度1:平面向量數(shù)量積的定義及辨析角度2:平面向量數(shù)量積的幾何意義高頻考點(diǎn)二:平面向量數(shù)量積的運(yùn)算角度1:用定義求數(shù)量積角度2:向量模運(yùn)算角度3:向量的夾角角度4:已知模求數(shù)量積角度5:已知模求參數(shù)高頻考點(diǎn)三:平面向量的綜合應(yīng)用高頻考點(diǎn)四:極化恒等式第四部分:高考真題感悟第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶1、平面向量數(shù)量積有關(guān)概念1.1向量的夾角已知兩個(gè)非零向量SKIPIF1<0和SKIPIF1<0,如圖所示,作SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(SKIPIF1<0)叫做向量SKIPIF1<0與SKIPIF1<0的夾角,記作SKIPIF1<0.(2)范圍:夾角SKIPIF1<0的范圍是SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),兩向量SKIPIF1<0,SKIPIF1<0共線且同向;當(dāng)SKIPIF1<0時(shí),兩向量SKIPIF1<0,SKIPIF1<0相互垂直,記作SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),兩向量SKIPIF1<0,SKIPIF1<0共線但反向.1.2數(shù)量積的定義:已知兩個(gè)非零向量SKIPIF1<0與SKIPIF1<0,我們把數(shù)量SKIPIF1<0叫做SKIPIF1<0與SKIPIF1<0的數(shù)量積(或內(nèi)積),記作SKIPIF1<0,即SKIPIF1<0,其中θ是SKIPIF1<0與SKIPIF1<0的夾角,記作:SKIPIF1<0.規(guī)定:零向量與任一向量的數(shù)量積為零.記作:SKIPIF1<0.1.3向量的投影①定義:在平面內(nèi)任取一點(diǎn)SKIPIF1<0,作SKIPIF1<0.過(guò)點(diǎn)SKIPIF1<0作直線SKIPIF1<0的垂線,垂足為SKIPIF1<0,則SKIPIF1<0就是向量SKIPIF1<0在向量SKIPIF1<0上的投影向量.②投影向量計(jì)算公式:當(dāng)SKIPIF1<0為銳角(如圖(1))時(shí),SKIPIF1<0與SKIPIF1<0方向相同,SKIPIF1<0,所以SKIPIF1<0;當(dāng)SKIPIF1<0為直角(如圖(2))時(shí),SKIPIF1<0,所以SKIPIF1<0;當(dāng)SKIPIF1<0為鈍角(如圖(3))時(shí),SKIPIF1<0與SKIPIF1<0方向相反,所以SKIPIF1<0,即SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0綜上可知,對(duì)于任意的SKIPIF1<0,都有SKIPIF1<0.2、平面向量數(shù)量積的性質(zhì)及其坐標(biāo)表示已知向量SKIPIF1<0,SKIPIF1<0為向量SKIPIF1<0和SKIPIF1<0的夾角:2.1數(shù)量積SKIPIF1<02.2模:SKIPIF1<02.3夾角:SKIPIF1<02.4非零向量SKIPIF1<0的充要條件:SKIPIF1<02.5三角不等式:SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立)SKIPIF1<0SKIPIF1<03、平面向量數(shù)量積的運(yùn)算①SKIPIF1<0②SKIPIF1<0③SKIPIF1<04、極化恒等式①平行四邊形形式:若在平行四邊形SKIPIF1<0中,則SKIPIF1<0②三角形形式:在SKIPIF1<0中,SKIPIF1<0為SKIPIF1<0的中點(diǎn),所以SKIPIF1<05、常用結(jié)論①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0第二部分:課前自我評(píng)估測(cè)試第二部分:課前自我評(píng)估測(cè)試一、判斷題1.(2022·全國(guó)·高一專題練習(xí))判斷(正確的填“正確”,錯(cuò)誤的填“錯(cuò)誤”)(1)兩個(gè)向量的數(shù)量積仍然是向量.()(2)若SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0.()(3)SKIPIF1<0,SKIPIF1<0共線?SKIPIF1<0·SKIPIF1<0=|SKIPIF1<0||SKIPIF1<0|.()(4)若SKIPIF1<0·SKIPIF1<0=SKIPIF1<0·SKIPIF1<0,則一定有SKIPIF1<0=SKIPIF1<0.()(5)兩個(gè)向量的數(shù)量積是一個(gè)實(shí)數(shù),向量的加法?減法?數(shù)乘運(yùn)算的運(yùn)算結(jié)果是向量.()2.(2021·全國(guó)·高二課前預(yù)習(xí))已知兩個(gè)向量SKIPIF1<0的夾角為60°,則∠NMP=60°.()二、單選題3.(2022·河南安陽(yáng)·高一階段練習(xí))已知向量SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(
)A.1 B.SKIPIF1<0 C.SKIPIF1<0 D.24.(2022·全國(guó)·模擬預(yù)測(cè)(文))在邊長(zhǎng)為2的正三角形SKIPIF1<0中,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.1 D.25.(2022·廣東·深圳市龍崗區(qū)德琳學(xué)校高一期中)在SKIPIF1<0中,若SKIPIF1<0,則SKIPIF1<0-定是(
)A.銳角三角形 B.直角三角形 C.鈍角三角形 D.等邊三角形第三部分:典型例題剖析第三部分:典型例題剖析高頻考點(diǎn)一:平面向量數(shù)量積的定義角度1:平面向量數(shù)量積的定義及辨析例題1.(2022·河北武強(qiáng)中學(xué)高一期中)已知向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(
)A.0 B.2 C.3 D.4例題2.(2022·山西太原·高一期中)給出以下結(jié)論,其中正確結(jié)論的個(gè)數(shù)是(
)①SKIPIF1<0
②SKIPIF1<0
③SKIPIF1<0
④SKIPIF1<0A.1 B.2 C.3 D.4例題3.(2022·江蘇·漣水縣第一中學(xué)高一階段練習(xí))在銳角SKIPIF1<0中,關(guān)于向量夾角的說(shuō)法,正確的是(
)A.SKIPIF1<0與SKIPIF1<0的夾角是銳角 B.SKIPIF1<0與SKIPIF1<0的夾角是銳角C.SKIPIF1<0與SKIPIF1<0的夾角是銳角 D.SKIPIF1<0與SKIPIF1<0的夾角是鈍角例題4.(2022·寧夏·平羅中學(xué)模擬預(yù)測(cè)(理))已知向量SKIPIF1<0的夾角為SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0方向上的投影為_(kāi)__________.角度2:平面向量數(shù)量積的幾何意義例題1.(2022·江西撫州·高一期中)已知向量SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0方向上的投影數(shù)量為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題2.(2022·全國(guó)·高三專題練習(xí)(理))在圓SKIPIF1<0中弦SKIPIF1<0的長(zhǎng)度為8,則SKIPIF1<0=(
)A.8 B.16 C.24 D.32例題3.(2022·甘肅·高臺(tái)縣第一中學(xué)高一階段練習(xí))已知SKIPIF1<0,SKIPIF1<0與SKIPIF1<0的夾角為120°,則向量SKIPIF1<0在SKIPIF1<0方向上的投影為()A.4 B.-4 C.2 D.-2例題4.(2022·吉林一中高一期中)在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0為邊上SKIPIF1<0的動(dòng)點(diǎn),則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0例題5.(2022·江西景德鎮(zhèn)·三模(理))窗花是貼在窗紙或窗戶玻璃上的剪紙,它是中國(guó)古老的傳統(tǒng)民間藝術(shù)之一.在2022年虎年新春來(lái)臨之際,人們?cè)O(shè)計(jì)了一種由外圍四個(gè)大小相等的半圓和中間正方形所構(gòu)成的剪紙窗花(如圖1).已知正方形SKIPIF1<0的邊長(zhǎng)為2,中心為SKIPIF1<0,四個(gè)半圓的圓心均在正方形SKIPIF1<0各邊的中點(diǎn)(如圖2,若點(diǎn)SKIPIF1<0在四個(gè)半圓的圓弧上運(yùn)動(dòng),則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0題型歸類練1.(2022·黑龍江·佳木斯一中高一期中)已知△ABC的外接圓圓心為O,且SKIPIF1<0,SKIPIF1<0,則向量SKIPIF1<0在向量SKIPIF1<0上的投影向量為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·內(nèi)蒙古呼和浩特·二模(理))非零向量SKIPIF1<0,SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上的正射影的數(shù)量為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·北京市第十九中學(xué)高一期中)如圖,已知四邊形ABCD為直角梯形,SKIPIF1<0,SKIPIF1<0,AB=1,AD=3,SKIPIF1<0,設(shè)點(diǎn)P為直角梯形ABCD內(nèi)一點(diǎn)(不包含邊界),則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·全國(guó)·高三專題練習(xí))在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,與SKIPIF1<0方向相同的單位向量為SKIPIF1<0,則向量SKIPIF1<0在SKIPIF1<0上的投影向量為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<05.(2022·河南河南·三模(理))在△SKIPIF1<0中,“SKIPIF1<0”是“△SKIPIF1<0為鈍角三角形”的(
)A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件6.(2022·四川·宜賓市敘州區(qū)第一中學(xué)校高一期中)在圓SKIPIF1<0中弦SKIPIF1<0,則SKIPIF1<0__________.7.(2022·四川·樹(shù)德中學(xué)高一階段練習(xí))如圖,直徑SKIPIF1<0的半圓,D為圓心,點(diǎn)C在半圓弧上,SKIPIF1<0,線段SKIPIF1<0上有動(dòng)點(diǎn)P,則SKIPIF1<0的取值范圍為_(kāi)________.高頻考點(diǎn)二:平面向量數(shù)量積的運(yùn)算角度1:用定義求數(shù)量積例題1.(2022·全國(guó)·華中師大一附中模擬預(yù)測(cè))正六邊形SKIPIF1<0的邊長(zhǎng)為2,則SKIPIF1<0=(
)A.-6 B.SKIPIF1<0 C.SKIPIF1<0 D.6例題2.(2022·廣東·東莞市東方明珠學(xué)校高一期中)已知正方形SKIPIF1<0的邊長(zhǎng)為2,SKIPIF1<0為SKIPIF1<0的中點(diǎn),則SKIPIF1<0(
)A.SKIPIF1<0 B.0 C.SKIPIF1<0 D.2例題3.(2022·北京·中關(guān)村中學(xué)高一期中)已知SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0的夾角為SKIPIF1<0,則SKIPIF1<0(
)A.1 B.SKIPIF1<0 C.2 D.SKIPIF1<0例題4.(2022·安徽·高二階段練習(xí))已知平面向量SKIPIF1<0,單位向量SKIPIF1<0滿足SKIPIF1<0,則向量SKIPIF1<0與SKIPIF1<0夾角為_(kāi)__________.例題5.(2022·上海奉賢區(qū)致遠(yuǎn)高級(jí)中學(xué)高一期中)在SKIPIF1<0中,SKIPIF1<0,則SKIPIF1<0_______角度2:向量模運(yùn)算例題1.(2022·山東濰坊·高一期中)已知SKIPIF1<0,SKIPIF1<0是平面內(nèi)的兩個(gè)向量,SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題2.(2022·四川綿陽(yáng)·高一期中)已知向量SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題3.(2022·河南安陽(yáng)·高一階段練習(xí))已知向量SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.1 C.2 D.4例題4.(2022·河南新鄉(xiāng)·高一期中)已知向量SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題5.(2022·河南·模擬預(yù)測(cè)(理))已知平面向量SKIPIF1<0,SKIPIF1<0的夾角為SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0______.例題6.(2022·河南·模擬預(yù)測(cè)(文))已知向量SKIPIF1<0,SKIPIF1<0,且向量SKIPIF1<0與SKIPIF1<0的夾角為SKIPIF1<0,則SKIPIF1<0______.角度3:向量的夾角例題1.(2022·內(nèi)蒙古赤峰·模擬預(yù)測(cè)(理))若向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0的夾角為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題2.(2022·山東濟(jì)南·三模)已知單位向量SKIPIF1<0、SKIPIF1<0、SKIPIF1<0,滿足SKIPIF1<0,則向量SKIPIF1<0和SKIPIF1<0的夾角為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題3.(2022·河北邯鄲·二模)若向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則向量SKIPIF1<0與SKIPIF1<0夾角的余弦值為(
).A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題4.(2022·河南·扶溝縣第二高中高一階段練習(xí))已知向量SKIPIF1<0,SKIPIF1<0是單位向量,若SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0的夾角為_(kāi)____.例題5.(2022·山東煙臺(tái)·高一期中)若SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0的夾角大小為_(kāi)_____.例題6.(2022·安徽·巢湖市第一中學(xué)模擬預(yù)測(cè)(文))已知向量SKIPIF1<0,SKIPIF1<0,則“SKIPIF1<0”是“SKIPIF1<0與SKIPIF1<0的夾角為銳角”的(
)A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件例題7.(2022·遼寧·東北育才學(xué)校高一期中)已知向量SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0與SKIPIF1<0的夾角為銳角,則實(shí)數(shù)SKIPIF1<0的取值范圍是______.例題8.(2022·黑龍江·勃利縣高級(jí)中學(xué)高一期中)已知向量SKIPIF1<0與向量SKIPIF1<0所成角為鈍角.則SKIPIF1<0的取值范圍是______.例題9.(2022·河北·高一期中)已知向量SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0的夾角為鈍角,則SKIPIF1<0的取值范圍為_(kāi)_____角度4:已知模求數(shù)量積例題1.(2022·吉林長(zhǎng)春·模擬預(yù)測(cè)(文))已知向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題2.(2022·全國(guó)·模擬預(yù)測(cè)(文))已知向量SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0(
)A.6 B.SKIPIF1<0 C.SKIPIF1<0 D.-2例題3.(2022·北京十五中高一期中)若向量SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0_____.例題4.(2022·安徽馬鞍山·三模(文))設(shè)向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0___________.例題5.(2022·貴州貴陽(yáng)·二模(理))已知向量SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0________.角度5:已知模求參數(shù)例題1.(2022·全國(guó)·高三專題練習(xí))已知SKIPIF1<0,向量SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.-2 D.2例題2.(2022·廣東·高一階段練習(xí))已知單位向量SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題3.(2022·湖北鄂州·高二期末)已知向量SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題4.(2022·安徽·高二階段練習(xí)(文))已知向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的值為_(kāi)_____.題型歸類練1.(2022·北京·潞河中學(xué)三模)已知菱形SKIPIF1<0的邊長(zhǎng)為SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·河南·方城第一高級(jí)中學(xué)模擬預(yù)測(cè)(理))已知向量SKIPIF1<0,SKIPIF1<0為單位向量,SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0的夾角為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·全國(guó)·高一單元測(cè)試)在SKIPIF1<0中,角SKIPIF1<0、SKIPIF1<0、SKIPIF1<0所對(duì)的邊分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<04.(2022·四川省內(nèi)江市第六中學(xué)高一期中(理))如圖,SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,P為CD上一點(diǎn),且滿足SKIPIF1<0,若AC=3,AB=4,則SKIPIF1<0的值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<05.(2022·湖南·長(zhǎng)沙市明德中學(xué)二模)已知非零向量SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,則向量SKIPIF1<0與向量SKIPIF1<0夾角的余弦值為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<06.(2022·廣東·模擬預(yù)測(cè))已知單位向量SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,則向量SKIPIF1<0,SKIPIF1<0的夾角為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<07.(2022·安徽師范大學(xué)附屬中學(xué)模擬預(yù)測(cè)(文))設(shè)SKIPIF1<0為非零向量,且SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0的夾角為_(kāi)__________.8.(2022·廣東廣州·三模)已知SKIPIF1<0為單位向量,若SKIPIF1<0,則SKIPIF1<0__________.9.(2022·山東濟(jì)寧·三模)在邊長(zhǎng)為SKIPIF1<0的等邊SKIPIF1<0中,已知SKIPIF1<0,點(diǎn)SKIPIF1<0在線段SKIPIF1<0上,且SKIPIF1<0,則SKIPIF1<0________.高頻考點(diǎn)三:平面向量的綜合應(yīng)用例題1.(2022·湖南·高二階段練習(xí))“趙爽弦圖”是中國(guó)古代數(shù)學(xué)的圖騰,它是由四個(gè)全等的直角三角形與一個(gè)小正方形拼成的一個(gè)大正方形.如圖,某人仿照趙爽弦圖,用四個(gè)三角形和一個(gè)小的平行四邊形拼成一個(gè)大平行四邊形,其中SKIPIF1<0分別是SKIPIF1<0的中點(diǎn),若SKIPIF1<0,則SKIPIF1<0(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題2.(2022·河南·唐河縣第一高級(jí)中學(xué)高一階段練習(xí))2022年北京冬奧會(huì)開(kāi)幕式中,當(dāng)《雪花》這個(gè)節(jié)目開(kāi)始后,一片巨大的“雪花”呈現(xiàn)在舞臺(tái)中央,十分壯觀.理論上,一片雪花的周長(zhǎng)可以無(wú)限長(zhǎng),圍成雪花的曲線稱作“雪花曲線”,又稱“科赫曲線”,是瑞典數(shù)學(xué)家科赫在1904年研究的一種分形曲線.如圖是“雪花曲線”的一種形成過(guò)程:從一個(gè)正三角形開(kāi)始,把每條邊分成三等份,然后以各邊的中間一段為底邊分別向外作正三角形,再去掉底邊,重復(fù)進(jìn)行這一過(guò)程.已知圖①中正三角形的邊長(zhǎng)為6,則圖③中SKIPIF1<0的值為(
)A.24 B.6 C.SKIPIF1<0 D.SKIPIF1<0例題4.(2022·江蘇·常州市第二中學(xué)高一階段練習(xí))如圖,已知平行四邊形SKIPIF1<0的對(duì)角線相交于點(diǎn)SKIPIF1<0,過(guò)點(diǎn)SKIPIF1<0的直線與SKIPIF1<0所在直線分別交于點(diǎn)SKIPIF1<0,SKIPIF1<0,滿足SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0的值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0例題5.(2022·江蘇·常州市第二中學(xué)高一階段練習(xí))在梯形SKIPIF1<0中,SKIPIF1<0分別為線段SKIPIF1<0,SKIPIF1<0上的動(dòng)點(diǎn).(1)求SKIPIF1<0;(2)若SKIPIF1<0,求SKIPIF1<0;(3)若SKIPIF1<0,求SKIPIF1<0的最小值;題型歸類練1.(2022·浙江·高一階段練習(xí))已知P是SKIPIF1<0的外心,且SKIPIF1<0,則cosC=(
)A.-SKIPIF1<0 B.-SKIPIF1<0 C.SKIPIF1<0或-SKIPIF1<0 D.SKIPIF1<0或-SKIPIF1<02.(2022·河南洛陽(yáng)·高二階段練習(xí)(文))在△SKIPIF1<0中,點(diǎn)D滿足SKIPIF1<0=SKIPIF1<0,直線SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0,則SKIPIF1<0的值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·山東淄博·高一期中)如圖,SKIPIF1<0,則SKIPIF1<0_________4.(2022·湖南·模擬預(yù)測(cè))在三角形ABC中,點(diǎn)D在邊BC上,若SKIPIF1<0,SKIPIF1<0SKIPIF1<0,則SKIPIF1<0______.5.(2022·浙江·高一階段練習(xí))平面內(nèi)的三個(gè)向量SKIPIF1<0.(1)若SKIPIF1<0,求實(shí)數(shù)k的值;(2)若SKIPIF1<0,求實(shí)數(shù)k的值.6.(2022·重慶市二0三中學(xué)校高一階段練習(xí))已知平面向量SKIPIF1<0.(1)若SKIPIF1<0,求SKIPIF1<0;(2)若SKIPIF1<0,求SKIPIF1<0與SKIPIF1<0夾角的余弦值.7.(2022·湖北·高一階段練習(xí))已知平行四邊形SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,AE和BF交于點(diǎn)P.(1)試用SKIPIF1<0,SKIPIF1<0表示向量SKIPIF1<0.(2)若SKIPIF1<0的面積為SKIPIF1<0,SKIPIF1<0的面積為SKIPIF1<0,求SKIPIF1<0的值.(3)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0的余弦值.8.(2022·四川省內(nèi)江市第六中學(xué)高一期中(文))如圖,設(shè)△ABC中角A,B,C所對(duì)的邊分別為a,b,c,AD為BC邊上的中線,已知SKIPIF1<0,c=1且SKIPIF1<0.(1)求b邊的長(zhǎng);(2)求△ABC的面積;(3)設(shè)點(diǎn)E,F(xiàn)分別為邊AB,AC上的動(dòng)點(diǎn),線段EF交AD于G,且△AEF的面積為△ABC面積的一半,求SKIPIF1<0的最小值.高頻考點(diǎn)四:極化恒等式例題1.(2021·全國(guó)·高一課時(shí)練習(xí))閱讀一下一段文字:SKIPIF1<0,SKIPIF1<0,兩式相減得:SKIPIF1<0,我們把這個(gè)等式稱作“極化恒等式”,它實(shí)現(xiàn)了在沒(méi)有夾角的參與下將兩個(gè)向量的數(shù)量積運(yùn)算化為“?!钡倪\(yùn)算.試根據(jù)上面的內(nèi)容解決以下問(wèn)題:如圖,在SKIPIF1<0中,SKIPIF1<0是SKIPIF1<0的中點(diǎn),SKIPIF1<0,SKIPIF1<0是SKIPIF1<0上的兩個(gè)三等分點(diǎn).(1)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0的值;(2)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0的值.例題2.(2022·河北唐山·高三期末)SKIPIF1<0中,SKIPIF1<0為SKIPIF1<0的中點(diǎn),SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0______.
溫馨提示
- 1. 本站所有資源如無(wú)特殊說(shuō)明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒(méi)有圖紙預(yù)覽就沒(méi)有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。
最新文檔
- 課件:《躺在波浪上看書(shū)》
- 2022高考英語(yǔ)唐山市路南區(qū)閱讀短文復(fù)習(xí)單詞暑假練習(xí)(12)及答案
- 【原創(chuàng)】江蘇省2020-2021學(xué)年高二第一學(xué)期第五次周練語(yǔ)文試題
- 山東省德州市2025屆高三上學(xué)期期中考試數(shù)學(xué)試卷(含解析)
- 【名師一號(hào)】2020-2021學(xué)年高中地理湘教版必修二-雙基限時(shí)練3
- 2024-2025學(xué)年部編版歷史七年級(jí)下冊(cè)期中綜合評(píng)估卷(第1-11課)(含答案)
- 【名師一號(hào)】2020-2021學(xué)年高中生物必修三:第二章-動(dòng)物和人體生命活動(dòng)的調(diào)節(jié)-單元檢測(cè)
- 《金版學(xué)案》2022屆高考數(shù)學(xué)理科一輪復(fù)習(xí)課時(shí)作業(yè)-3-8解三角形的應(yīng)用-
- 一年級(jí)數(shù)學(xué)計(jì)算題專項(xiàng)練習(xí)1000題匯編
- 【2020秋備課】高中生物學(xué)案新人教版必修3-2.1-通過(guò)神經(jīng)系統(tǒng)的調(diào)節(jié)
- 人體解剖上肢血管
- 精品解析浙教版科學(xué) 九年級(jí)上冊(cè) 3.43 簡(jiǎn)單機(jī)械之機(jī)械效率 同步練習(xí)
- 六棱塊護(hù)坡施工方案
- 機(jī)械制圖課件(完整版)
- 夸美紐斯-大教學(xué)論-文本細(xì)讀
- 《行政組織學(xué)小抄》word版
- 日立多聯(lián)機(jī)系統(tǒng)調(diào)試培訓(xùn)教材
- (完整版)環(huán)境科學(xué)與工程-專業(yè)英語(yǔ)詞匯必備(免費(fèi))
- 交通管理與控制課件(全)全書(shū)教學(xué)教程完整版電子教案最全幻燈片
- 小學(xué)鋼筆字寫(xiě)字課教案(20課時(shí)完整版)
- 紅金簡(jiǎn)約風(fēng)教師退休歡送會(huì)PPT通用模板
評(píng)論
0/150
提交評(píng)論