重难点15平面向量中的最值与范围问题十大题型（原卷版）_11.4-2_2026-11-04

      
      
重难点15 平面向量中的最值与范围问题【十大题型】

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3
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X7𝜆,𝜇∈𝑅FÂ2𝜆−𝜇4¬#%&C$    '
Aj[−2,1]
Bj[−1,1]
Cj[−1,2]
Dj[−2,2]
`< 7-3$23-24 ­õ·ö÷øù·Ă㥹'bzņ>𝐴𝑂𝐵”¤Ģ4Ģüd?ôs
2π
3 ,𝑃Cņ>ÕŇ$ė
Ę¿ń'Ö½­EFÝ𝑂𝑃= 𝑥𝑂𝐴+𝑦𝑂𝐵Fňʼn2(𝑥+ 𝑦)4"?#C$    '
Aj2
Bj 3
Cj4
Dj2 3
 8  9:;<
í 8$2024·ûĎ·6ĬĭĮ'[R △𝑂𝐴𝐵4/5s1,𝐴𝐵= 2FĒE𝑃,𝑄¤Aă𝐴𝐵ÄŊĒFđ|𝑃𝑄| = 1FÂ
𝑂𝑃⋅𝑂𝑄4"ô#s      
.
`< 8-1$2024·ŋï·6ĬĭĮ'¿Îs14ì‹>Õ3­ÕŌĢF𝑀𝑁CÕŌĢ4­\ōFE𝑃sì‹
>¾\¿Ä4ĒEF¢ō𝑀𝑁4ÎÔ"?éF𝑃𝑀⋅𝑃𝑁4¬#%&C      
.
`< 8-2$2024·ùĜŎ}ŏŐőòó·6ĬĭĮ'bz}dŅ> ABCD 7FEF C CD ¿ÄÎs 6 4œ
ŒĒ4AăF𝐴𝐷= 4F𝐴𝐵= 8 3F𝐵𝐶= 12 FÂ𝐵𝐸⋅𝐵𝐹4¬#%&s      
.  
      
      
`< 8-3$23-24 ­õ·îïœĞ·Ă㥹'ŔŕÇõŖŗFgŽ2
åæRŘ①(𝑎+ 𝑏)
2 = 𝑎2 +2𝑎
⋅𝑏+ 𝑏
2W②(𝑎−𝑏)
2 = 𝑎2−2𝑎⋅𝑏+ 𝑏
2.Ò①-②•(𝑎+ 𝑏)
2−(𝑎−𝑏)
2 = 4𝑎⋅𝑏⇔𝑎⋅𝑏= (𝑎+𝑏)
2−(𝑎−𝑏)
2
4
Fåæ…"
x۝4<řçs“9:;<”FŚĻPQś3cd8-4Ŝŝõqk_014
15s“6”4ƒ„.b
z”É4¾¿>𝐴𝐵𝐶𝐷7F𝐵𝐷= 8,𝐴𝐵⋅𝐴𝐷= 48F𝐸s𝐵𝐷7E.
(1)Ýcos∠𝐵𝐴𝐷=
12
13F! △𝐴𝐵𝐷4/5W
(2)Ý2𝐴𝐸= 𝐸𝐶F!𝐶𝐵⋅𝐶𝐷4#W
(3)Ý𝑃s./𝐴𝐵𝐶𝐷Õ­EF!𝑃𝐴⋅𝑃𝐵+ 𝑃𝐷4"ô#.
 9  =>? 
í 9$2024·ēĔŞş·­6'[R01𝑎F𝑏F𝑐Ĉĉ|𝑎| = |𝑏| = 𝑎⋅𝑏= 2F(𝑎−𝑐) ⋅(𝑏−2𝑐) = 0FÂ|𝑏−𝑐|4
"ô#s
Aj 3−1
2
Bj 7−3
2
Cj 3
2
Dj 7
2
`< 9-1$23-24 «õ·¾ģäĤ·Ă㥹'[Ròó01𝑎F𝑏Ĉĉ|2𝑎−𝑏| = 2FÝà¤01𝑐Fá•
𝑐−2𝑎⋅𝑐−𝑏= 0FÂ|𝑐|4¬#%&C$    '
      
      
Aj
6
2 ,
6
2 + 1
Bj
6
2 −1,
6
2
Cj
6
2 −1,
6
2 + 1
Dj[ 6−1, 6 + 1]
`< 9-2$23-24 «Ä·¾ģŠħ·Ă㥹'[R𝑒sòó01F01𝑎Ĉĉ
(𝑎−𝑒) ⋅(𝑎−5𝑒) = 0FÂ|𝑎+ 𝑒|
4"?#s$    '
Aj4
Bj5
Cj6
Dj7
`< 9-3$23-24 «Ä·ĦĞĦħ·Ă㥹'[R./01𝑎F𝑏F𝑐FĈĉ|𝑎| = |𝑏| = 𝑎⋅𝑏= 2Fđ
(𝑎−2𝑐) ⋅𝑏−𝑐= 0FÂ|𝑎−𝑐|4"ô#s$    '
Aj 3−1
2
Bj 7−3
2
Cj 3
2
Dj 7
2
 10  ;@()A
í 10$23-24 ­õ·ûĎ·Ă㥹'¤.€¾¿>𝐴𝐵𝐶𝐷7F𝐸s𝐶𝐷47EF𝐵𝐹=
1
3𝐵𝐶F𝐴𝐹-𝐵𝐸T·
E𝐺F¼E𝐺4}Aþÿ-įA𝐵𝐴F𝐵𝐶T·E𝑀F𝑁F𝐵𝑀= 𝜆𝐵𝐴F𝐵𝑁= 𝜇𝐵𝐶FÂ𝜆+ 2𝜇4"ô#s$    '
Aj1
Bj
8
7
Cj
9
7
Dj
9
5
`< 10-1$23-24 «Ä·ĖĐ·Ă㥹'ÀçMC
š4­_ûŢ{|Ftu74ţÀçF7üÀ
çĤŤťŦÇšŧF¤ĵ>𝐴𝐵𝐶𝐷7F∠𝐴𝐵𝐶＝120°FÇĵ>𝐴𝐵𝐶𝐷4¾\¿s}ġ0Ũý¾_ÑĢFP CÊ
¾_ÑĢĩÄ4­ĒEFÝ𝐷𝑃= 𝜆𝐷𝐴+𝜇𝐷𝐶FÂ𝜆+ 𝜇4"?#s$ũũ'
Aj5
Bj3
Cj
3
2
Dj
5
2
`< 10-2$23-24 ­õ·¾ģŪħ·ú7'¤ņ>𝑂𝐴𝐵7F∠𝐴𝑂𝐵= 60∘F𝐶sĩ𝐴𝐵Ä4­ĒEFÝ𝑂𝐶= 𝑥
𝑂𝐴+𝑦𝑂𝐵FÂ3𝑥+ 𝑦4¬#%&C      
j
`< 10-3$23-24 šÄ·ÄĆūï·Ă㥹'ìı¿> ABCDEF 7FP C△CDE Õ$ėĘ¿ń'4
ĒEF¸𝐴𝑃= 𝑚𝐴𝐵+𝑛𝐴𝐹F(𝑚,𝑛∈𝑅)FÂ𝑚+ 𝑛4¬#%&C      
.
­Bò
1j$2024·ĔŬŭĞ·6ĬĭĮ'¤.€¾¿>𝐴𝐵𝐶𝐷7𝐴= 45∘,𝐴𝐵= 1,𝐴𝐷=
2FÝ𝐴𝑃= 𝐴𝐵+𝑥𝐴𝐷(𝑥∈𝑅),
Â|𝐴𝑃|4"ô#s$    '
      
      
Aj
1
2
Bj 2
2
Cj1
Dj 2
2j$2024·īŮůģ·6ĬĭĮ'¤ △𝐴𝐵𝐶7F𝐵𝐷= 2𝐷𝐶F¼E𝐷4}AþÿT}A𝐴𝐵B𝐴𝐶·E𝐸B𝐹Fđ
𝐴𝐸= 𝑚𝐴𝐵,𝐴𝐹= 𝑛𝐴𝐶FX7𝑚> 0F𝑛> 0FÂ𝑚+ 2𝑛4"ô#s$    '
Aj2
Bj 2
Cj3
Dj
8
3
3j$2024·îïïŰ·6ĬĭĮ'[R¤İ­./Õ4«_E AFBFC Ĉĉ|𝐴𝐵| = 2F|
𝐶𝐴
|𝐶𝐴| −𝐶𝐵
|𝐶𝐵|| ≥1FÂ|𝐴𝐶+ 𝐵𝐶|
4¬#%&C$    '
Aj[0,1]
Bj[0,2]
Cj[0, 3]
Dj[0,2 3]
4j$2024·ĹĺĖĜ·š6' △𝐴𝐵𝐶C;ğ}d«d>FX7𝐴𝐵⊥𝐴𝐶,∣𝐴𝐵∣= 1F𝑃C △𝐴𝐵𝐶”¤./Õ4­
EFÝ𝐶𝑃= 𝜆𝐶𝐴+𝜇𝐶𝐵$𝜆≥0,𝜇≥0đ𝜆+ 2𝜇= 2'FÂ𝐶𝐴¤𝐶𝑃Ä4űŲ014ÎÔ4¬#%&C$    '
Aj 0,
2
2
Bj
2
2 ,1
Cj[1, 2]
Dj[ 2,2]
5j$2024·ö÷øù·«6'[R ⊙𝐶:𝑥2 + 𝑦2−10𝑥+ 9 = 0-}A𝑙T·𝐴,𝐵kEFđ ⊙𝐶ų𝑙Ŵ•kăĢĩ4
ÎÔ¯as1:3FÝ𝐷s ⊙𝐶Ä­EFÂ𝐷𝐴⋅𝐷𝐵4"?#s$    '
Aj18 2 +12
Bj16 2 +16
Cj12 2 +20
Dj10 2 +24
6j$2024·ĖĜĝĞ·«6'ÀçšC
š4ûŢVäŇþFŵŶŷषŸ;
@;
4Ź_þ
ź7F¤
ŻÄF
šC
żŽ4Ēž.bzF¤;¿ △𝐴𝐵𝐶7F𝐴𝐵= 2FÇ«\¿s}ġ0Ũý«
_ÑĢF𝑀C«_ÑĢĩÄ4­ĒEFÝ𝐵𝑀= 𝜆𝐴𝐵+𝜇𝐴𝐶FÂ𝜆+ 𝜇4"?#s$    '
Aj
1
2
Bj 3
3
Cj1
Dj
3
2
7j$2024·ùĜ·6ĬĭĮ'01𝑎F𝑏Ĉĉ〈𝑎,𝑏〉=
π
6F|𝑏| =
4
3 3Fđ∀𝑡∈RF‰;<|𝑏+𝑡𝑎| ≥|𝑏−𝑎|:ä .‡

𝑓(𝑥) = |𝑥𝑏−𝑎| + |𝑥𝑏−1
2𝑎|(𝑥∈𝑅)4"ô#s$    '
Aj
1
2
Bj1
Cj 3
Dj 5
8j$2024·¾ģäĤ·«6'[Rì‹> 𝐴𝐵𝐶𝐷 4¿Îs 1，𝑀，𝑁 þÿC¿ 𝐴𝐵，𝐴𝐷 Ä4E (ſ‰-×
      
      
EûL)Fę △𝐴𝑀𝑁，△𝐶𝑀𝑁 4/5þÿs 𝑆1，𝑆2 . Ý 𝑆1 = |𝐶𝑀⋅𝐴𝐵| ⋅|𝐶𝑁⋅𝐴𝐷| FÂ 𝑆2 4¬#%&
C$    '
Aj 1
4 ，
3
4
Bj
2−1，
3
4
Cj 1
4 ，
1
2
Dj
2−1，
1
2
šBƀ
9j$2024·ēĔīƁ·š6'Ý./01𝑎,𝑏,𝑐Ĉĉ|𝑎| = 1,|𝑏| = 1,|𝑐| = 3đ𝑎⋅𝑐= 𝑏⋅𝑐FÂ$    '
Aj|𝑎+ 𝑏+ 𝑐|4"ô#s 2
Bj|𝑎+ 𝑏+ 𝑐|4"?#s 5
Cj|𝑎−𝑏+ 𝑐|4"ô#s 2
Dj|𝑎−𝑏+ 𝑐|4"?#s 13
10j$2024·ŋāƂ7·6ĬĭĮ'¤ △𝐴𝐵𝐶7F𝐷s¿𝐴𝐶Ä­EđĈĉ𝐴𝐷=
1
2𝐷𝐶FÝ𝑃s¿𝐵𝐷Ä­EFđ
Ĉĉ𝐴𝑃= 𝜆𝐴𝐵+𝜇𝐴𝐶F𝜆F𝜇sìĻ
FÂõĊyžìČ4C$    '
Aj𝜆𝜇4"ô#s 1
Bj𝜆𝜇4"?#s
1
12
Cj
1
𝜆+
1
3𝜇4"?#s 12
Dj
1
𝜆+
1
3𝜇4"ô#s 4
11j$2024·ŋïƃƄ·š6'[R01𝑎F𝑏F𝑐s./01F|𝑎| = 1F|𝑏| = 2F𝑎⋅𝑏= 0F|𝑐−𝑎| =
1
2FÂ
$   '
Aj1 ≤|𝑐| ≤
3
2
Bj 𝑐−𝑎⋅𝑐−𝑏4"?#s1+2 5
4
Cj−1 ≤𝑏⋅𝑐≤1
DjÝ𝑐= 𝜆𝑎+𝜇𝑏FÂ𝜆+ 𝜇4"ô#s1−5
4
«BƅƆ
12j$2024·¾ģƇƈ·6ĬĭĮ'[RE𝑂,𝐴,𝐵¤İ­./Õđ𝐴sEF𝑂𝐴⋅𝐴𝐵= −2,𝑂𝐵⋅𝐴𝐵= 2,𝐶,𝐷þÿ
CE𝐵ƉƊÄ4Eđ𝐵𝐶= 2FÂ𝐵𝐷⋅𝐶𝐷4"?#-"ô#¯@C      
.
13j$2024·ö÷Ƌƌŋ·6ĬĭĮ'[R △𝐴𝐵𝐶7Fd𝐴 , 𝐵 , 𝐶”À4¿þÿs𝑎 , 𝑏 ,𝑐F∠𝐵𝐴𝐶=
π
6F𝑏= 1F
𝑐=
3FÝ𝐴𝐷=
𝑚
2(𝑚+𝑛)𝐴𝐵+
2𝑛
𝑚+𝑛𝐴𝐶FÂ|𝐴𝐷|4"ô#s      
j
14j$2024·ĹĺŃĆ·«6'¤.€¾¿>𝐴𝐵𝐶𝐷7F∠𝐴= 60°F𝐴𝐷=
2
3𝐴𝐵FE𝐸¤¿𝐷𝐶ÄFĈĉ𝐷𝐸=
1
3
𝐷𝐶FÂ01𝐴𝐸¤01𝐴𝐷Ä4űŲ01s      
$ƍ𝐴𝐷ÈÉ'WÝ𝐴𝐵= 3FE𝑀F𝑁þÿsAă𝐴𝐵F𝐵𝐶
Ä4ĒEFĈĉ𝐵𝑀+ 𝐵𝑁= 1FÂ𝐸𝑀⋅𝐸𝑁4"ô#s      
j
¾BgƎ
      
      
15j$23-24 ­õ·ĔŬĐƏ·Ă㥹'bzF¤ △𝐴𝐵𝐶7FE𝑃Ĉĉ𝐵𝑃= 2𝑃𝐶F¼E𝑃4}A-𝐴𝐵,𝐴𝐶”
¤4}AþÿT·E𝑀,𝑁FÝ𝐴𝑀= 𝜆𝐴𝐵,𝐴𝑁= 𝜇𝐴𝐶,(𝜆> 0,𝜇> 0).
(1)𝜆-𝜇4eW
(2)!𝜆+ 𝜇4"ô#
16j$23-24 ­·ēĔ·ú7'¤△𝐴𝐵𝐶7F[R𝐴𝐵= 3F𝐴𝐶= 1F𝐴𝐵⋅𝐴𝐶= −1F¸E𝑃s¿𝐵𝐶Ä­EFE
𝑄sAă𝐶𝐴ƐÎAÄ4­EFđ𝐴𝑄= 𝑡𝐴𝐶(𝑡< 0).
(1)¢𝑡= −1đ𝑃C¿𝐵𝐶Ä47EéF¸𝑃𝑄-𝐴𝐵T·E𝑀F!Aă𝐶𝑀4ÎW
(2)Ý𝑃𝐴⋅𝑃𝑄+3 = 𝐴𝑃⋅𝐴𝐵F!|𝐵𝑄|4"ô#.
17j$23-24 ­õ·ùĐÎķ·úć'bzF¸𝑂𝑥,𝑂𝑦C./ÕpTä60°d4k\
ţF𝑒1,𝑒2þÿC- x ţB
y ţì‹0İ04òó01.Ý01𝑂𝑀= 𝑥𝑒1 +𝑦𝑒2F…3ƑĻ
À(𝑥,𝑦)ƒƓ01𝑂𝑀¤+,e𝑂𝑥𝑦74+
,Fęý𝑂𝑀= (𝑥,𝑦).¤I+,e𝑂𝑥𝑦7FÝ𝑂𝐴= (3,0),𝑂𝐵= (0,2),𝑂𝑃= (3,2)F𝐸,𝐹þÿC𝑂𝐵,𝐵𝑃47EF𝐴𝐸,𝐴𝐹
þÿ-𝑂𝑃T·𝑅,𝑇kE.
      
      
(1)!
|𝑂𝑃|W
(2)!𝑂𝑅,𝑂𝑇4+,W
(3)ÝE M ¤Aă𝐴𝐹ăĒF¸𝑂𝑀= (𝑥,𝑦)F!𝑥𝑦4"?#.
18j$23-24 šÄ·ÄĆƔƕ·ú7'¤𝛥𝐴𝐵𝐶7FĈĉ
𝐴𝐵⊥𝐴𝐶FM C𝐵𝐶47E.
$1'Ý|𝐴𝐵| = |𝐴𝐶|F!01𝐴𝐵+2𝐴𝐶-012𝐴𝐵+ 𝐴𝐶4cd4Ɩō#W
$2'Ý O CAă𝐴𝑀ÄÖ½­EFđ|𝐴𝐵| = |𝐴𝐶| =
2F!𝑂𝐴⋅𝑂𝐵+ 𝑂𝐶⋅𝑂𝐴4"ô#
$3'ÝE P C∠𝐵𝐴𝐶Õ­EFđ|𝐴𝑃| = 2F𝐴𝑃⋅𝐴𝐶= 2F𝐴𝑃⋅𝐴𝐵= 1F!|𝐴𝐵+ 𝐴𝐶+ 𝐴𝑃|4"ô#.
19j$23-24 ­õ·ĔŬŬĞ·ú7'¤ěd △𝐴𝐵𝐶7Fcos𝐵=
2
2 FE𝑂s △𝐴𝐵𝐶4Ũü.
(1)Ý𝐵𝑂= 𝑥𝐵𝐴+𝑦𝐵𝐶F!𝑥+ 𝑦4"?#W
(2)Ý𝑏=
2F
$i'!º
𝑂𝐵+ sin2𝐴⋅𝑂𝐴−cos2𝐴⋅𝑂𝐶= 0W
$ii'!|3𝑂𝐵+ 2𝑂𝐴+ 𝑂𝐶|4¬#%&.