专题5.3平面向量的数量积及其应用八大题型（原卷版）_11.4-2_2026-11-04

      
      
专题5.3 平面向量的数量积及其应用【八大题型】

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= đ    ę
A −1
B 1
C 2
D −2
ÿS 1-1đ2024·ĚēěĜ·)ĖėĘęр !𝑎= (𝑚,2),𝑏= (𝑛,−1)(𝑛> 0)T⟨𝑎,𝑐⟩=
2π
3 T𝑎⋅𝑏= 0Tµ
|𝑐| =
3TÜ𝑏⋅𝑐= 3 2
2 T© 𝑎+ 𝑏⋅𝑐= đ    ę
A 3 2−2 6
2
B −6
2
C 3 2+2 6
2
D 3 6
2
ÿS 1-2đ2023·ĝĞğĠ·)ęрĂġÊ9 ABCDEF #Ê*Î 2TP jĂġÊ9 ABCDEF ÊÌĢL
?T©𝑃𝐴⋅𝑃𝐵#ģÙĤÎđ    ę
A 13
B 12
C 8
D 2 3
ÿS 1-3đ2024·hi·8)ęр?𝑁ŒÊ*Î 2 #ĂĥÊ9𝐴1,𝐴2,⋯,𝐴8#ÊÌT?𝑀ŒÊ𝐴1𝐴2ÌT©
𝐴1𝑀⋅𝐴1𝑁#óĤĦħjđ    ę
A [−4−2 2,2 2]
B [−4,4 + 2 2]
C [−2 2,4 + 2 2]
D [−2 2,4]
      
      
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A 30°
B 60°
C 120°
D 150°
ÿS 2-1đ2024·Ēēä·ĭ)ęр𝑎= ( 3,2 3)T𝑏= (−3,𝜆)TÜ𝑎+ 𝑏7𝑏#&'Î
2π
3 T©𝜆=
đ    ę
A -1
B 1
C  ± 1
D  ± 2
ÿS 2-2đ2024·Įh·ĭ)ęр !𝑎= (1−𝑥,−𝑥−3)T𝑏= (1 + 𝑥,2)T𝑎⋅𝑏= −4T©𝑎+2𝑏7𝑏#
&'Îđ    ę
A 
π
3
B 
π
4
C 
2π
3
D 
3π
4
ÿS 2-3đ2024·įh·)ĖėĘęâÊ9𝐴𝐵𝐶𝐷„T?𝐸､𝐹F±Î𝐴𝐷,𝐵𝐶#„?T|𝐶𝐷| = 2|𝐴𝐵| = 8,|𝐸𝐹|
= 5T©cos⟨𝐴𝐵,𝐷𝐶⟩= đ    ę
A 
5
16
B 
55
64
C −55
8
D −
23
40
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π
3T𝑎= −
3
2 , 1
2 T|𝑎−𝑏| = 1T©|𝑎+ 𝑏| = đ    ę
A 1
B  3
2
C  2
D  3
ÿS 3-1đ2024·ĝĞİı·8)ęр !𝑎T𝑏īĬ|𝑎| = 4T𝑏Œ𝑎^ Ì#-. !Î
1
2𝑎Tµ𝑏⊥
2𝑎−𝑏T©|𝑎+ 𝑏|#ĤÎđ    ę
A 4
B 4 3
C 16
D 48
ÿS 3-2đ2024·ĮIJ*ij·8)ęŒWâÊ9𝐴𝐵𝐶𝐷„T𝐴𝐶= 2𝐵𝐷= 4T?𝑃ÎĴWâÊ9ĵŒ
l#ĢL?T©|𝑃𝐴|2 +|𝑃𝐵|2 +|𝑃𝐶|2 +|𝑃𝐷|2#ģĶĤÎđ    ę
A 6
B 8
C 10
D 12
ÿS 3-3đ2024·ĮIJķĪ·8)ęŒ △𝐴𝐵𝐶„T∠𝐴𝐶𝐵= 120∘T|𝐴𝐶| = 3T|𝐵𝐶| = 4T𝐷𝐶⋅𝐷𝐵= 0T©|𝐴𝐵+ 𝐴𝐷|
#ģĶĤÎđ    ę
A 6 3−2
B 2 19−4
C 3 3−1
D  19−2
 4   !#+,(
Đ 4đ2024·ēĸĹĺ·)ĖėĘęр !𝑎= (𝑥,3),𝑏= (2,𝑥+ 5)TÜ𝑎⊥(𝑎−𝑏)T©𝑥= đ    ę
A 2 ³ 3
B −2³−3
C 1 ³−6
D −1³ 6
      
      
ÿS 4-1đ2024·ĻļĽľ·ĭ)ęр !𝑎= (2,4),𝑏= (3,−1)T©“𝑘=
2”j“⃗𝑎+ k⃗𝑏⊥⃗𝑎−k⃗𝑏”#
đ   ę
A ĿFćŀ@}~
B ŀ@ćĿF}~
C ĿFŀ@}~
D ŁćĿFłćŀ@}~
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đ    ę
A  ± 1
B  ±
2
C  ± 2
D  ± 2 3
ÿS 4-3đ2023·de·Bęр !𝑎= (1,1),𝑏= (1,−1)TÜ 𝑎+ 𝜆𝑏⊥𝑎+ 𝜇𝑏T©đ    ę
A 𝜆+ 𝜇= 1
B 𝜆+ 𝜇= −1
C 𝜆𝜇= 1
D 𝜆𝜇= −1
 5   !#-.
Đ 5đ2024·ŃĒńŅ·8)ęÜ£¤ !𝑎T𝑏īĬ|𝑎| = |𝑏| = |𝑎+ 𝑏|T©𝑎+2𝑏Œ𝑏^ Ì#-. !Î
đ    ę
A 2𝑏
B 
3
2𝑏
C 𝑏
D 
1
2𝑏
ÿS 5-1đ2024·ĝĞņŇ·ĭ)ęр !𝑎= (−1,2)T𝑏= (−3,1)T©𝑎Œ𝑏Ì#-. !Îđ   ę
A (−
3
2,
1
2)
B (−
1
2,1)
C (−5
5 ,2 5
5 )
D (−3 10
10 , 10
10 )
ÿS 5-2đ2024·ĒĨ·)ĖėĘęрZ[£¤ !𝑎、𝑏īĬ|𝑎+ 𝑏| = |𝑎−𝑏|T©𝑎−𝑏Œ𝑏Ì#-. !
Îđ    ę
A 𝑏
B −𝑏
C 
1
2𝑏 
D −
1
2𝑏 
ÿS 5-3đ2024·Įhňʼn·ĭ)ęр𝑥∈𝐑T !𝑎= (𝑥,2),𝑏= (2,−1)Tµ𝑎⊥𝑏T©𝑎+ 𝑏Œ𝑎Ì#-.
 !Îđ    ę
A  5
B 5
C (1,2)
D (2,−1)
 6  /0123 !(
Đ 6đ2024·ĒĨĩĪ·)ĖėĘęрŊ9𝐴𝐵𝐶𝐷#Ê*Î2,∠𝐴𝐵𝐶= 60∘Tŋ?𝑃Œ𝐵𝐶ÊÌđŌōŎ
?ęT©𝐴𝐷⋅𝐴𝑃#óĤĦħjđ    ę
A [0,1]
B [−1,2]
C [−2,2]
D [−1,1]
ÿS 6-1đ2024·âŏŐľ·)ĖėĘęőÁÔĵYT8[Ê*Î 2 #Ê8'9…}ʌ§,¸ÌT
Ê𝐵3𝐶3̅ 10 [ć§#?𝑃1T𝑃2T…T𝑃10TŒ𝑀𝑖= 𝐴𝐵2 ⋅𝐴𝑃𝑖(𝑖= 1,2, ⋅⋅⋅,10)T©𝑀1 + 𝑀2 + ⋅⋅⋅+ 𝑀10 =
      
      
đ    ę
A 18
B 180
C −18
D −180
ÿS 6-2đ2024·ĚēěĜ·)ĖėĘęőÔTр AB jœ𝑂#,ŔT𝐶jœ𝑂̍?T𝐴𝐶= 𝐶𝐵T?𝑃j
¸ŕ BC Ì#ŋ?Tµ △𝑃𝐴𝐵#"ŒÎ𝑆1Tœ𝑂#"ŒÎ𝑆2T­𝑃𝐴⋅𝑃𝐵óñģÙĤ†T
𝑆1
𝑆2 = đ   ę
A 
1
π
B 
2
π
C 
3
π
D 
4
π
ÿS 6-3đ2024·ŖĪŖľ·)ęőÔTŒÊ*Î 2 #Ă^9𝐴𝐵𝐶𝐷„ u𝐶ÎœŗT1 ÎŘŔ#œF±
Å𝐶𝐷,𝐵𝐶Ä?𝐸,𝐹 ­?𝑃ŒřŚ𝐸𝐹Ì$ŋ†T𝐵𝑃⋅𝐷𝑃#óĤĦħÎđ    ę
A  1−2 2,−1
2
B [1−2 2,−1]
C [−1,1−2]
D [1−2 2,1−2]
 7   !#456
Đ 7đ2024·śĹ*ĕ·)ę*ĒŜŝl_²IJhZŞWTőÔĵYрşŠŒšŢ„#ţWŤƒ𝑣1
#ÙĶ|𝑣1| = 10km/hTŢŜ#Ťƒ𝑣2#ÙĶ|𝑣2| = 4km/hT¢𝑣1˜𝑣2ĵ»'Î𝜃(0 < 𝜃< 𝜋)TÜşŠ@n𝐴
ţWòĂh^ ̨ÄhŞ#ťŦ𝐵ŧT©cos𝜃Äđ    ę
      
      
A −21
5
B −
2
5
C −
3
5
D −
4
5
ÿS 7-1đ2024·ŃĒŨĪ·ĭ)ę„Tőũ[ìŪòū#ÍTŬŒū#^ ÌÀ‹Mŕ
¨ŭT٦èůí[ūÃìŰMűTű#D%S
𝑊= 𝐹⋅𝑆đ>„𝑊jűT𝐹jūT𝑆j¨ŭęì
Œū𝐹1 = (2,4)˜𝐹2 = (−5,3)#ÍÁT¾?𝐴(1,0)ŭŋò?𝐵(2,4)TŒí[Ćü„íZ[ū#;ūÃì
ĵÍ#űÄđ    ę
A 25
B 5
C −5
D −25
ÿS 7-2đ2024·ĝĞŲų·ĭ)ęőÔĵYT[ìŴZÚŵžŶŷŸŹTµŧÄźŻż.рZ}
ŷÌ#ŸūF±j𝐹1,𝐹2Tµ𝐹1,𝐹27Ţ&'ŽÎ45°T|𝐹1| = |𝐹2| = 10 2NT©ì#žūÙĶÎ      
N.
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żTр𝐹1 = 1NT𝐹2 = 2NT𝐹17𝐹2#&'Î120°T©𝐹3#ÙĶÎ      
 đſƀNj#ūb¨ę
 8   !
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A −6
B −2 7
C 2 7³−2 7
D −6³ 6
ÿS 8-1đ2024·ŖĪƁƂ·8)ęŒ △𝐴𝐵𝐶„Tl' ATBTC ĵÃ#ÊF±Î aTbTcT𝐴=
2π
3 TÜ? D
īĬ𝐴𝐷⋅𝐴𝐵= 0Tµ𝐴𝐷=
4
5𝐴𝐶+
1
5𝐴𝐵T©
𝑏
𝑐= đ    ę
A 
1
2
B 2
C 
1
4
D 4
      
      
ÿS 8-2đ2024·ĝĞƃƄ·)ĖėĘęŒ △𝐴𝐵𝐶„T'𝐴,𝐵,𝐶ĵÃ#ÊF±Î𝑎,𝑏,𝑐.р𝐴𝐵⋅𝐴𝐶−𝐵𝐴⋅𝐵𝐶= 𝜆
𝐴𝐵
2
(1)Ü𝜆= 1T\] △𝐴𝐵𝐶#9Ż®
(2)Ü𝜆=
1
2TAtan(𝐵−𝐴)#ģÙĤ.
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? A Œ,¸ l ÌT? B<C Œƅ¸Ì.
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(2)Ü △𝐴𝐵𝐶ÎÊ8'9TA △𝐴𝐵𝐶"#Ħħ.
<b
1 đ2024·ƇƈĒ·)ĖėĘęр !|𝑎| = 3,|𝑎−𝑏| = |𝑎+2𝑏|T©|𝑎+ 𝑏| = đ    ę
A  3
B 2
C  5
D 3
2 đ2024·Ēēśě·)ĖėĘęÜ𝑎= (1,𝑥),𝑏= ( 2,−2)Tµ|𝑎+ 𝑏| = |𝑎−𝑏|T©𝑥= đ    ę
A  2
B −2
C −2
2
D  2
2
      
      
3 đ2024·Ļļ·)ĖėĘęÜ𝑎T𝑏j&'Î60°#Z[b¨ !T𝜆𝑎+ 𝑏72𝑎−𝑏+,T©𝜆= đ   ę
A 0
B 2
C −1
D −2
4 đ2023·ƇƈĒƉƊƋ·)ĖėĘęр !𝑎,𝑏īĬ|𝑎| = 2,𝑏= (3,0),|𝑎−𝑏| =
10T© !𝑎Œ !𝑏^ 
Ì#-. !Îđ    ę
A  1
6 ,0
B  1
3 ,0
C  1
2 ,0
D (1,0)
5 đ2024·ĚēěĜ·)ĖėĘęÜ !𝑎,𝑏īĬ|𝑎| =
2,|𝑏| = 1,|𝑎+ 𝑏| =
5T© !𝑎,𝑏&'#äåĤ
Îđ    ę
A  2
2
B −2
2
C 
1
2
D −
1
2
6 đ2024·ļƌƍƎĝ·8)ęр !𝑎= (−3,1),𝑏= (2,1)T©uÁů1ĂƏ#jđ   ę
A |𝑏−𝑎| =
5
B 𝑎^ Ì#b¨ !Î 3 10
10 ,
10
10
C  !𝑏Œ !𝑎Ì#-.Î 10
2
D Ü𝑐=
5
5 ,−2 5
5
T©𝑏⊥𝑐
7 đ2024·âŏŐľ·)ĖėĘę_Ɛ¢D#[œ9ƑƒƓŝőÔĵYT>„ŗƔF΍[Ê8'9
ƕƖTF±uÊ8'9#8}ÊÍÎĂ^9#}Êùú8[Ă^9ƓŝÄçƗƑƒƘƙT>„ƚ[
Ă^9…Z[ƛ?ƜƝŒœÌ Ü𝐴𝐵= 2𝑎T©𝐵𝐷⋅𝐶𝐸= đ    ę
A −4(2 +
3)𝑎2 B −2(2 +
3)𝑎2
C −2(3 +
3)𝑎2
D −2(1 +
3)𝑎2
8 đ2024·âŏ»ƞ·8)ęŒƟ9𝐴𝐵𝐶𝐷„T𝐴𝐵= 5T𝐴𝐷= 4T?𝐸īĬ2𝐴𝐸= 3𝐸𝐵TŒ𝐴𝐵𝐶𝐷„Tŋ?
𝑃īĬ𝑃𝐸⋅𝑃𝐵= 0T©𝐷𝑃⋅𝐴𝐶#ģÙĤÎđ    ę
A  41 +4
B  41−6
C 2 13 +4
D 2 13−6
ĭ<Ơ
9 đ2024·ĝĞ·)ĖėĘęр !𝑎= (1, 3)T𝑏= (−2,0)T©ÁÂů1ĂƏ#jđ    ę
A 𝑎⋅𝑏= 2
B 𝑎7𝑏#&'Î
π
3
C 𝑎⊥𝑎+ 2𝑏
D 𝑎+ 𝑏Œ𝑏Ì#-. !Î
1
2𝑏
      
      
10 đ2023·ĝĞŲų·)ĖėĘęр£¤ !𝑎≠𝑒T|
→
𝑒| = 1TÃĢL𝑡∈𝑅Tþ…|𝑎−𝑡𝑒| ≥|𝑎−𝑒|T©
đ    ę
A 𝑎Œ𝑒Ì#-.#
!Î 1
B |𝑎+ 𝑒| ≥|𝑎−2𝑒|
C 𝑎⊥𝑎−𝑒
D 𝑒⊥𝑎−𝑒
11 đ2024·įhơ—·)ęр𝑃Î △𝐴𝐵𝐶ĵŒl?T©ÁÂĂƏ#jđ    ę
A Ü𝑃𝐴+3𝑃𝐵+2𝑃𝐶= 0T©?𝑃Œ △𝐴𝐵𝐶#„¨¸Ì
B Ü𝑃𝐴+ 𝑃𝐵+ 𝑃𝐶= 0T©𝑃Î △𝐴𝐵𝐶#žŗ
C Ü𝐴𝐵⋅𝐴𝐶> 0T© △𝐴𝐵𝐶ÎĎ'8'9
D Ü𝐴𝑃=
1
3𝐴𝐵+
2
3𝐴𝐶T© △𝐴𝐵𝐶7 △𝐴𝐵𝑃#"ƢÎ3:2
8<wx
12 đ2024·ĚēěĜ·)ĖėĘęр !𝑎= (−2,𝑘−3)T𝑏= (𝑘,𝑘−2)Tµ𝑎⊥𝑏T©𝑘=      
.
13 đ2024·Ìƣ·)ĖėĘęр !𝑎T𝑏T𝑐īĬ|𝑎| = |𝑏| = 1T|𝑐| =
2Tµ𝑎+ 𝑏+ 𝑐= 0T©cos⟨𝑎−𝑐,𝑏−𝑐⟩
=
      
 
14 đ2024·Ƥƥįē·ĭ)ęőÔT,'Ʀ9 ABCD „T𝐴𝐵∥𝐶𝐷T𝐴𝐵⊥𝐴𝐷T𝐴𝐵= 2𝐶𝐷= 2𝐴𝐷= 2TŒ
Ƨ,'8'9 CDE „T∠𝐶= 90°T© !𝐴𝐸Œ !𝐶𝐵Ì#-. !#)Î      
®Ü MTN F±Î¸ŕ
BCTCE Ì#ŋ?Tµ𝐴𝑀⋅𝐴𝑁=
5
2T©𝑀𝐷⋅𝐷𝑁#ģĶĤÎ      
.
â<2ƨ
15 đ2024·Ƥƥįh·)ĖėĘęр !𝑎= (3,4)T𝑏= (1,𝑥)T𝑐= (1,2) 
(1)Ü𝑎⊥𝑏TA|𝑏|#Ĥ®
(2)Ü𝑐∥𝑎−2𝑏TA !𝑎−2𝑏7𝑎#&'#äåĤ 
      
      
16 đ23-24 Á·hiĞƩ·ƪ„ęр !𝑎,𝑏#&'Î
2π
3 Tµ|𝑎| = 2,|𝑏| = 4TA
(1)𝑎⋅𝑏;
(2)|𝑎−𝑏|;
(3)𝑎7𝑎−𝑏&'#äåĤ.
17 đ2024·ĮIJƫľ·)ęŒ △𝐴𝐵𝐶„Tl'𝐴īĬ 3sin2𝐴−cos2𝐴= 2.
(1)A'𝐴#ÙĶ®
(2)Ü𝐷𝐶= 2𝐵𝐷TA
|𝐴𝐷|
|𝐵𝐷|#ģÙĤ.
18 đ2024·ĮIJźľ·)ĖėĘęŒ △𝐴𝐵𝐶„Tl'𝐴,𝐵,𝐶ĵÃ#ÊF±Î𝑎,𝑏,𝑐Tр !𝑚,𝑛īĬ𝑚=
(2𝑎,−6)T𝑛= ( 2sin𝐵,𝑏)Tµ𝑚⊥𝑛 
(1)A'𝐴®
(2)Ü △𝐴𝐵𝐶jĎ'8'9Tµ𝑎= 3TA △𝐴𝐵𝐶Ƭ*#óĤĦħ 
19 đ23-24 Á·ŃĒļƭ·ƪƮęŒ,'Ʀ9𝐴𝐵𝐶𝐷„T𝐴𝐵//𝐶𝐷T∠𝐷𝐴𝐵= 90∘T𝐴𝐵= 2𝐴𝐷= 2𝐷𝐶= 4T
?𝐹j𝐵𝐶ÊÌ#„?.
(1)Ü?𝐸īĬ𝐷𝐸= 2𝐸𝐶Tµ𝐸𝐹= 𝜆𝐴𝐵+𝜇𝐴𝐷TA𝜆+ 𝜇#Ĥ®
      
      
(2)Ü?𝑃j¸ŕ𝐴𝐹Ì#ŋ?đHŎ?ęTA𝐴𝑃⋅𝐷𝑃#óĤĦħ.