专题7.5 空间向量的概念与运算【六大题型】
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2 '# !(") ....................................................................................................................5
3 !
!*+,) ....................................................................................................................6
4 !-.(+,) ................................................................................................................6
5 /012'#342'5 ....................................................................................................................7
6 !"67%& ............................................................................................................................9
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(3)[\Ą<Ċċ(w)²né !"E !"
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2|42'5
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1 !"#$%&
Ė 1ė2024·ĘęĚě·ĜĝĞğ¥ĠE}Z𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄEý𝐴𝐵−𝐴𝐷+ 𝐶𝐶1 = ė ğ
A|𝐵𝐷1
B|𝐷𝐵1
C|𝐴𝐶1
D|𝐶𝐴1
ür 1-1ė2024·WX·ĜĝĞğĒ A3B3C3D Ą"4Â2EГ𝐴𝐷= 𝐴𝐵+ 𝐴𝐶”]“A3B3C3D
42'ġ”"ė ğ
A|ÃAÍĢ;ÄÅ
B|Ģ;ÍÃAÄÅ
C|Ã;ÄÅ
D|ģÍÃAÍĢ;ÄÅ
ür 1-2ė23-24 ĤW·ĥĦħ0·ĨĩğÌy45Z𝐴𝐵𝐶𝐷ĄE𝐺]𝐵𝐷"Ą2EÐ𝐶𝐴+
1
2 𝐴𝐵+ 𝐴𝐷=
ė ğ
A|𝐴𝐺
B|𝐶𝐺
C|𝐵𝐺
D|𝐶𝐵
ür 1-3ė23-24 Ĥā·ĪīĬĭ·ĨĄğ}4Įį𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄE𝐶𝑀= 𝑀𝐷1E𝐶𝑄= 4𝑄𝐴1EÐ
ė ğ
A|𝐴𝑀=
1
3𝐴𝐵+
2
3𝐴𝐷+ 𝐴𝐴1
B|𝐴𝑀=
1
2𝐴𝐵+
1
3𝐴𝐷+
1
2𝐴𝐴1
C|𝐴𝑄=
1
4𝐴𝐵+
1
4𝐴𝐷+
3
4𝐴𝐴1
D|𝐴𝑄=
1
5𝐴𝐵+
1
5𝐴𝐷+
4
5𝐴𝐴1
2 '# !(")
Ė 2ė2024·İĭıIJij·ĜĝĞğÌy𝑒1E𝑒2E𝑒3ì'5E𝐴𝐵= 𝑒1 + 𝑒2 + 𝑒3E𝐵𝐶= 𝑒1 +𝜆𝑒2 +𝜇𝑒3E
µ𝐴,𝐵,𝐶12'#EÐ𝜆+ 𝜇= ė ğ
A|0
B|1
C|2
D|3
ür 2-1ė23-24 ĤW·Ĵĵ·ĨĄğÌy𝑀𝐴,𝑀𝐵]ÁÂì'#" !E𝑀𝐶= 5𝑀𝐴−3𝑀𝐵E«¬Ģ
{ė ğ
A|𝑀𝐴,𝑀𝐶'#
B|𝑀𝐵,𝑀𝐶'#
C|𝑀𝐴,𝑀𝐵,𝑀𝐶'5
D|𝑀𝐴,𝑀𝐵,𝑀𝐶ì'5
ür 2-2ė23-24 ĤW·Ķķ·Ĩĩğ}SË67ďĄEÌy2𝐴(0,0,1)E𝐵(1,2,3)E𝐶(𝑚,𝑛,2)E
!𝐴𝐵: !𝐵𝐶'#EÐ𝑚"ĸė ğ
A|0
B|
1
2
C|1
D|
3
2
ür 2-3ė23-24 ĤW·ĹĺĻ·Ĩĩğ}45Z𝐴𝐵𝐶𝐷ĄEE 𝐴𝐷"Ą2EG £5𝐵𝐶𝐷"^ļ|
𝐴𝐺:£5𝐵𝐶𝐸K°2 FEÐ
|𝐴𝐹|
|𝐴𝐺| = ė ğ
A|
1
2
B|
2
3
C|
3
4
D|
4
5
3 !
!*+,)
Ė 3ė2023·ĪīĽĶ·ĜĝĞğ}45Z𝐴𝐵𝐶𝐷ĄE𝐴𝐵= 3E𝐵𝐶= 4E𝐶𝐷= 5E𝐷𝐴= 6EÐ𝐴𝐶⋅𝐵𝐷"
ĸė ğ
A|7
B|9
C|11
D|13
ür 3-1ė2024·ĪľĿĭ·ĤğÌyŀ O `ٰJ4Įł𝑃−𝐴𝐵𝐶𝐷E𝑃𝐴= 𝐴𝐵= 2EEF ]ŀ O "ÏÄ
SŃE2 Q J4ĮłL5W"2EÐ𝑄𝐸⋅𝑄𝐹"ÎĸÔÕė ğ
A|[0,2]
B|[4−2 3,2]
C|[0,4−3]
D|[0,4−2 3]
ür 3-2ė2024·ńĦŅ·ĤğÌyġł𝑀𝑂"ó5ņŃ 3E 1E,Ą𝑂ó5ġļE𝐴𝐵]ó5
ġ"ÏÄSŃE2𝑃}ġł𝑀𝑂"Ň5W%ňEÐ𝑃𝐴⋅𝑃𝐵"þĸė ğ
A|−
9
4
B|−
3
2
C|−2
D|−1
ür 3-3ė2024·ʼnŊ·ĜĝĞğÌyġł𝑆𝑂"ó5ņŃ 2E2 P ó5ġŋW±FÏ2E2 Q Ň
5ėŌ°ō2Qó5ġŋğW±FÏ2EÐ𝑂𝑃⋅𝑂𝑄"ÎĸÔÕė ğ
A|(−4,4)
B|[−4,4]
C|(−2,2)
D|[−2,2]
4 !-.(+,)
Ė 4ė2023·ŎŏŎĭ·1ğ}1Įł P-ABC ĄE2 O △ABC "^ļE2 DEEEF A×ŇĮ PAE
PBEPC "Ą2E𝑎= 𝐴𝐹E𝑏= 𝐶𝐸E𝑐= 𝐵𝐷EÐ𝑂𝑃=ė ğ
A|
1
3𝑎+
1
3𝑏+
1
3𝑐
B|−
1
3𝑎−
1
3𝑏−
1
3𝑐
C|−
2
3𝑎−
1
3𝑏−
2
3𝑐
D|
2
3𝑎+
2
3𝑏+
2
3𝑐
ür 4-1ė23-24 Ĥā·4Őå²·ĨĩğÌy}45Z𝑂−𝐴𝐵𝐶ĄE 𝑎= 𝑂𝐴,𝑏= 𝑂𝐵,𝑐= 𝑂𝐶,𝑂𝑀=
1
3𝑀𝐴,𝑁
𝐵𝐶"Ą2E𝑀𝑁= 𝑥𝑎+𝑦𝑏+𝑧𝑐 EÐ 𝑥+ 𝑦+ 𝑧= ė ğ
A|3
B|
3
4
C|
1
2
D|
1
3
ür 4-2ė2024·őŒĪœŔŕ·ĜĝĞğ¥ĠE}¨{ĮŖ1"£¤ı5Z𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄE𝑀
𝐴1𝐶1:𝐵1𝐷1K2E∠𝐵𝐴𝐷= ∠𝐵𝐴𝐴1 = ∠𝐷𝐴𝐴1 = 60°EÐ𝐵𝑀"ė ğ
A| 5
4
B| 3
4
C| 5
2
D| 3
2
ür 4-3ė23-24 ĤW·ŗĴ·èŘğ}45Z𝐴𝐵𝐶𝐷Ąė¥ĠğE£5𝐴𝐵𝐷⊥£5𝐴𝐶𝐷E △𝐴𝐵𝐷]
uû1ËqE𝐴𝐷= 𝐶𝐷E𝐴𝐷⊥𝐶𝐷EM 𝐴𝐵"Ą2EN }Ň5𝐵𝐶𝐷WėřĆûŚğE𝑀𝑁= 𝑥𝐴𝐵+𝑦𝐴𝐶+𝑧
𝐴𝐷E(𝑥,𝑦,𝑧∈𝑅)ÐāśJà"]ė ğ
A|𝑥=
1
2EÐ𝑀𝑁∥£5𝐴𝐶𝐷
B|𝑧= 0EÐ𝑀𝑁⊥𝐶𝐷
C|»|𝑀𝑁|þ¼E𝑥=
1
4
D|»|𝑀𝑁|þ¼E𝑥= 0
5 /012'#342'5
Ė 5ė23-24 1W·4Őå²·èŘğ}4Įį𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄE𝐷1𝐸= 𝑘𝐷1𝐴,𝐷1𝐹= 𝑘𝐷1𝐵E
𝐷1𝐺= 𝑘𝐷1𝐶,𝐷1𝐻= 𝑘𝐷1𝐷|
(1)»𝑘=
3
4¼EŘ𝐴𝐵,𝐴𝐷,𝐴𝐴1LM𝐴𝐹
(2)/0
𝐸,𝐹,𝐺,𝐻42'5
ür 5-1ė2024 ĤW·ʼnŊ·ŜŝğÌy𝑂,𝐴,𝐵,𝐶,𝐷,𝐸,𝐹,𝐺,𝐻 9 Â2(¥Ġ)E޵𝑂𝐸= 𝑘𝑂𝐴,𝑂𝐹
= 𝑘𝑂𝐵E𝑂𝐻= 𝑘𝑂𝐷E𝐴𝐶= 𝐴𝐷+𝑚𝐴𝐵|𝐸𝐺= 𝐸𝐻+𝑚𝐸𝐹E</
(1)𝐴,𝐵,𝐶,𝐷42'5
(2)𝐴𝐶//𝐸𝐺
(3)𝑂𝐺= 𝑘𝑂𝐶|
ür 5-2ė23-24 ĤW·WX·şÿŠğ4Įį𝐴𝐵𝐶𝐷−𝐴′𝐵′𝐶′𝐷′"ıÂ5²]£¤4ûqE2𝑀}jË
#𝐴′𝐵WEµ|𝐴′𝑀| =
1
2|𝑀𝐵|E2𝑁}jË#𝐴′𝐶WEµ|𝐴′𝑁| =
1
3|𝑁𝐶||
(1)Ē !𝐴𝐵= 𝑎E𝐴𝐷= 𝑏E𝐴𝐴′ = 𝑐E𝑎3𝑏3𝑐LM !𝐷′𝑀3𝐷′𝑁
(2)</
𝑀3𝑁3𝐷′ 12'#|
ür 5-3ė23-24 Ĥā·ĪīĻĥš·ŢŜŝğÌy𝑎E𝑏E𝑐]Ąì'5" !E𝐴𝐵= 2𝑎−𝑏+
𝑐E𝐴𝐶= 𝑎+2𝑏−𝑐E𝐴𝐷= −𝑎+𝑚𝑏+𝑛𝑐.
(1)𝐵,𝐶,𝐷12'#E<𝑚,𝑛"ĸ
(2)𝐴,𝐵,𝐶,𝐷42'5E<𝑚𝑛"þĸ|
6 !"67%&
Ė 6ė2024·ńĦ·ĜĝĞğÌy !𝑎= (1,2,0),𝑏= (0,−1,1),𝑐= (2,3,𝑚)E𝑎,𝑏,𝑐'5EÐÇ
𝑚= ė ğ
A|1
B|2
C|3
D|4
ür 6-1ė2023·ľţŤťÐ·ÏğÌy !
→
𝑎= (−2,1,3),
→
𝑏= (−1,1,𝑥)E𝑎:𝑏RSEÐ|𝑎+ 2𝑏| =
ė ğ
A|2
B|5 2
C|2 13
D| 26
ür 6-2ė2024·4Ő`Ī·ĜĝĞğÌy𝑎= (2,−2,−3)E𝑏= (2,0,4)EÐcos〈𝑎,𝑏〉= ė ğ
A|4 85
85
B|−4 85
85
C|0
D|1
ür 6-3ė2024·ʼnŊ·ĜĝĞğĒ𝐴,𝐵,𝐶12}Į 2 "JZ"L5WEÐ𝐴𝐵⋅𝐴𝐶"þĸ
ė ğ
A|−
9
4
B|−2
C|−
3
2
D|−
4
3
Ï3
1|ė2024·ńĴ·ĜĝĞğ¥ĠE}45Z𝐴𝐵𝐶𝐷ĄE𝐺 △𝐴𝐶𝐷"^ļE𝐵𝐺= 𝑥𝐴𝐵+𝑦𝐴𝐶+𝑧𝐴𝐷EÐ
𝑥+ 𝑦+ 𝑧= ė ğ
A|−
1
3
B|
1
3
C|−
2
3
D|
2
3
2|ė2024·ŦĪŧŨ·ĜĝĞğĒ𝑥E𝑦∈𝑅,𝑎= (1,1,1),𝑏= (1,𝑦,𝑧),𝑐= (𝑥,−4,2)Eµ𝑎⊥𝑐,𝑏∥𝑐EÐ|2𝑎+ 𝑏| =
ė ğ
A|2 2
B|0
C|3
D|3 2
3|ė24-25 ĤW·WX·şÿŠğĒ𝑒1E𝑒2]ÁÂì'#"Í !EÌy𝐴𝐵= 2𝑒1 +𝑘𝑒2E𝐵𝐶= 𝑒1 +3
𝑒2E𝐷𝐶= 2𝑒1−𝑒2Eµ A3B3D 12'#EÐÇ
k "ĸė ğ
A|-8
B|-4
C|-2
D|8
4|ė2024·ŗĦũ·Ïğ}£¤ı5Z𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄEÌy𝐴𝐵= 4E𝐴𝐷= 3E𝐴𝐴1 = 5E
∠𝐵𝐴𝐷= 90°E∠𝐵𝐴𝐴1 = ∠𝐷𝐴𝐴1 = 60°EÐ𝐴𝐶⋅𝐵𝐷1"ĸė ğ
A|10.5
B|12.5
C|22.5
D|42.5
5|ė2024·ʼnŊ·ĜĝĞğ}Į 2 "JZ𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄEÌy𝐴𝑃= 𝐴𝐵+
1
2𝐴𝐷+
1
4𝐴𝐴1EŪ5
𝐴𝐷1𝑃:JZŇ5𝐵𝐶𝐶1𝐵1K°#𝑀𝑁EÐ#𝑀𝑁"ė ğ
A|1
B| 2
C|3 2
2
D|2 2
6|ė2024·ĘęŤū·ĤğÌyĮ 1 "JZ𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1EnJZĄļŀļ"ŀ𝑂:JZ
"äÄĮiŁE2𝑃}ŀ𝑂"JZŬŭėĆJZL5ğ%ňEÐ𝑃𝐴⋅𝑃𝐵"þĸė ğ
A|2
B|
7
4
C|
3
4
D|
1
4
7|ė2023·ńĦ·ĜĝĞğ¥ĠE}£¤ı5Z𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄEó5𝐴𝐵𝐶𝐷EŇ5𝐴1𝐴𝐷𝐷1²]J
qEµĤ5Ë𝐴1−𝐴𝐷−𝐵"120°E𝐴𝐵= 2E𝑃]𝐶1𝐷:𝐶𝐷1"K2EÐ𝐴𝑃= ė ğ
A| 3
B| 5
C| 7
D|3
8|ė2024·ĪīŮů·ĜĝĞğŰűŲ&ųŴĄãó5ŵqµ{ÏÄŇĮ:ó5RS"4Įł“Ŷ
ŷ”EŸ{ÏŶŷ𝑃−𝐴𝐵𝐶𝐷E𝑃𝐴⊥5𝐴𝐵𝐶𝐷E𝑃𝐴= 𝐴𝐵= 𝐴𝐷= 2E𝑀ó5𝐴𝐵𝐶𝐷+,`ŭ"ÏÂň2µŹź
𝑃𝑀=
5EÐ𝑃𝑀⋅𝐵𝑀"ÎĸÔÕ]ė ğ
A|[1−2 2,1 + 2 2]
B|[−1,2]
C|[1−2,−1]
D|[−1,1]
Ĥ3Ż
9|ė23-24 Ĥā·ʼnŊ·şÿŠğ¥Ġ¨ME}JZ𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄEāśärĄ%&¿¦ !𝐴𝐶1
"]ė
ğ
A| 𝐴𝐵+ 𝐵𝐶+ 𝐶𝐶1
B| 𝐴𝐴1 + 𝐴1𝐷1 + 𝐷1𝐶1
C| 𝐴𝐵+ 𝐵𝐵1 + 𝐵1𝐶1
D| 𝐴𝐴1 + 𝐴1𝐵1 + 𝐵1𝐶1|
10|ė2024·Ę꿎·Ĥğ¥ĠE}£¤ı5Z ABCD﹣A1B1C1D1 ĄEnō2 A ž2"1ÄĮ²]
1EµſñƀƁ"Ê˲]
π
3EM A1C1 : B1D1 "K2|𝐴𝐵= 𝑎E𝐴𝐷= 𝑏E𝐴𝐴1 = 𝑐EÐāśƂJà"
]ėƃƃğ
A|𝐶𝑀= −
1
2𝑎−
1
2𝑏+ 𝑐
B|⟨𝐶𝑀,𝐴𝐶1⟩=
π
3
C|𝐵𝐷1 = 𝑎+ 𝑏+ 𝑐
D|𝐴𝐷⋅𝐵𝐷1 = 1
11|ė2024·ĪīĦĵ·ĤğÌy£¤ı5Z𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1"ĮŖ 2E∠𝐴1𝐴𝐵= ∠𝐴1
𝐴𝐷= ∠𝐵𝐴𝐷= 60°E2𝑃} △𝐴1𝐵𝐷`EÐė ğ
A|𝐴1𝑃//£5𝐵1𝐶𝐷1
B|𝐴1𝑃⊥𝐴𝐶1
C||𝑃𝐶1| ≥
6|𝐴𝑃|
D||𝐴𝑃| + |𝑃𝐶1| ≥2 6
13p
12|ė2024·WX·1ğÌy !𝑎= (1,−1,0)E𝑏= (0,1,1)E𝑐= (1,2,𝑚)'5EÐÇ
𝑚=
.
13|ė2024·ĘęƄĦ·Ïğ}1Įį𝐴𝐵𝐶−𝐴1𝐵1𝐶1ĄE𝐴𝑀= 2𝑀𝐵E𝐴1𝑁= 𝑚𝐴1𝐶1Eµ𝐵𝑁//£5𝐴1𝐶𝑀E
Ð𝑚"ĸ
.
14|ė2024·Ĺĺ·ÏğÌy𝑒1,𝑒2] !E⟨e1,e2⟩= 105°E !𝑎Źź𝑎⋅𝑒1 = 1E𝑎⋅𝑒2 =
2Eµj°±F𝑥,𝑦∈RE²{|𝑎−(𝑥𝑒1 + 𝑦𝑒2)| ≥|𝑎−(𝑥0𝑒1 + 𝑦0𝑒2)| = 1ė,Ą𝑥0,𝑦0 ∈RğEÐ|𝑎| =
|
43Dƅ
15|ė23-24 Ĥā·Īī·şƆĝŝğÌy£¤ı5Z𝐴𝐵𝐶𝐷−𝐴′𝐵′𝐶′𝐷′Eýāś !LƇrEŞ}ĠĄ7
Āýïƈ" !
(1)𝐴𝐵+ 𝐴𝐷+ 𝐴𝐴′
(2)𝐷𝐷′−𝐴𝐵+ 𝐵𝐶
(3)𝐴𝐵+ 𝐴𝐷+
1
2 𝐷𝐷′−𝐵𝐶.
16|ė23-24 Ĥā·Īīmĭ·ĨĄğÌy12𝐴(−2,0,2)E𝐵(−1,1,2)E𝐶(−3,0,4)EĒ𝑎= 𝐴𝐵E𝑏= 𝐴𝐶|
(1)𝑘𝑎+ 𝑏:𝑘𝑎−2𝑏©iRSE<Ç
𝑘"ĸ
(2)|𝑐| = 3E𝑐//𝐵𝐶E<𝑐|
17|ė2024 1·ʼnŊ·Ŝŝğ¥ĠEÌy4Įł𝑃−𝐴𝐵𝐶𝐷"ó5]ƉqEjË#𝐴𝐶,𝐵𝐷K°2𝑂E
𝑂𝐴= 4E𝑂𝐵= 3E𝑂𝑃= 4,𝑂𝑃⊥ó5𝐴𝐵𝐶𝐷E𝐸,𝐹A×ŇĮ𝑃𝐵,𝑃𝐷"Ą2E2𝑀}𝐶𝑃Wµ𝐶𝑀= 2𝑀𝑃|</
𝐴,𝐸,𝑀,𝐹42'5.
18|ė2024·İĭıIJij·ĜĝĞğ¥ĠE}Į 4 "JZ𝐴𝐵𝐶𝐷−𝐴1𝐵1𝐶1𝐷1ĄE𝐷1𝐸= 𝐵𝐹= 1EĒ
𝐷𝐴= 𝑎E𝐷𝐶= 𝑏E𝐷𝐷1 = 𝑐.
(1)Ř𝑎E𝑏E𝑐LM𝐸𝐹
(2)<𝐸𝐹".
19|ė2024·ĥĦ·ĜĝĞğ1ޤśr]DČƊƋê
%&"&E,%&Ð¥ā
|
𝑎1
𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2
𝑐3| = 𝑎1
𝑏2𝑐3 + 𝑎2𝑏3𝑐1 + 𝑎3𝑏1𝑐2−𝑎3𝑏2𝑐1−𝑎2𝑏1𝑐3−𝑎1𝑏3𝑐2.𝑎× 𝑏=|
𝑖
𝑗
𝑘
𝑥1
𝑦1
𝑧1
𝑥2
𝑦2
𝑧2|EÐ𝑎× 𝑏 !𝑎:𝑏"ƌºE
,Ą𝑎= 𝑥1𝑖+ 𝑦1𝑗+ 𝑧1𝑘(𝑥1,𝑦1,𝑧1 ∈𝑅)E𝑏= 𝑥2𝑖+ 𝑦2𝑗+ 𝑧2𝑘(𝑥2,𝑦2,𝑧2 ∈𝑅)E 𝑖,𝑗,𝑘JK-ó.n𝑂67
ƍ2EA×n𝑖,𝑗,𝑘" 𝑥Ǝ、𝑦Ǝ、𝑧Ǝ"J ŏYSË67ďEÌy𝐴,𝐵]SË67ďĄŌ°𝑂
"ì´Á2.
(1)①𝐴(0,2,1),𝐵(−1,3,2)E<𝑂𝐴× 𝑂𝐵
②/0
𝑂𝐴× 𝑂𝐵+ 𝑂𝐵× 𝑂𝐴= 0.
(2) △𝐴𝑂𝐵"5*𝑆△𝐴𝑂𝐵E/0
𝑆△𝐴𝑂𝐵=
1
2|𝑂𝐴× 𝑂𝐵|
(3)Ĉ
(𝑂𝐴× 𝑂𝐵)
2"[\FGLMn △𝐴𝑂𝐵ó5、|𝑂𝐴× 𝑂𝐵|"1ĮłZ*"ŻlƏƐ