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Задачи од векторски производ на вектори

Module by: Beti Andonovic. E-mail the author

Summary: решени задачи од векторски производ на вектори solved problems of vector product

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Решени задачи од векторски производ на вектори

1. Да се определи параметарот λλ size 12{λ} {}, така што векторите p→=λa→−5b→p→=λa→−5b→ size 12{ {p} cSup { size 8{ rightarrow } } =λ {a} cSup { size 8{ rightarrow } } - 5 {b} cSup { size 8{ rightarrow } } } {} и q→=3a→−b→q→=3a→−b→ size 12{ {q} cSup { size 8{ rightarrow } } =3 {a} cSup { size 8{ rightarrow } } - {b} cSup { size 8{ rightarrow } } } {} да бидат колинеарни, ако a→a→ size 12{ {a} cSup { size 8{ rightarrow } } } {} и b→b→ size 12{ {b} cSup { size 8{ rightarrow } } } {} не се колинеарни.

Решение.

За да бидат колинеарни векторите p→p→ size 12{ {p} cSup { size 8{ rightarrow } } } {} и q→q→ size 12{ {q} cSup { size 8{ rightarrow } } } {} , потребно е p→×q→=0→p→×q→=0→ size 12{ {p} cSup { size 8{ rightarrow } } times {q} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}. Односно,

(λa→−5b→)×(3a→−b→)=0→(λa→−5b→)×(3a→−b→)=0→ size 12{ \( λ {a} cSup { size 8{ rightarrow } } - 5 {b} cSup { size 8{ rightarrow } } \) times \( 3 {a} cSup { size 8{ rightarrow } } - {b} cSup { size 8{ rightarrow } } \) = {0} cSup { size 8{ rightarrow } } } {}.

3λa→×a→−λa→×b→−15b→×a→+5b→×b→=0→3λa→×a→−λa→×b→−15b→×a→+5b→×b→=0→ size 12{3λ {a} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } - λ {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } - "15" {b} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } +5 {b} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}.

Бидејќи a→×a→=0→a→×a→=0→ size 12{ {a} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {} и b→×b→=0→b→×b→=0→ size 12{ {b} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}, имаме

−λa→×b→+15a→×b→=0→−λa→×b→+15a→×b→=0→ size 12{ - λ {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } +"15" {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}.

(15−λ)a→×b→=0→(15−λ)a→×b→=0→ size 12{ \( "15" - λ \) {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}.

a→a→ size 12{ {a} cSup { size 8{ rightarrow } } } {} и b→b→ size 12{ {b} cSup { size 8{ rightarrow } } } {} не се колинеарни вектори, па мора 15−λ=015−λ=0 size 12{"15" - λ=0} {}, т.е. λ=15λ=15 size 12{λ="15"} {}.

2. Да се докаже дека ако a→×b→+b→×c→+c→×a→=0→a→×b→+b→×c→+c→×a→=0→ size 12{ {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } + {b} cSup { size 8{ rightarrow } } times {c} cSup { size 8{ rightarrow } } + {c} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}, тогаш a→a→ size 12{ {a} cSup { size 8{ rightarrow } } } {}, b→b→ size 12{ {b} cSup { size 8{ rightarrow } } } {} и c→c→ size 12{ {c} cSup { size 8{ rightarrow } } } {} се компланарни.

Решение.

a → × b → + b → × c → + c → × a → = a → × b → + b → × c → + c → × a → = size 12{ {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } + {b} cSup { size 8{ rightarrow } } times {c} cSup { size 8{ rightarrow } } + {c} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } ={}} {} (1)
= a → × b → − c → × b → + c → × a → = = a → × b → − c → × b → + c → × a → = size 12{ {}= {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } + {c} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } ={}} {} (2)
= ( a → − c → ) × b → + c → × a → = = ( a → − c → ) × b → + c → × a → = size 12{ {}= \( {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } \) times {b} cSup { size 8{ rightarrow } } + {c} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } ={}} {} (3)

= ( a → − c → ) × b → + c → × a → − a → × a → = = ( a → − c → ) × b → + c → × a → − a → × a → = size 12{ {}= \( {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } \) times {b} cSup { size 8{ rightarrow } } + {c} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } - {a} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } ={}} {}

= ( a → − c → ) × b → − ( a → − c → ) × a → = = ( a → − c → ) × b → − ( a → − c → ) × a → = size 12{ {}= \( {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } \) times {b} cSup { size 8{ rightarrow } } - \( {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } \) times {a} cSup { size 8{ rightarrow } } ={}} {} (4)

=(a→−c→)×(b→−a→)=0→=(a→−c→)×(b→−a→)=0→ size 12{ {}= \( {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } \) times \( {b} cSup { size 8{ rightarrow } } - {a} cSup { size 8{ rightarrow } } \) = {0} cSup { size 8{ rightarrow } } } {}.

Оттука следува дека мора a→−c→a→−c→ size 12{ {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } } {} и b→−a→b→−a→ size 12{ {b} cSup { size 8{ rightarrow } } - {a} cSup { size 8{ rightarrow } } } {} да бидат колинеарни, т.е.

a→−c→=λ(b→−a→)a→−c→=λ(b→−a→) size 12{ {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } =λ \( {b} cSup { size 8{ rightarrow } } - {a} cSup { size 8{ rightarrow } } \) } {}.

a→−c→=λb→−λa→a→−c→=λb→−λa→ size 12{ {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } =λ {b} cSup { size 8{ rightarrow } } - λ {a} cSup { size 8{ rightarrow } } } {}.

(1+λ)a→−λb→−c→=0→(1+λ)a→−λb→−c→=0→ size 12{ \( 1+λ \) {a} cSup { size 8{ rightarrow } } - λ {b} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {}, од каде следува дека мора a→a→ size 12{ {a} cSup { size 8{ rightarrow } } } {}, b→b→ size 12{ {b} cSup { size 8{ rightarrow } } } {} и c→c→ size 12{ {c} cSup { size 8{ rightarrow } } } {} да бидат компланарни.

3. Да се пресмета плоштината на паралелограмот конструиран над векторите:

a→=2i→+3j→−k→a→=2i→+3j→−k→ size 12{ {a} cSup { size 8{ rightarrow } } =2 {i} cSup { size 8{ rightarrow } } +3 {j} cSup { size 8{ rightarrow } } - {k} cSup { size 8{ rightarrow } } } {} и b→=−3i→−j→+k→b→=−3i→−j→+k→ size 12{ {b} cSup { size 8{ rightarrow } } = - 3 {i} cSup { size 8{ rightarrow } } - {j} cSup { size 8{ rightarrow } } + {k} cSup { size 8{ rightarrow } } } {}.

Решение.

Table 1
graphics1.png
Слика 1
a → = 2,3, − 1 a → = 2,3, − 1 size 12{ {a} cSup { size 8{ rightarrow } } = left lbrace 2,3, - 1 right rbrace } {} (5)
b → = − 3, − 1,1 b → = − 3, − 1,1 size 12{ {b} cSup { size 8{ rightarrow } } = left lbrace - 3, - 1,1 right rbrace } {} (6)

a→×b→=∣3−1−11∣,∣−121−3∣,∣23−31∣=2,1,7a→×b→=∣3−1−11∣,∣−121−3∣,∣23−31∣=2,1,7 size 12{ {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } = left lbrace lline matrix { 3 {} # - 1 {} ## - 1 {} # 1{} } rline , lline matrix { - 1 {} # 2 {} ## 1 {} # - 3{} } rline , lline matrix { 2 {} # 3 {} ## - 3 {} # 1{} } rline right rbrace = left lbrace 2,1,7 right rbrace } {}.

P=∣a→×b→∣=22+12+72=54=36P=∣a→×b→∣=22+12+72=54=36 size 12{P= lline {a} cSup { size 8{ rightarrow } } times {b} cSup { size 8{ rightarrow } } rline = sqrt {2 rSup { size 8{2} } +1 rSup { size 8{2} } +7"" lSup { size 8{2} } } = sqrt {"54"} =3 sqrt {6} } {}.

4. Да се пресмета плоштината на триаголникот ABC со дадени темиња:

A(4,−1,2)A(4,−1,2) size 12{A \( 4, - 1,2 \) } {}, B(−8,0,4)B(−8,0,4) size 12{B \( - 8,0,4 \) } {} и C(8,2,3)C(8,2,3) size 12{C \( 8,2,3 \) } {}.

Решение.

P ABC = 1 2 P ABDC = 1 2 ∣ AB → × AC → ∣ P ABC = 1 2 P ABDC = 1 2 ∣ AB → × AC → ∣ size 12{P rSub { size 8{ ital "ABC"} } = { {1} over {2} } P rSub { size 8{ ital "ABDC"} } = { {1} over {2} } lline { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } rline } {} (7)
AB → = − 12 , 1,2 AB → = − 12 , 1,2 size 12{ { ital "AB"} cSup { size 8{ rightarrow } } = left lbrace - "12",1,2 right rbrace } {} (8)
Table 2
graphics2.png
Слика 2

AB→×AC→=∣1231∣,∣2−1214∣,∣−12143∣=−5,20,−40AB→×AC→=∣1231∣,∣2−1214∣,∣−12143∣=−5,20,−40 size 12{ { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } = left lbrace lline matrix { 1 {} # 2 {} ## 3 {} # 1{} } rline , lline matrix { 2 {} # - "12" {} ## 1 {} # 4{} } rline , lline matrix { - "12" {} # 1 {} ## 4 {} # 3{} } rline right rbrace = left lbrace - 5,"20", - "40" right rbrace } {}.

PABC=12∣AB→×AC→∣=12(−5)2+202+(-40)2=25+400+16002=20252=452PABC=12∣AB→×AC→∣=12(−5)2+202+(-40)2=25+400+16002=20252=452 size 12{P rSub { size 8{ ital "ABC"} } = { {1} over {2} } lline { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } rline = { {1} over {2} } sqrt { \( - 5 \) rSup { size 8{2} } +20 rSup { size 8{2} } + \( "-40" \) "" lSup { size 8{2} } } = { { sqrt {"25"+4"00+1600"} } over {2} } = { { sqrt {"2025"} } over {2} } = { {"45"} over {2} } } {}.

5. Да се пресмета должината на висината повлечена од темето B кон страната AC во триаголникот ABC, ако неговите темиња се:

A(1,−1,2)A(1,−1,2) size 12{A \( 1, - 1,2 \) } {}, B(5,−6,2)B(5,−6,2) size 12{B \( 5, - 6,2 \) } {} и C(1,3,−1)C(1,3,−1) size 12{C \( 1,3, - 1 \) } {}.

Решение.

Table 3
graphics3.png
Слика 3
P ABC = 1 2 ∣ AB → × AC → ∣ P ABC = 1 2 ∣ AB → × AC → ∣ size 12{P rSub { size 8{ ital "ABC"} } = { {1} over {2} } lline { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } rline } {} (9)
AB → = 4, − 5,0 AB → = 4, − 5,0 size 12{ { ital "AB"} cSup { size 8{ rightarrow } } = left lbrace 4, - 5,0 right rbrace } {} (10)
AC → = 0,4, − 3 AC → = 0,4, − 3 size 12{ { ital "AC"} cSup { size 8{ rightarrow } } = left lbrace 0,4, - 3 right rbrace } {} (11)

AB→×AC→=∣−504−3∣,∣04−30∣,∣4−504∣=15,12,16AB→×AC→=∣−504−3∣,∣04−30∣,∣4−504∣=15,12,16 size 12{ { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } = left lbrace lline matrix { - 5 {} # 0 {} ## 4 {} # - 3{} } rline , lline matrix { 0 {} # 4 {} ## - 3 {} # 0{} } rline , lline matrix { 4 {} # - 5 {} ## 0 {} # 4{} } rline right rbrace = left lbrace "15","12","16" right rbrace } {}.

PABC=12∣AB→×AC→∣=225+144+2562=6252=252PABC=12∣AB→×AC→∣=225+144+2562=6252=252 size 12{P rSub { size 8{ ital "ABC"} } = { {1} over {2} } lline { ital "AB"} cSup { size 8{ rightarrow } } times { ital "AC"} cSup { size 8{ rightarrow } } rline = { { sqrt {"225"+1"44+256"} } over {2} } = { { sqrt {"625"} } over {2} } = { {"25"} over {2} } } {}.

Од друга страна, имаме PABC=∣AC→∣⋅∣hb→∣2PABC=∣AC→∣⋅∣hb→∣2 size 12{P rSub { size 8{ ital "ABC"} } = { { lline { ital "AC"} cSup { size 8{ rightarrow } } rline cdot lline {h rSub { size 8{b} } } cSup { size 8{ rightarrow } } rline } over {2} } } {}, односно ∣hb→∣=2PABC∣AC→∣∣hb→∣=2PABC∣AC→∣ size 12{ lline {h rSub { size 8{b} } } cSup { size 8{ rightarrow } } rline = { {2P rSub { size 8{ ital "ABC"} } } over { lline { ital "AC"} cSup { size 8{ rightarrow } } rline } } } {}.

∣hb→∣=2⋅2520+16+9=2525=255=5∣hb→∣=2⋅2520+16+9=2525=255=5 size 12{ lline {h rSub { size 8{b} } } cSup { size 8{ rightarrow } } rline = { {2 cdot { {"25"} over {2} } } over { sqrt {"0+"1"6+9"} } } = { {"25"} over { sqrt {"25"} } } = { {"25"} over {5} } =5} {}.

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