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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 on vector product

Решени задачи од векторски производ на вектори

1. Да се определи параметарот λλ size 12{λ} {}, така што векторите p=λa5bp=λa5b size 12{ {p} cSup { size 8{ rightarrow } } =λ {a} cSup { size 8{ rightarrow } } - 5 {b} cSup { size 8{ rightarrow } } } {} и q=3abq=3ab size 12{ {q} cSup { size 8{ rightarrow } } =3 {a} cSup { size 8{ rightarrow } } - {b} cSup { size 8{ rightarrow } } } {} да бидат колинеарни, ако aa size 12{ {a} cSup { size 8{ rightarrow } } } {} и bb size 12{ {b} cSup { size 8{ rightarrow } } } {} не се колинеарни.

Решение.

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

(λa5b)×(3ab)=0(λa5b)×(3ab)=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 } } } {}.

a×aλa×b15b×a+5b×b=0a×aλa×b15b×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=0a×a=0 size 12{ {a} cSup { size 8{ rightarrow } } times {a} cSup { size 8{ rightarrow } } = {0} cSup { size 8{ rightarrow } } } {} и b×b=0b×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=0λa×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 } } } {}.

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

aa size 12{ {a} cSup { size 8{ rightarrow } } } {} и bb size 12{ {b} cSup { size 8{ rightarrow } } } {} не се колинеарни вектори, па мора 15λ=015λ=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=0a×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 aa size 12{ {a} cSup { size 8{ rightarrow } } } {}, bb size 12{ {b} cSup { size 8{ rightarrow } } } {} и cc 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)

=(ac)×(ba)=0=(ac)×(ba)=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 acac size 12{ {a} cSup { size 8{ rightarrow } } - {c} cSup { size 8{ rightarrow } } } {} и baba size 12{ {b} cSup { size 8{ rightarrow } } - {a} cSup { size 8{ rightarrow } } } {} да бидат колинеарни, т.е.

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

ac=λbλaac=λbλa 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)aλbc=0(1+λ)aλbc=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 aa size 12{ {a} cSup { size 8{ rightarrow } } } {}, bb size 12{ {b} cSup { size 8{ rightarrow } } } {} и cc size 12{ {c} cSup { size 8{ rightarrow } } } {} да бидат компланарни.

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

a=2i+3jka=2i+3jk 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=3ij+kb=3ij+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=3111,1213,2331=2,1,7a×b=3111,1213,2331=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=1231,21214,12143=5,20,40AB×AC=1231,21214,12143=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=12AB×AC=12(5)2+202+(-40)2=25+400+16002=20252=452PABC=12AB×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=5043,0430,4504=15,12,16AB×AC=5043,0430,4504=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=12AB×AC=225+144+2562=6252=252PABC=12AB×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=AChb2PABC=AChb2 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=2PABCAChb=2PABCAC 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=22520+16+9=2525=255=5hb=22520+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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