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Задачи од мешан производ на вектори

Module by: Beti Andonovic. E-mail the author

Summary: решени задачи од мешан производ на вектори solved problems of triple scalar product of vectors

Решени задачи од мешан производ на вектори

1. Да се провери дали се компланарни векторите:

а) a={1,2,1}a={1,2,1} size 12{ {a} cSup { size 8{ rightarrow } } = lbrace 1,2,1 rbrace } {}, b={2,3,4}b={2,3,4} size 12{ {b} cSup { size 8{ rightarrow } } = lbrace 2,3,4 rbrace } {} и c={4,5,10}c={4,5,10} size 12{ {c} cSup { size 8{ rightarrow } } = lbrace 4,5,"10" rbrace } {}.

б) a={2,1,3}a={2,1,3} size 12{ {a} cSup { size 8{ rightarrow } } = lbrace 2,1, - 3 rbrace } {}, b={1,4,1}b={1,4,1} size 12{ {b} cSup { size 8{ rightarrow } } = lbrace 1, - 4,1 rbrace } {} и c={3,2,2}c={3,2,2} size 12{ {c} cSup { size 8{ rightarrow } } = lbrace 3, - 2,2 rbrace } {}.

Решение.

а) ( a , b , c ) = 1 2 1 2 3 4 4 5 10 = ( a , b , c ) = 1 2 1 2 3 4 4 5 10 = size 12{ \( {a} cSup { size 8{ rightarrow } } , {b} cSup { size 8{ rightarrow } } , {c} cSup { size 8{ rightarrow } } \) = lline matrix { 1 {} # 2 {} # 1 {} ## 2 {} # 3 {} # 4 {} ## 4 {} # 5 {} # "10"{} } rline ={}} {}

= 1 3 10 + 2 4 4 + 2 5 1 4 3 1 5 4 1 2 2 10 = = 1 3 10 + 2 4 4 + 2 5 1 4 3 1 5 4 1 2 2 10 = size 12{ {}=1 cdot 3 cdot "10"+2 cdot 4 cdot 4+2 cdot 5 cdot 1 - 4 cdot 3 cdot 1 - 5 cdot 4 cdot 1 - 2 cdot 2 cdot "10"={}} {}
(1)

=30+32+10122040=7272=0=30+32+10122040=7272=0 size 12{ {}="30"+"32"+"10" - "12" - "20" - "40"="72" - "72"=0} {}.

Значи, векторите 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.

б) ( a , b , c ) = 2 1 3 1 4 1 3 2 2 = ( a , b , c ) = 2 1 3 1 4 1 3 2 2 = size 12{ \( {a} cSup { size 8{ rightarrow } } , {b} cSup { size 8{ rightarrow } } , {c} cSup { size 8{ rightarrow } } \) = lline matrix { 2 {} # 1 {} # - 3 {} ## 1 {} # - 4 {} # 1 {} ## 3 {} # - 2 {} # 2{} } rline ={}} {}

= 2 ( 4 ) 2 + 1 1 3 + 1 ( 2 ) ( 3 ) 3 ( 4 ) ( 3 ) 2 1 ( 2 ) 2 1 1 = = 2 ( 4 ) 2 + 1 1 3 + 1 ( 2 ) ( 3 ) 3 ( 4 ) ( 3 ) 2 1 ( 2 ) 2 1 1 = size 12{ {}=2 cdot \( - 4 \) cdot 2+1 cdot 1 cdot 3+1 cdot \( - 2 \) cdot \( - 3 \) - 3 cdot \( - 4 \) cdot \( - 3 \) - 2 cdot 1 cdot \( - 2 \) - 2 cdot 1 cdot 1={}} {}
(2)

=(16)+3+636+42=(7)34=(41)0=(16)+3+636+42=(7)34=(41)0 size 12{ {}= \( - "16" \) +3+6 - "36"+4 - 2= \( - 7 \) - "34"= \( - "41" \) <> 0} {}.

Следува векторите 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 } } } {} не се компланарни.

2. Да се пресмета волуменот на паралелопипедот конструиран над векторите a={2,3,4}a={2,3,4} size 12{ {a} cSup { size 8{ rightarrow } } = lbrace 2,3,4 rbrace } {}, b={4,5,0}b={4,5,0} size 12{ {b} cSup { size 8{ rightarrow } } = lbrace 4,5,0 rbrace } {} и c={3,4,1}c={3,4,1} size 12{ {c} cSup { size 8{ rightarrow } } = lbrace 3,4,1 rbrace } {}.

Решение.

V=(a,b,c)=234450341=10+646012=42=2=2V=(a,b,c)=234450341=10+646012=42=2=2 size 12{V= \lline \( {a} cSup { size 8{ rightarrow } } , {b} cSup { size 8{ rightarrow } } , {c} cSup { size 8{ rightarrow } } \) \lline = \lline lline matrix { 2 {} # 3 {} # 4 {} ## 4 {} # 5 {} # 0 {} ## 3 {} # 4 {} # 1{} } rline \lline = \lline "10"+"64" - "60" - "12" \lline = \lline 4 - 2 \lline = \lline 2 \lline =2} {}.

3. Да се покаже дека точките A(1,2,1)A(1,2,1) size 12{A \( 1,2, - 1 \) } {}, B(0,1,5)B(0,1,5) size 12{B \( 0,1,5 \) } {}, C(1,2,1)C(1,2,1) size 12{C \( - 1,2,1 \) } {} и D(2,1,3)D(2,1,3) size 12{D \( 2,1,3 \) } {} лежат во иста рамнина.

Решение.

AB={1,1,6}AB={1,1,6} size 12{ { ital "AB"} cSup { size 8{ rightarrow } } = lbrace - 1, - 1,6 rbrace } {}, AC={2,0,2}AC={2,0,2} size 12{ { ital "AC"} cSup { size 8{ rightarrow } } = lbrace - 2,0,2 rbrace } {}, AD={1,1,4}AD={1,1,4} size 12{ { ital "AD"} cSup { size 8{ rightarrow } } = lbrace 1, - 1,4 rbrace } {}.

(AB,AC,AD)=116202114=(2)+1282=0(AB,AC,AD)=116202114=(2)+1282=0 size 12{ \( { ital "AB"} cSup { size 8{ rightarrow } } , { ital "AC"} cSup { size 8{ rightarrow } } , { ital "AD"} cSup { size 8{ rightarrow } } \) = lline matrix { - 1 {} # - 1 {} # 6 {} ## - 2 {} # 0 {} # 2 {} ## 1 {} # - 1 {} # 4{} } rline = \( - 2 \) +"12" - 8 - 2=0} {}.

Следува векторите AB,AC,ADAB,AC,AD size 12{ { ital "AB"} cSup { size 8{ rightarrow } } ,` { ital "AC"} cSup { size 8{ rightarrow } } , {` ital "AD"} cSup { size 8{ rightarrow } } } {} се компланарни, па и точките A, B, C и D лежат во иста рамнина.

4. Волуменот на тетраедар е 5. Три негови темиња се во точките:

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

Да се определат координатите на четвртото теме D, ако се знае дека тоа лежи на y-оската.

Решение.

Темето D има координати: D(0,d,0)D(0,d,0) size 12{D \( 0,d,0 \) } {}, бидејќи лежи на y-оската.

VT=5VT=5 size 12{V rSub { size 8{T} } =5} {}. Следува

16(AB,AC,AD)=516(AB,AC,AD)=5 size 12{ { {1} over {6} } \lline \( { ital "AB"} cSup { size 8{ rightarrow } } , { ital "AC"} cSup { size 8{ rightarrow } } , { ital "AD"} cSup { size 8{ rightarrow } } \) \lline =5} {}.

AB={1,1,2}AB={1,1,2} size 12{ { ital "AB"} cSup { size 8{ rightarrow } } = lbrace 1, - 1,2 rbrace } {},

AC={0,2,4}AC={0,2,4} size 12{ { ital "AC"} cSup { size 8{ rightarrow } } = lbrace 0,2, - 4 rbrace } {},

AD={2,d1,1}AD={2,d1,1} size 12{ { ital "AD"} cSup { size 8{ rightarrow } } = lbrace - 2,d - 1,1 rbrace } {}.

16(AB,AC,AD)=161120242d11=516(AB,AC,AD)=161120242d11=5 size 12{ { {1} over {6} } \lline \( { ital "AB"} cSup { size 8{ rightarrow } } , { ital "AC"} cSup { size 8{ rightarrow } } , { ital "AD"} cSup { size 8{ rightarrow } } \) \lline = { {1} over {6} } \lline lline matrix { 1 {} # - 1 {} # 2 {} ## 0 {} # - 2 {} # 4 {} ## - 2 {} # d - 1 {} # 1{} } rline \lline =5} {}.

16(2+884d+4)=516(2+884d+4)=5 size 12{ { {1} over {6} } \lline \( - 2+8 - 8 - 4d+4 \) \lline =5} {}.

16(24d)=516(24d)=5 size 12{ { {1} over {6} } \lline \( 2 - 4d \) \lline =5} {}.

24d=3024d=30 size 12{ \lline 2 - 4d \lline ="30"} {}. Оттука, добиваме

[ 2 4d = 30 2 4d = 30 [ d 1 = 28 4 d 2 = 32 4 [ d 1 = 7 d 2 = 8 [ 2 4d = 30 2 4d = 30 [ d 1 = 28 4 d 2 = 32 4 [ d 1 = 7 d 2 = 8 size 12{\[ matrix { 2 - 4d="30"`` {} ## 2 - 4d= - "30" } dlrarrow \[ matrix { d rSub { size 8{1} } = { {"28"} over { - 4} } `` {} ## d rSub { size 8{2} } = { { - "32"} over { - 4} } } dlrarrow \[ matrix { d rSub { size 8{1} } = - 7 {} ## d rSub { size 8{2} } =8`` } } {}
(3)

Има две можности: D1(0,7,0)D1(0,7,0) size 12{D rSub { size 8{1} } \( 0, - 7,0 \) } {} и D2(0,8,0)D2(0,8,0) size 12{D rSub { size 8{2} } \( 0,8,0 \) } {}.

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