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Задачи од мешан производ на вектори

Module by: Beti Andonovic. E-mail the author

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

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

Значи, векторите 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.

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

Следува векторите 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 } } } {} 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+64−60−12∣=∣4−2∣=∣2∣=2V=∣(a→,b→,c→)∣=∣∣234450341∣∣=∣10+64−60−12∣=∣4−2∣=∣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→)=∣−1−16−2021−14∣=(−2)+12−8−2=0(AB→,AC→,AD→)=∣−1−16−2021−14∣=(−2)+12−8−2=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→,AD→AB→,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,d−1,1}AD→={−2,d−1,1} size 12{ { ital "AD"} cSup { size 8{ rightarrow } } = lbrace - 2,d - 1,1 rbrace } {}.

16∣(AB→,AC→,AD→)∣=16∣∣1−120−24−2d−11∣∣=516∣(AB→,AC→,AD→)∣=16∣∣1−120−24−2d−11∣∣=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+8−8−4d+4)∣=516∣(−2+8−8−4d+4)∣=5 size 12{ { {1} over {6} } \lline \( - 2+8 - 8 - 4d+4 \) \lline =5} {}.

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

∣2−4d∣=30∣2−4d∣=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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