Израчунавање висине и запремине паралелопипеда помоћу мешовитог производа вектора. Решени задаци.
Задаци Текст задатака објашњених у видео лекцији.
Пр.4) Одреди висину која одговара основи ( a , → b → ) \left( {\overrightarrow {a,} \overrightarrow b } \right) ( a , b ) a → = 3 i → + 2 j → − k → \overrightarrow a = 3\overrightarrow i + 2\overrightarrow j - \overrightarrow k a = 3 i + 2 j − k b → = 2 i → + j → − k → \overrightarrow b = 2\overrightarrow i + \overrightarrow j - \overrightarrow k b = 2 i + j − k c → = i → − 2 j → + 3 k → \overrightarrow c = \overrightarrow i - 2\overrightarrow j + 3\overrightarrow k c = i − 2 j + 3 k
Пр.5) Тачке A ( 2 , 3 , 1 ) A\left( {2,3,1} \right) A ( 2 , 3 , 1 ) B ( 4 , 1 , − 2 ) B\left( {4,1, - 2} \right) B ( 4 , 1 , − 2 ) C ( 6 , 3 , 7 ) C\left( {6,3,7} \right) C ( 6 , 3 , 7 ) D ( − 5 , − 4 , 8 ) D\left( { - 5, - 4,8} \right) D ( − 5 , − 4 , 8 )
одговара темену D D D
Пр.4)
a → = ( 3 , 2 , − 1 ) b → = ( 2 , 1 , − 1 ) c → = ( 1 , − 2 , 3 ) V = B ⋅ H V = ∣ [ a → , b → , c → ] ∣ = ∣ ∣ 3 2 − 1 2 1 − 1 1 − 2 3 ∣ 3 2 1 2 1 − 2 ∣ = ∣ 9 − 2 + 4 + 1 − 6 − 12 ∣ = ∣ − 6 ∣ = 6 B = ∣ a → × b → ∣ a → × b → = ∣ i → j → k → 3 2 − 1 2 1 − 1 ∣ i → 3 2 j → 2 1 = − 2 i → − 2 j → + 3 k → − 4 k → + i → + 3 j → = − i → + j → − k → = ( − 1 , 1 , − 1 ) B = ∣ a → × b → ∣ = ( − 1 ) 2 + 1 2 + ( − 1 ) 2 = 3 6 = 3 H ⇒ H = 6 3 ⋅ 3 3 = 2 3 \begin{aligned}&\overrightarrow a = \left( {3,2, - 1} \right) \\ &\overrightarrow b = \left( {2,1, - 1} \right) \\ &\overrightarrow c = \left( {1, - 2,3} \right) \\ &V = B \cdot H \\ &V = \left| {\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]} \right| = \left| {\left| {\begin{array}{cccccccccccccccccccc} 3&2&{ - 1} \\ 2&1&{ - 1} \\ 1&{ - 2}&3 \end{array}} \right|\begin{array}{cccccccccccccccccccc} 3 \\ 2 \\ 1 \end{array}\begin{array}{cccccccccccccccccccc} {} \\ {} \\ {} \end{array}\begin{array}{cccccccccccccccccccc} 2 \\ 1 \\ { - 2} \end{array}} \right| = \left| {9 - 2 + 4 + 1 - 6 - 12} \right| = \left| { - 6} \right| = 6 \\ &B = \left| {\overrightarrow a \times \overrightarrow b } \right| \\ &\overrightarrow a \times \overrightarrow b = \left| {\begin{array}{cccccccccccccccccccc} {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\ 3&2&{ - 1} \\ 2&1&{ - 1} \end{array}} \right|\begin{array}{cccccccccccccccccccc} {\overrightarrow i } \\ 3 \\ 2 \end{array}\begin{array}{cccccccccccccccccccc} {} \\ {} \\ {} \end{array}\begin{array}{cccccccccccccccccccc} {\overrightarrow j } \\ 2 \\ 1 \end{array} = - 2\overrightarrow i - 2\overrightarrow j + 3\overrightarrow k - 4\overrightarrow k + \overrightarrow i + 3\overrightarrow j = - \overrightarrow i + \overrightarrow j - \overrightarrow k = \left( { - 1,1, - 1} \right) \\ &B = \left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} = \sqrt 3 \\ &6 = \sqrt 3 H \Rightarrow H = \frac{6}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{{\sqrt 3 }} = 2\sqrt 3\end{aligned} a = ( 3 , 2 , − 1 ) b = ( 2 , 1 , − 1 ) c = ( 1 , − 2 , 3 ) V = B ⋅ H V = [ a , b , c ] = 3 2 1 2 1 − 2 − 1 − 1 3 3 2 1 2 1 − 2 = ∣ 9 − 2 + 4 + 1 − 6 − 12 ∣ = ∣ − 6 ∣ = 6 B = a × b a × b = i 3 2 j 2 1 k − 1 − 1 i 3 2 j 2 1 = − 2 i − 2 j + 3 k − 4 k + i + 3 j = − i + j − k = ( − 1 , 1 , − 1 ) B = a × b = ( − 1 ) 2 + 1 2 + ( − 1 ) 2 = 3 6 = 3 H ⇒ H = 3 6 ⋅ 3 3 = 2 3
Пр.5)
A B → = B − A = ( 2 , − 2 , − 3 ) A C → = C − A = ( 4 , 0 , 6 ) A D → = D − A = ( − 7 , − 7 , 7 ) [ A B → , A C → , A D → ] = ∣ 2 − 2 − 3 4 0 6 − 7 − 7 7 ∣ 2 4 − 7 − 2 0 − 7 = 0 + 84 + 84 − 0 + 84 + 56 = 308 V = 1 6 ∣ [ A B → , A C → , A D → ] ∣ = 1 6 ⋅ 308 = 154 3 ∣ A B → × A C → ∣ = ( − 12 ) 2 + ( − 24 ) 2 + 8 2 = 28 B = 1 2 ∣ A B → × A C → ∣ = 14 V = 1 3 B H 154 3 = 1 3 ⋅ 14 ⋅ H ⇒ H = 11 \begin{aligned}&\overrightarrow {AB} = B - A = \left( {2, - 2, - 3} \right) \\ &\overrightarrow {AC} = C - A = \left( {4,0,6} \right) \\ &\overrightarrow {AD} = D - A = \left( { - 7, - 7,7} \right) \\ &\left[ {\overrightarrow {AB} ,\overrightarrow {AC} ,\overrightarrow {AD} } \right] = \left| {\begin{array}{cccccccccccccccccccc} 2&{ - 2}&{ - 3} \\ 4&0&6 \\ { - 7}&{ - 7}&7 \end{array}} \right|\begin{array}{cccccccccccccccccccc} 2 \\ 4 \\ { - 7} \end{array}\begin{array}{cccccccccccccccccccc} {} \\ {} \\ {} \end{array}\begin{array}{cccccccccccccccccccc} { - 2} \\ 0 \\ { - 7} \end{array} = 0 + 84 + 84 - 0 + 84 + 56 = 308 \\ &V = \frac{1}{6}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} ,\overrightarrow {AD} } \right]} \right| = \frac{1}{6} \cdot 308 = \frac{{154}}{3} \\ & \\ &\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \sqrt {{{\left( { - 12} \right)}^2} + {{\left( { - 24} \right)}^2} + {8^2}} = 28 \\ &B = \frac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = 14 \\ &V = \frac{1}{3}BH \\ &\frac{{154}}{3} = \frac{1}{3} \cdot 14 \cdot H \Rightarrow H = 11\end{aligned} A B = B − A = ( 2 , − 2 , − 3 ) A C = C − A = ( 4 , 0 , 6 ) A D = D − A = ( − 7 , − 7 , 7 ) [ A B , A C , A D ] = 2 4 − 7 − 2 0 − 7 − 3 6 7 2 4 − 7 − 2 0 − 7 = 0 + 84 + 84 − 0 + 84 + 56 = 308 V = 6 1 [ A B , A C , A D ] = 6 1 ⋅ 308 = 3 154 A B × A C = ( − 12 ) 2 + ( − 24 ) 2 + 8 2 = 28 B = 2 1 A B × A C = 14 V = 3 1 B H 3 154 = 3 1 ⋅ 14 ⋅ H ⇒ H = 11
У к р а т к о Висина паралелепипеда над основом коју граде a → \overrightarrow{a} a b → \overrightarrow{b} b V = B ⋅ H V = B \cdot H V = B ⋅ H V = ∣ [ a → , b → , c → ] ∣ V = |[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}]| V = ∣ [ a , b , c ] ∣ B = ∣ a → × b → ∣ B = |\overrightarrow{a} \times \overrightarrow{b}| B = ∣ a × b ∣ H = ∣ [ a → , b → , c → ] ∣ ∣ a → × b → ∣ H = \frac{|[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}]|}{|\overrightarrow{a} \times \overrightarrow{b}|} H = ∣ a × b ∣ ∣ [ a , b , c ] ∣
За тространу пирамиду A B C D ABCD A B C D V = 1 6 ∣ [ A B → , A C → , A D → ] ∣ V = \frac{1}{6}|[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}]| V = 6 1 ∣ [ A B , A C , A D ] ∣ B = 1 2 ∣ A B → × A C → ∣ B = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| B = 2 1 ∣ A B × A C ∣ V = 1 3 B H V = \frac{1}{3}BH V = 3 1 B H D D D D D D A B C ABC A B C 1 6 \frac{1}{6} 6 1 1 2 \frac{1}{2} 2 1
М е њ а ј т е в и с и н у а н т е н е x y z A B C D H 2 4 2 4 2 4 Превуците да окренете Висина врха антене 7 m
Запремина пирамиде V 10 m³
Површина основе B 15 m²
Растојање од крова H 2 m Слика је кров из примера у размери 1 : 2, а бројеви уз њу су у метрима. Испрекидана дуж је висина пирамиде A B C D ABCD A B C D D D D H = 3 V B H = \frac{3V}{B} H = B 3 V H = 2 H = 2 H = 2 П р и м е р и з ж и в о т а У пројектовању кровова и у 3D моделовању често треба растојање тачке од косе равни, на пример врха антене или носача соларног панела од кровне равни. Рачуна се управо овако: H = ∣ [ A B → , A C → , A D → ] ∣ ∣ A B → × A C → ∣ H = \frac{|[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}]|}{|\overrightarrow{AB} \times \overrightarrow{AC}|} H = ∣ A B × A C ∣ ∣ [ A B , A C , A D ] ∣
Антена на косом крову Раван косог крова пролази кроз тачке A ( 0 , 0 , 3 ) A(0, 0, 3) A ( 0 , 0 , 3 ) B ( 6 , 0 , 3 ) B(6, 0, 3) B ( 6 , 0 , 3 ) C ( 0 , 4 , 6 ) C(0, 4, 6) C ( 0 , 4 , 6 ) z z z D ( 3 , 2 , 7 ) D(3, 2, 7) D ( 3 , 2 , 7 )
Решење Вектори из темена A A A A B → = ( 6 , 0 , 0 ) \overrightarrow{AB} = (6, 0, 0) A B = ( 6 , 0 , 0 ) A C → = ( 0 , 4 , 3 ) \overrightarrow{AC} = (0, 4, 3) A C = ( 0 , 4 , 3 ) A D → = ( 3 , 2 , 4 ) \overrightarrow{AD} = (3, 2, 4) A D = ( 3 , 2 , 4 ) A B → × A C → = ( 0 ⋅ 3 − 0 ⋅ 4 , 0 ⋅ 0 − 6 ⋅ 3 , 6 ⋅ 4 − 0 ⋅ 0 ) = ( 0 , − 18 , 24 ) \overrightarrow{AB} \times \overrightarrow{AC} = (0 \cdot 3 - 0 \cdot 4,\ 0 \cdot 0 - 6 \cdot 3,\ 6 \cdot 4 - 0 \cdot 0) = (0, -18, 24) A B × A C = ( 0 ⋅ 3 − 0 ⋅ 4 , 0 ⋅ 0 − 6 ⋅ 3 , 6 ⋅ 4 − 0 ⋅ 0 ) = ( 0 , − 18 , 24 ) ∣ A B → × A C → ∣ = 30 |\overrightarrow{AB} \times \overrightarrow{AC}| = 30 ∣ A B × A C ∣ = 30 B = 15 B = 15 B = 15 [ A B → , A C → , A D → ] = ( 0 , − 18 , 24 ) ⋅ ( 3 , 2 , 4 ) = 0 − 36 + 96 = 60 [\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = (0, -18, 24) \cdot (3, 2, 4) = 0 - 36 + 96 = 60 [ A B , A C , A D ] = ( 0 , − 18 , 24 ) ⋅ ( 3 , 2 , 4 ) = 0 − 36 + 96 = 60 A B C D ABCD A B C D V = 60 6 = 10 V = \frac{60}{6} = 10 V = 6 60 = 10 Из V = 1 3 B H V = \frac{1}{3}BH V = 3 1 B H H = 3 V B = 30 15 = 2 H = \frac{3V}{B} = \frac{30}{15} = 2 H = B 3 V = 15 30 = 2 Одговор: Врх антене је 2 m од крова, мерено нормално на кров.
Задатак за вежбу Израчунајте висину паралелепипеда над векторима a → = ( 1 , 2 , 2 ) \overrightarrow{a} = (1, 2, 2) a = ( 1 , 2 , 2 ) b → = ( 2 , 1 , − 2 ) \overrightarrow{b} = (2, 1, -2) b = ( 2 , 1 , − 2 ) c → = ( 1 , 1 , 4 ) \overrightarrow{c} = (1, 1, 4) c = ( 1 , 1 , 4 ) a → \overrightarrow{a} a b → \overrightarrow{b} b
Прикажи решење a → × b → = ( − 6 , 6 , − 3 ) \overrightarrow{a} \times \overrightarrow{b} = (-6, 6, -3) a × b = ( − 6 , 6 , − 3 ) ∣ a → × b → ∣ = 9 |\overrightarrow{a} \times \overrightarrow{b}| = 9 ∣ a × b ∣ = 9 [ a → , b → , c → ] = − 6 + 6 − 12 = − 12 [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] = -6 + 6 - 12 = -12 [ a , b , c ] = − 6 + 6 − 12 = − 12 V = 12 V = 12 V = 12 H = 12 9 = 4 3 H = \frac{12}{9} = \frac{4}{3} H = 9 12 = 3 4
