Почетна Видео лекције Трећи разред средње школе Векторски производ вектора – примери 4 Векторски производ вектора – примери 4 Израчунавање површине применом векторског производа вектора
Задаци Текст задатака објашњених у видео лекцији.
Пр.6) Наћи површину паралелограма ако су његове дијагонале задате векторима 2 m → − n → 2\overrightarrow m - \overrightarrow n 2 m − n m → \overrightarrow m m n → \overrightarrow n n 45 ∘ {45^ \circ } 4 5 ∘
Пр.7) Израчунати висину h a {h_a} h a a → = 2 m → + 3 n → \overrightarrow a = 2\overrightarrow m + 3\overrightarrow n a = 2 m + 3 n b → = m → − 4 n → \overrightarrow b = \overrightarrow m - 4\overrightarrow n b = m − 4 n m → \overrightarrow m m n → \overrightarrow n n
Пр.6)
d 1 → = 2 m → − n → d 2 → = 4 m → − 5 n → d 1 → = a → + b → d 2 → = − a → + b → ‾ 2 b → = d 1 → + d 2 → 2 b → = 2 m → − n → + 4 m → − 5 n → 2 b → = 6 m → − 6 n → b → = 3 m → − 3 n → 2 m → − n → = a → + 3 m → − 3 n → a → = − m → + 2 n → P = ∣ a → × b → ∣ = ∣ ( − m → + 2 n → ) × ( 3 m → − 3 n → ) ∣ = = ∣ − 3 m → × m → + 3 m → × n → + 6 n → × m → − 6 n → × n → ∣ = = ∣ − 3 n → × m → + 6 n → × m → ∣ = ∣ 3 n → × m → ∣ = 3 ∣ n → ∣ ⋅ ∣ m → ∣ s i n ∡ ( n → , m → ) = = 3 ⋅ 1 ⋅ 1 ⋅ 2 2 = 3 2 2 \begin{aligned}&\overrightarrow {{d_1}} = 2\overrightarrow m - \overrightarrow n \\ &\overrightarrow {{d_2}} = 4\overrightarrow m - 5\overrightarrow n \\ &\overrightarrow {{d_1}} = \overrightarrow a + \overrightarrow b \\ &\underline {\overrightarrow {{d_2}} = - \overrightarrow a + \overrightarrow b } \\ &2\overrightarrow b = \overrightarrow {{d_1}} + \overrightarrow {{d_2}} \\ &2\overrightarrow b = 2\overrightarrow m - \overrightarrow n + 4\overrightarrow m - 5\overrightarrow n \\ &2\overrightarrow b = 6\overrightarrow m - 6\overrightarrow n \\ &\overrightarrow b = 3\overrightarrow m - 3\overrightarrow n \\ & \\ &2\overrightarrow m - \overrightarrow n = \overrightarrow a + 3\overrightarrow m - 3\overrightarrow n \\ &\overrightarrow a = - \overrightarrow m + 2\overrightarrow n \\ & \\ &P = \left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\left( { - \overrightarrow m + 2\overrightarrow n } \right) \times \left( {3\overrightarrow m - 3\overrightarrow n } \right)} \right| = \\ &= \left| { - 3\overrightarrow m \times \overrightarrow m + 3\overrightarrow m \times \overrightarrow n + 6\overrightarrow n \times \overrightarrow m - 6\overrightarrow n \times \overrightarrow n } \right| = \\ &= \left| { - 3\overrightarrow n \times \overrightarrow m + 6\overrightarrow n \times \overrightarrow m } \right| = \left| {3\overrightarrow n \times \overrightarrow m } \right| = 3\left| {\overrightarrow n } \right| \cdot \left| {\overrightarrow m } \right|sin\measuredangle \left( {\overrightarrow n ,\overrightarrow m } \right) = \\ &= 3 \cdot 1 \cdot 1 \cdot \frac{{\sqrt 2 }}{2} = \frac{{3\sqrt 2 }}{2}\end{aligned} d 1 = 2 m − n d 2 = 4 m − 5 n d 1 = a + b d 2 = − a + b 2 b = d 1 + d 2 2 b = 2 m − n + 4 m − 5 n 2 b = 6 m − 6 n b = 3 m − 3 n 2 m − n = a + 3 m − 3 n a = − m + 2 n P = a × b = ( − m + 2 n ) × ( 3 m − 3 n ) = = − 3 m × m + 3 m × n + 6 n × m − 6 n × n = = − 3 n × m + 6 n × m = 3 n × m = 3 n ⋅ m s in ∡ ( n , m ) = = 3 ⋅ 1 ⋅ 1 ⋅ 2 2 = 2 3 2
Пр.7)
P = ∣ a → × b → ∣ P = a ⋅ h a = ∣ a → ∣ ⋅ h a a → × b → = ( 2 m → + 3 n → ) × ( m → − 4 n → ) = 2 m → × m → − 8 m → × n → + 3 n → × m → − 12 n → × n → = = 8 n → × m → + 3 n → × m → = 11 n → × m → P = ∣ a → × b → ∣ = ∣ 11 n → × m → ∣ = 11 ∣ n → ∣ ∣ m → ∣ ⋅ sin ∡ ( m → , n → ) = 11 ∣ a → ∣ 2 = a → ⋅ a → = ( 2 m → + 3 n → ) ⋅ ( 2 m → + 3 n → ) = 4 m → × m → + 6 m → × n → + 6 n → × m → − 12 + 9 n → × n → = = 4 ∣ m → ∣ 2 + 12 m → n → + 9 ∣ n → ∣ 2 = m → n → = ∣ m → ∣ ∣ n → ∣ cos ∡ ( m → , n → ) = 1 ⋅ 1 ⋅ 0 = 0 = 4 + 9 = 13 ∣ a → ∣ = 13 11 = 13 ⋅ h a ⇒ h a = 11 13 13 \begin{aligned}&P = \left| {\overrightarrow a \times \overrightarrow b } \right| \\ &P = a \cdot {h_a} = \left| {\overrightarrow a } \right| \cdot {h_a} \\ & \\ &\overrightarrow a \times \overrightarrow b = \left( {2\overrightarrow m + 3\overrightarrow n } \right) \times \left( {\overrightarrow m - 4\overrightarrow n } \right) = 2\overrightarrow m \times \overrightarrow m - 8\overrightarrow m \times \overrightarrow n + 3\overrightarrow n \times \overrightarrow m - 12\overrightarrow n \times \overrightarrow n = \\ &= 8\overrightarrow n \times \overrightarrow m + 3\overrightarrow n \times \overrightarrow m = 11\overrightarrow n \times \overrightarrow m \\ &P = \left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {11\overrightarrow n \times \overrightarrow m } \right| = 11\left| {\overrightarrow n } \right|\left| {\overrightarrow m } \right| \cdot \sin \measuredangle \left( {\overrightarrow m ,\overrightarrow n } \right) = 11 \\ &{\left| {\overrightarrow a } \right|^2} = \overrightarrow a \cdot \overrightarrow a = \left( {2\overrightarrow m + 3\overrightarrow n } \right) \cdot \left( {2\overrightarrow m + 3\overrightarrow n } \right) = 4\overrightarrow m \times \overrightarrow m + 6\overrightarrow m \times \overrightarrow n + 6\overrightarrow n \times \overrightarrow m - 12 + 9\overrightarrow n \times \overrightarrow n = \\ &= 4{\left| {\overrightarrow m } \right|^2} + 12\overrightarrow m \overrightarrow n + 9{\left| {\overrightarrow n } \right|^2} = \\ & \\ &\overrightarrow m \overrightarrow n = \left| {\overrightarrow m } \right|\left| {\overrightarrow n } \right|\cos \measuredangle \left( {\overrightarrow m ,\overrightarrow n } \right) = 1 \cdot 1 \cdot 0 = 0 \\ & \\ &= 4 + 9 = 13 \\ & \\ &\left| {\overrightarrow a } \right| = \sqrt {13} \\ &11 = \sqrt {13} \cdot {h_a} \Rightarrow {h_a} = \frac{{11\sqrt {13} }}{{13}}\end{aligned} P = a × b P = a ⋅ h a = a ⋅ h a a × b = ( 2 m + 3 n ) × ( m − 4 n ) = 2 m × m − 8 m × n + 3 n × m − 12 n × n = = 8 n × m + 3 n × m = 11 n × m P = a × b = 11 n × m = 11 n m ⋅ sin ∡ ( m , n ) = 11 a 2 = a ⋅ a = ( 2 m + 3 n ) ⋅ ( 2 m + 3 n ) = 4 m × m + 6 m × n + 6 n × m − 12 + 9 n × n = = 4 m 2 + 12 m n + 9 n 2 = m n = m n cos ∡ ( m , n ) = 1 ⋅ 1 ⋅ 0 = 0 = 4 + 9 = 13 a = 13 11 = 13 ⋅ h a ⇒ h a = 13 11 13
У к р а т к о Ако паралелограм над a → \overrightarrow{a} a b → \overrightarrow{b} b d 1 → = a → + b → \overrightarrow{d_1} = \overrightarrow{a} + \overrightarrow{b} d 1 = a + b d 2 → = b → − a → \overrightarrow{d_2} = \overrightarrow{b} - \overrightarrow{a} d 2 = b − a d 1 → × d 2 → = 2 a → × b → \overrightarrow{d_1} \times \overrightarrow{d_2} = 2\,\overrightarrow{a} \times \overrightarrow{b} d 1 × d 2 = 2 a × b P = 1 2 ∣ d 1 → × d 2 → ∣ P = \frac{1}{2}|\overrightarrow{d_1} \times \overrightarrow{d_2}| P = 2 1 ∣ d 1 × d 2 ∣ b → = d 1 → + d 2 → 2 \overrightarrow{b} = \frac{\overrightarrow{d_1} + \overrightarrow{d_2}}{2} b = 2 d 1 + d 2 a → = d 1 → − d 2 → 2 \overrightarrow{a} = \frac{\overrightarrow{d_1} - \overrightarrow{d_2}}{2} a = 2 d 1 − d 2
Висину на страницу a a a P = ∣ a → ∣ ⋅ h a P = |\overrightarrow{a}| \cdot h_a P = ∣ a ∣ ⋅ h a h a = ∣ a → × b → ∣ ∣ a → ∣ h_a = \frac{|\overrightarrow{a} \times \overrightarrow{b}|}{|\overrightarrow{a}|} h a = ∣ a ∣ ∣ a × b ∣ ∣ a → ∣ |\overrightarrow{a}| ∣ a ∣ ∣ a → ∣ 2 = a → ⋅ a → |\overrightarrow{a}|^2 = \overrightarrow{a} \cdot \overrightarrow{a} ∣ a ∣ 2 = a ⋅ a a → = 2 m → + 3 n → \overrightarrow{a} = 2\overrightarrow{m} + 3\overrightarrow{n} a = 2 m + 3 n m → \overrightarrow{m} m n → \overrightarrow{n} n 4 + 9 = 13 4 + 9 = 13 4 + 9 = 13 a → ⋅ a → \overrightarrow{a} \cdot \overrightarrow{a} a ⋅ a
П о м е р и т е т е м е н а A , B и D −1 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 0 x y φ d ₁d ₂h ₐA B D C
Површина P 17
½ · d₁ · d₂ · sin φ 17
hₐ = P : |AB| 3,4 Површина паралелограма једнака је 1 2 d 1 d 2 sin φ \frac{1}{2}d_1 d_2 \sin\varphi 2 1 d 1 d 2 sin φ φ \varphi φ ∣ A B ∣ ⋅ h a |AB| \cdot h_a ∣ A B ∣ ⋅ h a A B AB A B П р и м е р и з ж и в о т а Геодете површину парцеле облика паралелограма могу добити из дијагонала ако знају њихове дужине и правце, а висина паралелограма је ширина парцеле мерена нормално на улицу.
Парцела облика паралелограма Парцела у Зрењанину има облик паралелограма A B C D ABCD A B C D A B AB A B m → \overrightarrow{m} m n → \overrightarrow{n} n A C → = 5 m → + 8 n → \overrightarrow{AC} = 5\overrightarrow{m} + 8\overrightarrow{n} A C = 5 m + 8 n B D → = − 3 m → + 2 n → \overrightarrow{BD} = -3\overrightarrow{m} + 2\overrightarrow{n} B D = − 3 m + 2 n
Решење За a → = A B → \overrightarrow{a} = \overrightarrow{AB} a = A B b → = A D → \overrightarrow{b} = \overrightarrow{AD} b = A D A C → = a → + b → \overrightarrow{AC} = \overrightarrow{a} + \overrightarrow{b} A C = a + b B D → = b → − a → \overrightarrow{BD} = \overrightarrow{b} - \overrightarrow{a} B D = b − a b → = A C → + B D → 2 = m → + 5 n → \overrightarrow{b} = \frac{\overrightarrow{AC} + \overrightarrow{BD}}{2} = \overrightarrow{m} + 5\overrightarrow{n} b = 2 A C + B D = m + 5 n a → = A C → − B D → 2 = 4 m → + 3 n → \overrightarrow{a} = \frac{\overrightarrow{AC} - \overrightarrow{BD}}{2} = 4\overrightarrow{m} + 3\overrightarrow{n} a = 2 A C − B D = 4 m + 3 n a → × b → = ( 4 m → + 3 n → ) × ( m → + 5 n → ) = 20 m → × n → + 3 n → × m → = 17 m → × n → \overrightarrow{a} \times \overrightarrow{b} = (4\overrightarrow{m} + 3\overrightarrow{n}) \times (\overrightarrow{m} + 5\overrightarrow{n}) = 20\,\overrightarrow{m} \times \overrightarrow{n} + 3\,\overrightarrow{n} \times \overrightarrow{m} = 17\,\overrightarrow{m} \times \overrightarrow{n} a × b = ( 4 m + 3 n ) × ( m + 5 n ) = 20 m × n + 3 n × m = 17 m × n ∣ m → × n → ∣ = 1 ⋅ 1 ⋅ sin 90 ∘ = 1 |\overrightarrow{m} \times \overrightarrow{n}| = 1 \cdot 1 \cdot \sin 90^\circ = 1 ∣ m × n ∣ = 1 ⋅ 1 ⋅ sin 9 0 ∘ = 1 P = 17 P = 17 P = 17 Провера преко дијагонала: 1 2 ∣ A C → × B D → ∣ = 1 2 ∣ 10 m → × n → − 24 n → × m → ∣ = 1 2 ⋅ 34 = 17 \frac{1}{2}|\overrightarrow{AC} \times \overrightarrow{BD}| = \frac{1}{2}|10\,\overrightarrow{m} \times \overrightarrow{n} - 24\,\overrightarrow{n} \times \overrightarrow{m}| = \frac{1}{2} \cdot 34 = 17 2 1 ∣ A C × B D ∣ = 2 1 ∣10 m × n − 24 n × m ∣ = 2 1 ⋅ 34 = 17 ∣ a → ∣ 2 = a → ⋅ a → = 16 + 24 m → ⋅ n → + 9 = 25 |\overrightarrow{a}|^2 = \overrightarrow{a} \cdot \overrightarrow{a} = 16 + 24\,\overrightarrow{m} \cdot \overrightarrow{n} + 9 = 25 ∣ a ∣ 2 = a ⋅ a = 16 + 24 m ⋅ n + 9 = 25 m → ⋅ n → = 0 \overrightarrow{m} \cdot \overrightarrow{n} = 0 m ⋅ n = 0 ∣ a → ∣ = 5 |\overrightarrow{a}| = 5 ∣ a ∣ = 5 h a = 17 5 = 3,4 h_a = \frac{17}{5} = 3{,}4 h a = 5 17 = 3 , 4 Јединица је 10 m, па је P = 17 ⋅ 100 = 1700 P = 17 \cdot 100 = 1700 P = 17 ⋅ 100 = 1700 h a = 3,4 ⋅ 10 = 34 h_a = 3{,}4 \cdot 10 = 34 h a = 3 , 4 ⋅ 10 = 34 Одговор: Парцела има површину 1700 m², а широка је 34 m, мерено нормално на улицу.
Задатак за вежбу Дијагонале паралелограма су d 1 → = m → + 2 n → \overrightarrow{d_1} = \overrightarrow{m} + 2\overrightarrow{n} d 1 = m + 2 n d 2 → = 3 m → − n → \overrightarrow{d_2} = 3\overrightarrow{m} - \overrightarrow{n} d 2 = 3 m − n m → \overrightarrow{m} m n → \overrightarrow{n} n 60 ∘ 60^\circ 6 0 ∘
Прикажи решење d 1 → × d 2 → = − m → × n → + 6 n → × m → = − 7 m → × n → \overrightarrow{d_1} \times \overrightarrow{d_2} = -\overrightarrow{m} \times \overrightarrow{n} + 6\,\overrightarrow{n} \times \overrightarrow{m} = -7\,\overrightarrow{m} \times \overrightarrow{n} d 1 × d 2 = − m × n + 6 n × m = − 7 m × n P = 1 2 ⋅ 7 ⋅ sin 60 ∘ = 7 3 4 ≈ 3,03 P = \frac{1}{2} \cdot 7 \cdot \sin 60^\circ = \frac{7\sqrt{3}}{4} \approx 3{,}03 P = 2 1 ⋅ 7 ⋅ sin 6 0 ∘ = 4 7 3 ≈ 3 , 03
