Почетна Видео лекције Трећи разред средње школе Векторски производ вектора – примери 3 Векторски производ вектора – примери 3 Израчунавање површине применом векторског производа вектора. Услов колинеарности.
Задаци Текст задатака објашњених у видео лекцији.
Пр.4) Ако је u → = 2 a → + b , → v → = a → − 2 b → , ∣ a → ∣ = 3 , ∣ b → ∣ = 2 , ∠ ( a → , b → ) = π 6 \overrightarrow u = 2\overrightarrow a + \overrightarrow {b,} \overrightarrow v = \overrightarrow a - 2\overrightarrow b ,\left| {\overrightarrow a } \right| = 3,\left| {\overrightarrow b } \right| = 2,\angle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{\pi }{6} u = 2 a + b , v = a − 2 b , a = 3 , b = 2 , ∠ ( a , b ) = 6 π ∣ u → × v → ∣ \left| {\overrightarrow u \times \overrightarrow v } \right| u × v
Пр.5) Наћи површину паралелограма конструисаног над векторима p → = 2 b → − a → \overrightarrow p = 2\overrightarrow b - \overrightarrow a p = 2 b − a q → = 3 a → + 2 b → \overrightarrow q = 3\overrightarrow a + 2\overrightarrow b q = 3 a + 2 b ∣ a → ∣ = ∣ b → ∣ = 5 , ∠ ( a → , b → ) = π 4 \left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 5,\angle \left( {\overrightarrow a ,\overrightarrow b } \right) = \frac{\pi }{4} a = b = 5 , ∠ ( a , b ) = 4 π
Пр.6) Који услов треба да задовољавају вектори a → \overrightarrow a a b → \overrightarrow b b 2 a → − b → 2\overrightarrow a - \overrightarrow b 2 a − b 2 a → + b → 2\overrightarrow a + \overrightarrow b 2 a + b
Пр.4)
∣ u → × v → ∣ = ∣ ( 2 a → + b → ) × ( a → − 2 b → ) ∣ = ∣ 2 a → × a → − 4 a → × b → + b → × a → − 2 b → × b → ∣ = = ∣ − 4 a → × b → − a → × b → ∣ = ∣ − 5 a → × b → ∣ = − 5 ∣ a → × b → ∣ = − 5 ∣ a → ∣ ⋅ ∣ b → ∣ s i n ∡ ( a → , b → ) = 5 ⋅ 3 ⋅ 2 ⋅ 1 2 = 15 \begin{aligned}&\left| {\overrightarrow u \times \overrightarrow v } \right| = \left| {\left( {2\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a - 2\overrightarrow b } \right)} \right| = \left| {2\overrightarrow a \times \overrightarrow a - 4\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow a - 2\overrightarrow b \times \overrightarrow b } \right| = \\ &= \left| { - 4\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow b } \right| = \left| { - 5\overrightarrow a \times \overrightarrow b } \right| = - 5\left| {\overrightarrow a \times \overrightarrow b } \right| = - 5\left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|sin\measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = 5 \cdot 3 \cdot 2 \cdot \frac{1}{2} = 15\end{aligned} u × v = ( 2 a + b ) × ( a − 2 b ) = 2 a × a − 4 a × b + b × a − 2 b × b = = − 4 a × b − a × b = − 5 a × b = − 5 a × b = − 5 a ⋅ b s in ∡ ( a , b ) = 5 ⋅ 3 ⋅ 2 ⋅ 2 1 = 15
Пр.5)
P = ∣ p → × q → ∣ = ∣ ( 2 b → − a → ) × ( 3 a → + 2 b → ) ∣ = ∣ 6 b → × a → + 4 b → × b → − 3 a → × a → − 2 a → × b → ∣ = = ∣ 6 b → × a → + 2 b → × a → ∣ = ∣ 8 b → × a → ∣ = 8 ∣ b → × a → ∣ = 8 ∣ b → ∣ ⋅ ∣ a → ∣ ⋅ s i n ∡ ( a → , b → ) = 8 ⋅ 5 ⋅ 5 ⋅ 2 2 = 100 2 \begin{aligned}&P = \left| {\overrightarrow p \times \overrightarrow q } \right| = \left| {\left( {2\overrightarrow b - \overrightarrow a } \right) \times \left( {3\overrightarrow a + 2\overrightarrow b } \right)} \right| = \left| {6\overrightarrow b \times \overrightarrow a + 4\overrightarrow b \times \overrightarrow b - 3\overrightarrow a \times \overrightarrow a - 2\overrightarrow a \times \overrightarrow b } \right| = \\ &= \left| {6\overrightarrow b \times \overrightarrow a + 2\overrightarrow b \times \overrightarrow a } \right| = \left| {8\overrightarrow b \times \overrightarrow a } \right| = 8\left| {\overrightarrow b \times \overrightarrow a } \right| = 8\left| {\overrightarrow b } \right| \cdot \left| {\overrightarrow a } \right| \cdot sin\measuredangle \left( {\overrightarrow a ,\overrightarrow b } \right) = 8 \cdot 5 \cdot 5 \cdot \frac{{\sqrt 2 }}{2} = 100\sqrt 2\end{aligned} P = p × q = ( 2 b − a ) × ( 3 a + 2 b ) = 6 b × a + 4 b × b − 3 a × a − 2 a × b = = 6 b × a + 2 b × a = 8 b × a = 8 b × a = 8 b ⋅ a ⋅ s in ∡ ( a , b ) = 8 ⋅ 5 ⋅ 5 ⋅ 2 2 = 100 2
Пр.6)
2 a → − b → ∥ 2 a → + b → ⇔ ∣ ( 2 a → − b → ) × ( 2 a → + b → ) ∣ = 0 ∣ 4 a → × a → + 2 a → × b → − 2 b → × a → − b → × b → ∣ = 0 ∣ 2 a → × b → + 2 a → × b → ∣ = 0 ∣ 4 a → × b → ∣ = 0 4 ∣ a → × b → ∣ = 0 ∣ a → × b → ∣ = 0 ⇔ a → ∥ b → \begin{aligned}&2\overrightarrow a - \overrightarrow b \parallel 2\overrightarrow a + \overrightarrow b \Leftrightarrow \left| {\left( {2\overrightarrow a - \overrightarrow b } \right) \times \left( {2\overrightarrow a + \overrightarrow b } \right)} \right| = 0 \\ &\left| {4\overrightarrow a \times \overrightarrow a + 2\overrightarrow a \times \overrightarrow b - 2\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right| = 0 \\ &\left| {2\overrightarrow a \times \overrightarrow b + 2\overrightarrow a \times \overrightarrow b } \right| = 0 \\ &\left| {4\overrightarrow a \times \overrightarrow b } \right| = 0 \\ &4\left| {\overrightarrow a \times \overrightarrow b } \right| = 0 \\ &\left| {\overrightarrow a \times \overrightarrow b } \right| = 0 \Leftrightarrow \overrightarrow a \parallel \overrightarrow b\end{aligned} 2 a − b ∥ 2 a + b ⇔ ( 2 a − b ) × ( 2 a + b ) = 0 4 a × a + 2 a × b − 2 b × a − b × b = 0 2 a × b + 2 a × b = 0 4 a × b = 0 4 a × b = 0 a × b = 0 ⇔ a ∥ b
У к р а т к о Кад су вектори задати као комбинације вектора a → \overrightarrow{a} a b → \overrightarrow{b} b a → × a → = b → × b → = 0 → \overrightarrow{a} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{b} = \overrightarrow{0} a × a = b × b = 0 b → × a → = − a → × b → \overrightarrow{b} \times \overrightarrow{a} = -\overrightarrow{a} \times \overrightarrow{b} b × a = − a × b ( 2 a → + b → ) × ( a → − 2 b → ) = − 4 a → × b → + b → × a → = − 5 a → × b → (2\overrightarrow{a} + \overrightarrow{b}) \times (\overrightarrow{a} - 2\overrightarrow{b}) = -4\,\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{a} = -5\,\overrightarrow{a} \times \overrightarrow{b} ( 2 a + b ) × ( a − 2 b ) = − 4 a × b + b × a = − 5 a × b 5 ∣ a → ∣ ∣ b → ∣ sin φ 5|\overrightarrow{a}||\overrightarrow{b}|\sin\varphi 5∣ a ∣∣ b ∣ sin φ
Уопште је ( m a → + n b → ) × ( k a → + l b → ) = ( m l − n k ) a → × b → (m\overrightarrow{a} + n\overrightarrow{b}) \times (k\overrightarrow{a} + l\overrightarrow{b}) = (ml - nk)\,\overrightarrow{a} \times \overrightarrow{b} ( m a + n b ) × ( k a + l b ) = ( m l − nk ) a × b ∣ m l − n k ∣ |ml - nk| ∣ m l − nk ∣ a → \overrightarrow{a} a b → \overrightarrow{b} b m l − n k = 0 ml - nk = 0 m l − nk = 0 a → × b → \overrightarrow{a} \times \overrightarrow{b} a × b
М е њ а ј т е к о е ф и ц и ј е н т е Плави ромб је паралелограм над a → \overrightarrow{a} a b → \overrightarrow{b} b p → = m a → + n b → \overrightarrow{p} = m\overrightarrow{a} + n\overrightarrow{b} p = m a + n b q → = k a → + l b → \overrightarrow{q} = k\overrightarrow{a} + l\overrightarrow{b} q = k a + l b ∣ m l − n k ∣ |ml - nk| ∣ m l − nk ∣ m l = n k ml = nk m l = nk П р и м е р и з ж и в о т а Површина ћелије кристалне решетке рачуна се као ∣ a → × b → ∣ |\overrightarrow{a} \times \overrightarrow{b}| ∣ a × b ∣ a → \overrightarrow{a} a b → \overrightarrow{b} b
Тепих на поду од ромбова Под предсобља поплочан је једнаким плочицама облика ромба странице 20 cm, са оштрим углом од 60 ∘ 60^\circ 6 0 ∘ a → \overrightarrow{a} a b → \overrightarrow{b} b ∣ a → ∣ = ∣ b → ∣ = 20 |\overrightarrow{a}| = |\overrightarrow{b}| = 20 ∣ a ∣ = ∣ b ∣ = 20 p → = 4 a → + b → \overrightarrow{p} = 4\overrightarrow{a} + \overrightarrow{b} p = 4 a + b q → = a → + 3 b → \overrightarrow{q} = \overrightarrow{a} + 3\overrightarrow{b} q = a + 3 b
Решење p → × q → = ( 4 a → + b → ) × ( a → + 3 b → ) = 4 a → × a → + 12 a → × b → + b → × a → + 3 b → × b → \overrightarrow{p} \times \overrightarrow{q} = (4\overrightarrow{a} + \overrightarrow{b}) \times (\overrightarrow{a} + 3\overrightarrow{b}) = 4\,\overrightarrow{a} \times \overrightarrow{a} + 12\,\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{a} + 3\,\overrightarrow{b} \times \overrightarrow{b} p × q = ( 4 a + b ) × ( a + 3 b ) = 4 a × a + 12 a × b + b × a + 3 b × b Пошто је a → × a → = b → × b → = 0 → \overrightarrow{a} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{b} = \overrightarrow{0} a × a = b × b = 0 b → × a → = − a → × b → \overrightarrow{b} \times \overrightarrow{a} = -\overrightarrow{a} \times \overrightarrow{b} b × a = − a × b p → × q → = 11 a → × b → \overrightarrow{p} \times \overrightarrow{q} = 11\,\overrightarrow{a} \times \overrightarrow{b} p × q = 11 a × b Површина једне плочице: ∣ a → × b → ∣ = 20 ⋅ 20 ⋅ sin 60 ∘ = 200 3 ≈ 346,4 |\overrightarrow{a} \times \overrightarrow{b}| = 20 \cdot 20 \cdot \sin 60^\circ = 200\sqrt{3} \approx 346{,}4 ∣ a × b ∣ = 20 ⋅ 20 ⋅ sin 6 0 ∘ = 200 3 ≈ 346 , 4 Површина паралелограма: ∣ p → × q → ∣ = 11 ⋅ 200 3 = 2200 3 ≈ 3810,5 |\overrightarrow{p} \times \overrightarrow{q}| = 11 \cdot 200\sqrt{3} = 2200\sqrt{3} \approx 3810{,}5 ∣ p × q ∣ = 11 ⋅ 200 3 = 2200 3 ≈ 3810 , 5 Одговор: Паралелограм има површину 2200 3 ≈ 3810 2200\sqrt{3} \approx 3810 2200 3 ≈ 3810
Задатак за вежбу Нека је u → = a → + 3 b → \overrightarrow{u} = \overrightarrow{a} + 3\overrightarrow{b} u = a + 3 b v → = 2 a → − b → \overrightarrow{v} = 2\overrightarrow{a} - \overrightarrow{b} v = 2 a − b ∣ a → ∣ = 4 |\overrightarrow{a}| = 4 ∣ a ∣ = 4 ∣ b → ∣ = 3 |\overrightarrow{b}| = 3 ∣ b ∣ = 3 ∡ ( a → , b → ) = π 6 \measuredangle(\overrightarrow{a}, \overrightarrow{b}) = \frac{\pi}{6} ∡ ( a , b ) = 6 π ∣ u → × v → ∣ |\overrightarrow{u} \times \overrightarrow{v}| ∣ u × v ∣
Прикажи решење u → × v → = − a → × b → + 6 b → × a → = − 7 a → × b → \overrightarrow{u} \times \overrightarrow{v} = -\overrightarrow{a} \times \overrightarrow{b} + 6\,\overrightarrow{b} \times \overrightarrow{a} = -7\,\overrightarrow{a} \times \overrightarrow{b} u × v = − a × b + 6 b × a = − 7 a × b ∣ u → × v → ∣ = 7 ⋅ 4 ⋅ 3 ⋅ sin π 6 = 42 |\overrightarrow{u} \times \overrightarrow{v}| = 7 \cdot 4 \cdot 3 \cdot \sin\frac{\pi}{6} = 42 ∣ u × v ∣ = 7 ⋅ 4 ⋅ 3 ⋅ sin 6 π = 42
