Operaciones con vectores: Producto escalar Producto escalar de dos vectores El producto escalar de dos vectores u → \overrightarrow u u v → \overrightarrow v v
u → ⋅ v → = ∣ u → ∣ ⋅ ∣ v → ∣ ⋅ cos ( u → , v → ^ ) \overrightarrow u\cdot \overrightarrow v= |\overrightarrow u|\ \cdot |\overrightarrow v| \cdot \cos(\widehat{\overrightarrow u, \overrightarrow v}) u ⋅ v = ∣ u ∣ ⋅ ∣ v ∣ ⋅ cos ( u , v )
Recuerda que: si el producto escalar de dos vectores u → \overrightarrow u u v → \overrightarrow v v 0 0 0 cos ( 90 ) = 0 \cos(90)=0 cos ( 90 ) = 0
Ejemplo Calcula el producto escalar de los vectores u → ( 1 , 3 ) \overrightarrow u(1,3) u ( 1 , 3 ) v → ( 3 , 2 ) \overrightarrow v(3,2) v ( 3 , 2 ) 60 º 60º 60º
u → ⋅ v → = ∣ u → ∣ ⋅ ∣ v → ∣ ⋅ ( c o s u → , v → ^ ) = 1 2 + 3 2 ⋅ 3 2 + 2 2 ⋅ cos ( 60 ) = 10 ⋅ 13 ⋅ 0 , 5 = 5 , 7 ‾ \overrightarrow u\cdot \overrightarrow v= |\overrightarrow u|\ \cdot |\overrightarrow v| \cdot \ (cos\widehat{\overrightarrow u, \overrightarrow v})=\sqrt{1^2+3^2}\cdot\sqrt{3^2+2^2}\cdot\cos(60)=\sqrt{10}\cdot \sqrt{13}\cdot 0,5= \underline{5,7} u ⋅ v = ∣ u ∣ ⋅ ∣ v ∣ ⋅ ( cos u , v ) = 1 2 + 3 2 ⋅ 3 2 + 2 2 ⋅ cos ( 60 ) = 10 ⋅ 13 ⋅ 0 , 5 = 5 , 7
Propiedades del producto escalar 1.
Si
u → ≠ 0 ⟹ u → ⋅ u → = ∣ u → ∣ 2 > 0 \overrightarrow u\neq0\implies \overrightarrow u \cdot \overrightarrow u=|\overrightarrow u|^2 \gt0 u = 0 ⟹ u ⋅ u = ∣ u ∣ 2 > 0
2.
u → ⋅ v → = v → ⋅ u → \overrightarrow u \cdot \overrightarrow v=\overrightarrow v \cdot \overrightarrow u u ⋅ v = v ⋅ u , ya que
cos ( u → , v → ^ ) = cos ( v → , u → ^ ) \cos(\widehat{\overrightarrow u, \overrightarrow v})=\cos(\widehat{\overrightarrow v, \overrightarrow u}) cos ( u , v ) = cos ( v , u )
3.
λ ( u → ⋅ v → ) = ( λ u → ) ⋅ v → = u → ⋅ ( λ v → ) \lambda(\overrightarrow u\cdot\overrightarrow v)=(\lambda\overrightarrow u)\cdot \overrightarrow v=\overrightarrow u\cdot (\lambda\overrightarrow v) λ ( u ⋅ v ) = ( λ u ) ⋅ v = u ⋅ ( λ v )
4.
u → ⋅ ( v → + w → ) = u → ⋅ v → + u → ⋅ w → \overrightarrow u\cdot (\overrightarrow v+ \overrightarrow w)=\overrightarrow u\cdot \overrightarrow v+\overrightarrow u \cdot \overrightarrow w u ⋅ ( v + w ) = u ⋅ v + u ⋅ w
Ejemplo Sabiendo que v → ⋅ u → = 3 \overrightarrow v\cdot\overrightarrow u=3 v ⋅ u = 3 u → ⋅ w → = 5 \overrightarrow u\cdot \overrightarrow w=5 u ⋅ w = 5 4 u → ⋅ ( v → + w → ) 4\overrightarrow u\cdot(\overrightarrow v + \overrightarrow w) 4 u ⋅ ( v + w )
4 u → ⋅ ( v → + w → ) = 4 u → ⋅ v → + 4 u → ⋅ w → = 4 ( u → ⋅ v → ) + 4 ( u → ⋅ w → ) = = 4 ( v → ⋅ u → ) + 4 ( u → ⋅ w → ) = 4 ⋅ 3 + 4 ⋅ 5 = 32 ‾ 4\overrightarrow u\cdot(\overrightarrow v + \overrightarrow w)=4\overrightarrow u\cdot \overrightarrow v+4\overrightarrow u\cdot \overrightarrow w=4(\overrightarrow u\cdot \overrightarrow v)+4(\overrightarrow u\cdot \overrightarrow w)=\newline=4(\overrightarrow v\cdot \overrightarrow u)+4(\overrightarrow u\cdot \overrightarrow w)=4\cdot 3+4\cdot 5=\underline{32} 4 u ⋅ ( v + w ) = 4 u ⋅ v + 4 u ⋅ w = 4 ( u ⋅ v ) + 4 ( u ⋅ w ) = = 4 ( v ⋅ u ) + 4 ( u ⋅ w ) = 4 ⋅ 3 + 4 ⋅ 5 = 32
Expresión analítica del producto escalar en la base canónica 2 dimensiones Tomando como base la base canónica { i → , j → } \lbrace\overrightarrow i,\overrightarrow j\rbrace { i , j } u → \overrightarrow u u v → \overrightarrow v v
u → ⋅ v → = ( u 1 i → + u 2 j → ) ⋅ ( v 1 i → + v 2 j → ) = u 1 v 1 i → ⋅ i → + u 1 v 2 i → ⋅ j → + u 2 v 1 j → ⋅ i → + u 2 v 2 j → ⋅ j → = u 1 v 1 i → ⋅ i → + u 2 v 2 j → ⋅ j → = u 1 v 1 + u 2 v 2 ⏟ \overrightarrow u\cdot \overrightarrow v=(u_1\overrightarrow i+u_2\overrightarrow j)\cdot (v_1\overrightarrow i+v_2\overrightarrow j)=u_1v_1\overrightarrow i\cdot\overrightarrow i +u_1v_2\overrightarrow i\cdot\overrightarrow j +u_2v_1\overrightarrow j\cdot\overrightarrow i+u_2v_2\overrightarrow j\cdot\overrightarrow j =u_1v_1\overrightarrow i\cdot\overrightarrow i + u_2v_2\overrightarrow j\cdot\overrightarrow j=\underbrace{u_1v_1+u_2v_2} u ⋅ v = ( u 1 i + u 2 j ) ⋅ ( v 1 i + v 2 j ) = u 1 v 1 i ⋅ i + u 1 v 2 i ⋅ j + u 2 v 1 j ⋅ i + u 2 v 2 j ⋅ j = u 1 v 1 i ⋅ i + u 2 v 2 j ⋅ j = u 1 v 1 + u 2 v 2
3 dimensiones Para 3 dimensiones, la base canónica es { i → , j → , k → } \lbrace\overrightarrow i, \overrightarrow j, \overrightarrow k\rbrace { i , j , k }
u → ⋅ v → = u 1 v 1 + u 2 v 2 + u 3 v 3 ⏟ \overrightarrow u\cdot \overrightarrow v=\underbrace{u_1v_1+ u_2v_2 + u_3v_3} u ⋅ v = u 1 v 1 + u 2 v 2 + u 3 v 3
Ejemplo Calcula el producto escalar de los vectores u → = ( 2 , − 1 ) \overrightarrow u=(2,-1) u = ( 2 , − 1 ) v → = ( 1 , 1 ) \overrightarrow v=(1,1) v = ( 1 , 1 )
u → ⋅ v → = 2 ⋅ 1 + 1 ⋅ ( − 1 ) = 1 ‾ \overrightarrow u \cdot \overrightarrow v=2\cdot 1 + 1\cdot (-1)=\underline1 u ⋅ v = 2 ⋅ 1 + 1 ⋅ ( − 1 ) = 1
Interpretación geométrica del producto escalar
Geométricamente, el producto escalar se puede definir como el módulo de un vector por la proyección ortogonal del otro sobre él. Teniendo en cuenta que proy u → t → = O Q ′ → \operatorname {proy}_{\overrightarrow u} \overrightarrow t=\overrightarrow {OQ'} proy u t = O Q ′
∣ u → ⋅ t → ∣ = ∣ proy u → t → ∣ ⋅ u → = ∣ proy t → u → ∣ ⋅ t → |\overrightarrow u\cdot \overrightarrow t|=|\operatorname {proy}_{\overrightarrow u} \overrightarrow t|\cdot \overrightarrow u= |\operatorname {proy}_{\overrightarrow t} \overrightarrow u|\cdot \overrightarrow t ∣ u ⋅ t ∣ = ∣ proy u t ∣ ⋅ u = ∣ proy t u ∣ ⋅ t
Así, también podemos concluir que:
∣ proy v → u → ∣ = ∣ u → ⋅ v → ∣ ∣ v → ∣ ∣ proy u → v → ∣ = ∣ u → ⋅ v → ∣ ∣ u → ∣ |\operatorname{proy}_{\overrightarrow v}\overrightarrow u|=\cfrac{|\overrightarrow u\cdot \overrightarrow v|}{|\overrightarrow v|} \qquad \qquad |\operatorname{proy}_{\overrightarrow u}\overrightarrow v|=\cfrac{|\overrightarrow u\cdot \overrightarrow v|}{|\overrightarrow u|} ∣ proy v u ∣ = ∣ v ∣ ∣ u ⋅ v ∣ ∣ proy u v ∣ = ∣ u ∣ ∣ u ⋅ v ∣
