Este artículo presenta una deducción para la expresión del módulo resultante de dos vectores (véase vector (física) y módulo (vector) ) de un espacio vectorial (sobre los números reales ).
Sean dos vectores a → {\displaystyle {\vec {a}}} y b → {\displaystyle {\vec {b}}} que forman un ángulo θ {\displaystyle \theta } entre sí:
Imagen de vectores colocados La fórmula para calcular | a → + b → | {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|} se deduce observando los triángulos rectángulos que se forman, OCB y ACB, y aplicando el Teorema de Pitágoras . En el triángulo OCB:
OB ¯ 2 = OC ¯ 2 + CB ¯ 2 {\displaystyle {\overline {\text{OB}}}^{2}={\overline {\text{OC}}}^{2}+{\overline {\text{CB}}}^{2}} OB ¯ = | a → + b → | {\displaystyle {\overline {\text{OB}}}=|{\vec {a}}+{\vec {b}}|} OC ¯ = | a → | + A C {\displaystyle {\overline {\text{OC}}}=\left|{\vec {a}}\left|+AC\right.\right.} Resultando:
| a → + b → | 2 = ( | a → | + AC ¯ ) 2 + CB ¯ 2 {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=\left(|{\vec {a}}|+{\overline {\text{AC}}}\right)^{2}+{\overline {\text{CB}}}^{2}}
En el triángulo ACB :
AC ¯ = | b → | cos θ {\displaystyle {\overline {\text{AC}}}=|{\vec {b}}|\cos \theta } CB ¯ = | b → | sin θ {\displaystyle {\overline {\text{CB}}}=|{\vec {b}}|\sin \theta } Sustituyendo esto en la igualdad de antes resulta:
| a → + b → | 2 = ( | a → | + | b → | cos θ ) 2 + ( | b → | sin θ ) 2 {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=\left(|{\vec {a}}|+|{\vec {b}}|\cos \theta \right)^{2}+\left(|{\vec {b}}|\sin \theta \right)^{2}}
| a → + b → | 2 = | a → | 2 + 2 | a → | | b → | cos θ + | b → | 2 cos 2 θ + | b → | 2 sin 2 θ {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}\cos ^{2}\theta +|{\vec {b}}|^{2}\sin ^{2}\theta }
| a → + b → | 2 = | a → | 2 + 2 | a → | | b → | cos θ + | b → | 2 ( cos 2 θ + sin 2 θ ) {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}\left(\cos ^{2}\theta +\sin ^{2}\theta \right)}
cos 2 θ + sin 2 θ = 1 ⇒ | a → + b → | 2 = | a → | 2 + 2 | a → | | b → | cos θ + | b → | 2 {\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1\quad \Rightarrow \quad \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}}
| a → + b → | = | a → | 2 + 2 | a → | | b → | cos θ + | b → | 2 {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|={\sqrt {|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}}}}
| a → + b → | = | a → | 2 + | b → | 2 + 2 | a → | | b → | cos θ {\displaystyle \left|{\vec {a}}+{\vec {b}}\right|={\sqrt {|{\vec {a}}|^{2}+|{\vec {b}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta }}}