Deducción del módulo de la suma

Sean dos vectores ${\displaystyle {\vec {a}}}$ y ${\displaystyle {\vec {b}}}$ que forman un ángulo ${\displaystyle \theta }$ entre sí:

Imagen de vectores colocados

La fórmula para calcular ${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|}$ se deduce observando los triángulos rectángulos que se forman, OCB y ACB, y aplicando el Teorema de Pitágoras. En el triángulo OCB:

${\displaystyle {\overline {\text{OB}}}^{2}={\overline {\text{OC}}}^{2}+{\overline {\text{CB}}}^{2}}$

${\displaystyle {\overline {\text{OB}}}=|{\vec {a}}+{\vec {b}}|}$

${\displaystyle {\overline {\text{OC}}}=\left|{\vec {a}}\left|+AC\right.\right.}$

Resultando:

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=\left(|{\vec {a}}|+{\overline {\text{AC}}}\right)^{2}+{\overline {\text{CB}}}^{2}}$

En el triángulo ACB :

${\displaystyle {\overline {\text{AC}}}=|{\vec {b}}|\cos \theta }$

${\displaystyle {\overline {\text{CB}}}=|{\vec {b}}|\sin \theta }$

Sustituyendo esto en la igualdad de antes resulta:

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=\left(|{\vec {a}}|+|{\vec {b}}|\cos \theta \right)^{2}+\left(|{\vec {b}}|\sin \theta \right)^{2}}$

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}\cos ^{2}\theta +|{\vec {b}}|^{2}\sin ^{2}\theta }$

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}\left(\cos ^{2}\theta +\sin ^{2}\theta \right)}$

${\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1\quad \Rightarrow \quad \left|{\vec {a}}+{\vec {b}}\right|^{2}=|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}}$

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|={\sqrt {|{\vec {a}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta +|{\vec {b}}|^{2}}}}$

${\displaystyle \left|{\vec {a}}+{\vec {b}}\right|={\sqrt {|{\vec {a}}|^{2}+|{\vec {b}}|^{2}+2|{\vec {a}}||{\vec {b}}|\cos \theta }}}$