# Talk:QB/d Bell.polarization/components

The component, or vector projection of the vector A along a direction parallel to B is obtained by dropping a perpendicular to the line defined by B and measuring the length of side labeled ${\displaystyle |A|\cos \theta }$ as shown above. Note that the length of B is not important here. In application to polarization, the length and direction of A represents the electric field, and B indicates the direction of the polarizing filter.
${\displaystyle 1:1:{\sqrt {2}}{\text{ and }}{\tfrac {1}{\sqrt {2}}}:{\tfrac {1}{\sqrt {2}}}:1{\text{ (for 45-45-90 degree triangle)}}}$
${\displaystyle 1:2:{\sqrt {3}}{\text{ and }}{\tfrac {1}{2}}:{\tfrac {\sqrt {3}}{2}}:1{\text{ (for 30-60-90 degree triangle)}}}$
{\displaystyle {\begin{aligned}1^{2}+1^{2}=({\sqrt {2}})^{2}&{\text{, and }}&({\tfrac {1}{\sqrt {2}}})^{2}+({\tfrac {1}{\sqrt {2}}})^{2}={\tfrac {1}{2}}+{\tfrac {1}{2}}=1\\1^{2}+({\sqrt {3}})^{2}=2^{2}&{\text{, and }}&({\tfrac {1}{2}})^{2}+({\tfrac {\sqrt {3}}{2}})^{2}={\tfrac {1}{4}}+{\tfrac {3}{4}}=1\end{aligned}}}