Let $ABCD$ be a parallelogram such that $\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $ and $\angle BAD$ be an acute angle. If $\overrightarrow r $ is the vector that coincide with the altitude directed from the vertex $B$ to the side $AD,$ then $\overrightarrow r $ is given by :
$\overrightarrow r = 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p $
$\overrightarrow r = - \overrightarrow q + {{\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p $
$\vec r = \vec q - {{\left( {\vec p.\vec q} \right)} \over {\left( {\vec p.\vec p} \right)}}\vec p$
$\overrightarrow r = - 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}$
