Question

$a$ and $b$ are two vectors as adjacent sides of a parallelogram, then the vector which is the altitude of the parallelogram and is perpendicular to $a$ is$\frac{(a.b)}{\left|a\right|}a-b$ or $\frac{1}{{\left|a\right|}^2}\{{\left|a\right|}^{2}b-(a.b)a\}$.

Answer 1

$\begin{array}{c}\overrightarrow{OD}=tx\overrightarrow{a}\\ \overrightarrow{DB}=\overrightarrow{b}-t\overrightarrow{a}\\ OD\cdot DB=0\\ \left(b-ta\right)\cdot t\overrightarrow{a}=0\\ \left(b-t\overrightarrow{a}\right)\overrightarrow{a}=0\\ t=\frac{\overrightarrow{b}\overrightarrow{a}}{{\left|\overrightarrow{a}\right|}^2}\\ \overrightarrow{DB}=\overrightarrow{b}-\frac{\overrightarrow{b}\cdot \overrightarrow{a}}{{\left|\overrightarrow{a}\right|}^2}\end{array}$