$\overline{a}$ and  $\overline{b}$  are two non-collinear unit vectors, and  $\overline{u}=\overline{a}-\left(\overline{a}.\overline{b}\right)\overline{b}$ and  $\overline{v}=\overline{a}\times \overline{b}$. Then  $\left|\overline{v}\right|$ is

$\overline{u}=\overline{a}-\left(\overline{a}.\overline{b}\right)\overline{b}$

$\begin{array}{l}=\overline{a}\left(\overline{b}.\overline{b}\right)-\left(\overline{a}.\overline{b}\right)\overline{b}\\ =\overline{b}\times \left(\overline{a}\times \overline{b}\right)\end{array}$

$\left|\overline{u}\right|=\left|\overline{b}\times \left(\overline{a}\times \overline{b}\right)\right|=\left|\overline{b}\right|\left|\overline{a}\times \overline{b}\right|\sin {90}^{\circ }$

$=\left|\overline{b}\right|\left|\overline{a}\times \overline{b}\right|=\left|\overline{a}\times \overline{b}\right|=\left|\overline{v}\right|$

Also  $\overline{u}.\overline{b}=\left(\overline{b}\times \left(\overline{a}\times \overline{b}\right)\right).\overline{b}$

$\left[\begin{array}{lll}\overline{b}& \overline{a}\times \overline{b}& \overline{b}\end{array}\right]=0$

$\left|\overline{v}\right|=\left|\overline{u}\right|+\left|\overline{u}.\overline{b}\right|$