Let $\hat{\mathit{a}}$ and $\hat{\mathit{b}}$ are two non-collinear unit vectors. If $\overrightarrow{u}=\hat{a}-\left(\hat{a}\mathit{·}\hat{b}\right)\hat{b}$ and $\overrightarrow{v}=\hat{\mathit{a}}\mathit{\times }\hat{\mathit{b}}$, then $\left|\overrightarrow{v}\right|$ is equal to

We have, $\left|\hat{a}\right|=\left|\hat{b}\right|=1$

Given, $\overrightarrow{v}=\hat{a}\times \hat{\mathit{b}}=\left|\hat{\mathit{a}}\right|\left|\hat{b}\right|\sin \theta \hat{\mathit{n}}=\sin \theta \hat{n}$

where, $\theta$ is the angle between $\hat{a}$ and $\hat{b}$. 

$\therefore \left|\overrightarrow{v}\right|=\sin \theta$

Now, $\overrightarrow{u}=\hat{a}-\left(\hat{\mathit{a}}\mathit{·}\hat{\mathit{b}}\right)\hat{b}=\hat{a}-\left(\left|\hat{a}\right|\left|\hat{b}\right|\cos \theta \right)\hat{b}$

$\overrightarrow{u}=\hat{a}-\left(\cos \theta \right)\hat{b}$

$\therefore \left|\overrightarrow{u}\right|=\sqrt{{\left|\hat{a}\right|}^2+{\left|-\cos \theta \left|\hat{b}\right|\right|}^2-2\left|\hat{a}\right|\left|-\cos \theta \hat{b}\right|\cos \theta }$

$\Rightarrow \left|\overrightarrow{u}\right|=\sqrt{1+{\cos }^2\theta -2{\cos }^2\theta }$

$\Rightarrow \left|\overrightarrow{u}\right|=\sqrt{1-{\cos }^2\theta }=\sqrt{{\sin }^2\theta }$

$\therefore \left|\overrightarrow{u}\right|=\sin \theta$

So, $\left|\overrightarrow{v}\right|=\left|\overrightarrow{u}\right|$