Let ^ a and ^ b are& two non-collinear unit vectors. If → u = ^ a -...

We have,& \left|\overset{^}{a}\right| = \left|\overset{^}{b}\right| = 1

Given,~\overset{→}{v} = \overset{^}{a} \times \overset{^}{b} = \left|\overset{^}{a}\right|\left|\overset{^}{b}\right|sin\theta\overset{^}{n} = sin\theta\overset{^}{n}

where,~$\theta$~is& the angle between \overset{^}{a} and~\overset{^}{b}.~

$∴ \left|\overset{→}{v}\right| = sin\theta$

Now, \overset{→}{u} = \overset{^}{a} - \left(\overset{^}{a} \cdot \overset{^}{b}\right)\overset{^}{b} = \overset{^}{a} - \left(\left|\overset{^}{a}\right|\left|\overset{^}{b}\right|cos\theta\right)\overset{^}{b}

\overset{→}{u} = \overset{^}{a} - \left(cos\theta\right)\overset{^}{b}

∴ \left|\overset{→}{u}\right| = \sqrt{\left|\overset{^}{a}\right|2 + \left|- cos\theta\left|\overset{^}{b}\right|\right|2 - 2\left|\overset{^}{a}\right|\left|- cos\theta\overset{^}{b}\right|cos\theta}

$⇒ \left|\overset{→}{u}\right| = \sqrt{1 + \cos^{2}\theta - 2\cos^{2}\theta}$

$⇒ \left|\overset{→}{u}\right| = \sqrt{1 - \cos^{2}\theta} = \sqrt{\sin^{2}\theta}$

$∴ \left|\overset{→}{u}\right| = sin\theta$

So,~\left|\overset{→}{v}\right| = \left|\overset{→}{u}\right|