Let → a = ^ i - ^ j + 2 ^ k and let → b be a vector such that → a ×...

Given,

\overset{\to}{a} = \overset{^}{i} - \overset{^}{j} + 2\overset{^}{k} , \overset{\to}{a} \times \overset{\to}{b} = 2\overset{^}{i} - \overset{^}{k} and $\overset{\to}{a} \cdot \overset{\to}{b} = 3$

Now using the formula ${\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{a} \cdot \overset{\to}{b}\right|}^{2} = \left({\left|\overset{\to}{a}\right|}^{2} \cdot {\left|\overset{\to}{b}\right|}^{2}\sin^{2}\theta + {\left|\overset{\to}{a}\right|}^{2} \cdot {\left|\overset{\to}{b}\right|}^{2}\cos^{2}\theta\right)$

$⇒ {\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{a} \cdot \overset{\to}{b}\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2} \cdot {\left|\overset{\to}{b}\right|}^{2}\left(\sin^{2}\theta + \cos^{2}\theta\right) = {\left|\overset{\to}{a}\right|}^{2} \cdot {\left|\overset{\to}{b}\right|}^{2}$

We get,

⇒ {\left|2\overset{^}{i} - \overset{^}{k}\right|}^{2} + {\left|3\right|}^{2} = {\left|\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}

$⇒ {\left|\overset{\to}{b}\right|}^{2} = \dfrac{5 + 9}{6} = \dfrac{14}{6} = \dfrac{7}{3}$

Now finding \left|\overset{\to}{a} - \overset{\to}{b}\right| = \sqrt{{\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} - 2\left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|} = \sqrt{6 + \dfrac{7}{3} - 6} = \sqrt{\dfrac{7}{3}}$</p><p>Now projection of$\overset{\to}{b}$on$\overset{\to}{a} - \overset{\to}{b} = \dfrac{\overset{\to}{b} \cdot \left(\overset{\to}{a} - \overset{\to}{b}\right)}{\left|\overset{\to}{a} - \overset{\to}{b}\right|}$</p><p>$= \dfrac{\overset{\to}{b} \cdot \overset{\to}{a} - {\left|\overset{\to}{b}\right|}^{2}}{\left|\overset{\to}{a} - \overset{\to}{b}\right|} = \dfrac{3 - \dfrac{7}{3}}{\sqrt{\dfrac{7}{3}}}$</p><p>$= \dfrac{2}{\sqrt{21}}