The vector → a = - ^ i + 2 ^ j + ^ k is rotated through a right...

Given,

The vector \overset{\to}{a} = - \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k} is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\overset{\to}{b}$

Now taking, \overset{\to}{b} = \lambda\overset{\to}{a} \times \left(\overset{\to}{a} \times \overset{^}{j}\right) {note here we have taken cross vector with \overset{^}{j} as it is rotated along $y -$axis}

⇒ \overset{\to}{b} = \lambda\left(- \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right) \times \left(\left(- \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right) \times \overset{^}{j}\right)

⇒ \overset{\to}{b} = \lambda\left(- \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right) \times \left(- \overset{^}{i} - \overset{^}{k}\right)

⇒ \overset{\to}{b} = \lambda\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 1 & 2 & 1 \\ - 1 & 0 & - 1\end{cases}\right|

⇒ \overset{\to}{b} = \lambda\left(- 2\overset{^}{i} - 2\overset{^}{j} + 2\overset{^}{k}\right)

Also magnitude of both vector will be equal,

So,\left|\overset{\to}{b}\right| = \left|\overset{\to}{a}\right| ⇒ \sqrt{6} = \sqrt{12}\left|\lambda\right|$</p><p>$⇒ \lambda = \pm \dfrac{1}{\sqrt{2}}$</p><p>(Now$\lambda = \dfrac{1}{\sqrt{2}}$rejected$∵ \overset{\to}{b}$makes acute angle with y axis )</p><p>So,$\overset{\to}{b} = - \sqrt{2}\left(- \overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)$</p><p>Now projection of$3\overset{\to}{a} + \sqrt{2}\overset{\to}{b}$on$\overset{\to}{c}$is given by,</p><p>$\dfrac{\left(3\overset{\to}{a} + \sqrt{2b}\right) \cdot \overset{\to}{c}}{\left|\overset{\to}{c}\right|} = 3\sqrt{2}$