Let → a = 2 ^ i - ^ j + 5 ^ k and → b = α ^ i + β ^ j + 2 ^ k . If...

Given, \overset{\to}{a} = 2\overset{^}{i} - \overset{^}{j} + 5\overset{^}{k}, \overset{\to}{b} = \alpha\overset{^}{i} + \beta\overset{^}{j} + 2\overset{^}{k}

Also given \left(\left(\overset{\to}{a} \times \overset{\to}{b}\right) \times \overset{^}{i}\right) \cdot \overset{^}{k} = \dfrac{23}{2}, then \left|\overset{\to}{b} \times 2\overset{^}{j}\right| is

Now using triple cross product we have, \left(\left(\overset{\to}{a} \cdot \overset{^}{i}\right)\overset{\to}{b} - \left(\overset{\to}{b} \cdot \overset{^}{i}\right)\overset{\to}{a}\right) \cdot \overset{^}{k} = \dfrac{23}{2}$</p><p>$⇒ \left(2 . \overset{\to}{b} - \alpha \cdot \overset{\to}{a}\right) . \overset{^}{k} = \dfrac{23}{2}$</p><p>$⇒ 2 \times 2 - \alpha \times 5 = \dfrac{23}{2}$</p><p>$⇒ 5\alpha = 4 - \dfrac{23}{2}$</p><p>$⇒ \alpha = \dfrac{- 3}{2}$</p><p>Now finding,$\overset{\to}{b} \times 2\overset{^}{j} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ \alpha & \beta & 2 \\ 0 & 2 & 0\end{cases}\right| = - 4\overset{^}{i} + 2\alpha\overset{^}{k}$</p><p>$∴ \left|\overset{\to}{b} \times 2\overset{^}{j}\right| = \sqrt{16 + 4{\alpha}^{2}} = \sqrt{16 + 4 \times \dfrac{9}{4}} = 5