Let → a = ^ i + α ^ j + β ^ k , α, β ∈ R . Let a vector → b be such...

Given: {\left|\overset{\to}{b}\right|}^{2} = 6, \overset{\to}{a} \cdot \overset{\to}{b} = 3\sqrt{2}, \overset{\to}{a} = \overset{^}{i} + \alpha\overset{^}{j} + \beta\overset{^}{k}

We know that, $\overset{\to}{a} . \overset{\to}{b} = \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|cos\theta$

$⇒ 3\sqrt{2} = \sqrt{{1}^{2} + {\alpha}^{2} + {\beta}^{2}}\left(\sqrt{6}\right)cos\dfrac{\pi}{4}$

$⇒ 3\sqrt{2} = \sqrt{{1}^{2} + {\alpha}^{2} + {\beta}^{2}}\left(\sqrt{3}\right)$

$⇒ \sqrt{1 + {\alpha}^{2} + {\beta}^{2}} = \sqrt{6}$

$⇒ 1 + {\alpha}^{2} + {\beta}^{2} = 6$

$⇒ {\alpha}^{2} + {\beta}^{2} = 5$

Also, \left|\overset{\to}{a} \times \overset{\to}{b}\right| = \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|sin\theta$</p><p>$⇒ \left|\overset{\to}{a} \times \overset{\to}{b}\right| = \sqrt{6}\sqrt{6}sin\dfrac{\pi}{4} = \dfrac{6}{\sqrt{2}}$</p><p>$⇒ {\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = 18$</p><p>$⇒ \left({\alpha}^{2} + {\beta}^{2}\right){\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = 5 \times 18$</p><p>$⇒ \left({\alpha}^{2} + {\beta}^{2}\right){\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = 90