The least positive integral value of α , for which the angle...

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

⇒ cos\theta = \dfrac{( \alpha\overset{^}{i} - 2\overset{^}{j} + 2\overset{^}{k} ) \cdot ( \alpha\overset{^}{i} + 2\alpha\overset{^}{j} - 2\overset{^}{k} )}{\sqrt{{\alpha}^{2} + 4 + 4}\sqrt{{\alpha}^{2} + 4{\alpha}^{2} + 4}}

$⇒ cos\theta = \dfrac{{\alpha}^{2} - 4\alpha - 4}{\sqrt{{\alpha}^{2} + 8}\sqrt{5{\alpha}^{2} + 4}}$

Since $\theta$ is acute,

$⇒ {\alpha}^{2} - 4\alpha - 4 > 0$

$⇒ {\alpha}^{2} - 4\alpha + 4 > 8$

$⇒ ( \alpha - 2{)}^{2} > 8$

$⇒ \alpha - 2 > 2\sqrt{2}$ or $\alpha - 2 < - 2\sqrt{2}$

$\alpha > 2 + 2\sqrt{2}$ or $\alpha < 2 - 2\sqrt{2}$

So, the least positive integral value of $\alpha$ is $5$.