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Step-by-Step Solution
Step 1: Identify the given vectors
We have two vectors:
\vec{A} = \alpha \hat{i} \;-\; 2 \hat{j} \;+\; 2 \hat{k}
and
\vec{B} = \alpha \hat{i} \;+\; 2\alpha \hat{j} \;-\; 2 \hat{k}.
We want the angle between these two vectors to be acute.
Step 2: Condition for an acute angle
The angle \theta between two vectors is acute if and only if their dot product is positive:
\vec{A} \cdot \vec{B} > 0.
Step 3: Compute \vec{A} \cdot \vec{B}
To find the dot product, multiply corresponding components of \vec{A} and \vec{B} :
\vec{A} \cdot \vec{B} = (\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2).
So,
\vec{A} \cdot \vec{B} = \alpha^2 - 4 \alpha - 4.
Step 4: Inequality for an acute angle
For the angle to be acute,
\alpha^2 - 4 \alpha - 4 \;>\; 0.
Step 5: Solve the quadratic inequality
First, solve the equation
\alpha^2 - 4 \alpha - 4 = 0
to find critical points. Using the quadratic formula,
\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
where a = 1 , b = -4 , and c = -4 :
\alpha = \frac{4 \pm \sqrt{16 + 16}}{2}
= \frac{4 \pm \sqrt{32}}{2}
= \frac{4 \pm 4\sqrt{2}}{2}
= 2 \pm 2\sqrt{2}.
Step 6: Determine the valid interval
The roots are 2 - 2\sqrt{2} and 2 + 2\sqrt{2} . Since the coefficient of \alpha^2 is positive (1), the parabola opens upwards. Therefore,
\alpha^2 - 4 \alpha - 4 > 0
holds for
\alpha < 2 - 2\sqrt{2}
or
\alpha > 2 + 2\sqrt{2}.
Step 7: Find the least positive integral value
Note that 2 - 2\sqrt{2} is negative (approximately 2 - 2.828 \approx -0.828 ), so it does not help us find a positive solution. For positive \alpha , we need
\alpha > 2 + 2\sqrt{2}.
Since 2 + 2\sqrt{2} \approx 4.828 , the smallest positive integer greater than this is 5. Thus, the least positive integral value of \alpha is
\boxed{5}.
