Key Concepts and Formulas

• Dot Product and Angle Between Vectors: The angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
• Condition for Acute Angle: The angle $\theta$ is acute ($0 < \theta < \frac{\pi}{2}$) if and only if $\cos \theta > 0$. Since $|\vec{a}|$ and $|\vec{b}|$ are always positive, this implies $\vec{a} \cdot \vec{b} > 0$.
• Dot Product Calculation: For vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, their dot product is $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.
• Solving Quadratic Inequalities: For a quadratic $ax^2+bx+c$, if $a>0$, the inequality $ax^2+bx+c > 0$ holds for values of $x$ outside the roots of $ax^2+bx+c=0$.

Step-by-Step Solution

Step 1: Understand the Condition for an Acute Angle We are given that the angle between two vectors is acute. The angle $\theta$ between two vectors is acute if $0 < \theta < \frac{\pi}{2}$. This implies that $\cos \theta > 0$. From the formula for the angle between vectors, $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$. Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are always positive for non-zero vectors, the condition $\cos \theta > 0$ is equivalent to the dot product being positive: $\vec{a} \cdot \vec{b} > 0$ This is the fundamental condition we will use to solve the problem.

Step 2: Define the Given Vectors Let the two given vectors be $\vec{a}$ and $\vec{b}$. $\vec{a} = \alpha \hat{i}-2 \hat{j}+2 \hat{k}$ $\vec{b} = \alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ Clearly defining these vectors is crucial for accurate calculation of their dot product.

Step 3: Calculate the Dot Product of the Vectors We compute the dot product $\vec{a} \cdot \vec{b}$ using the components of $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b} = (\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2)$ $\vec{a} \cdot \vec{b} = \alpha^2 - 4\alpha - 4$ This calculation directly applies the dot product formula to the given vectors.

Step 4: Set Up and Solve the Inequality for $\alpha$} Using the condition from Step 1, we must have $\vec{a} \cdot \vec{b} > 0$. Substituting the result from Step 3: $\alpha^2 - 4\alpha - 4 > 0$ To solve this quadratic inequality, we first find the roots of the corresponding quadratic equation $\alpha^2 - 4\alpha - 4 = 0$. Using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\alpha = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}$ $\alpha = \frac{4 \pm \sqrt{16 + 16}}{2}$ $\alpha = \frac{4 \pm \sqrt{32}}{2}$ $\alpha = \frac{4 \pm 4\sqrt{2}}{2}$ $\alpha = 2 \pm 2\sqrt{2}$ The roots are $\alpha_1 = 2 - 2\sqrt{2}$ and $\alpha_2 = 2 + 2\sqrt{2}$. Since the coefficient of $\alpha^2$ in $\alpha^2 - 4\alpha - 4$ is positive (1), the parabola opens upwards. Therefore, the inequality $\alpha^2 - 4\alpha - 4 > 0$ is satisfied when $\alpha$ is outside the interval defined by the roots: $\alpha < 2 - 2\sqrt{2} \quad \text{or} \quad \alpha > 2 + 2\sqrt{2}$ We approximate the values of the roots: $\sqrt{2} \approx 1.414$. $2 - 2\sqrt{2} \approx 2 - 2(1.414) = 2 - 2.828 = -0.828$ $2 + 2\sqrt{2} \approx 2 + 2(1.414) = 2 + 2.828 = 4.828$ So, the inequality holds for $\alpha < -0.828$ or $\alpha > 4.828$.

Step 5: Determine the Least Positive Integral Value of $\alpha$} We need to find the least positive integral value of $\alpha$ that satisfies the conditions derived in Step 4. The two intervals are:

1. $\alpha < -0.828$: This interval contains negative numbers and zero, but no positive integers.
2. $\alpha > 4.828$: This interval contains positive integers. The integers in this interval are $5, 6, 7, \dots$.

The least positive integer in the interval $\alpha > 4.828$ is $5$. Therefore, the least positive integral value of $\alpha$ for which the angle between the vectors is acute is $5$.

Common Mistakes & Tips

• Misinterpreting "acute": Remember that "acute" means the dot product is strictly positive ($\vec{a} \cdot \vec{b} > 0$), not non-negative ($\vec{a} \cdot \vec{b} \ge 0$).
• Sign errors in dot product: Carefully multiply corresponding components, especially when negative signs are involved.
• Solving quadratic inequalities: For $ax^2+bx+c > 0$ with $a>0$, the solution lies outside the roots. For $a<0$, it lies between the roots. Always check the leading coefficient.
• Ignoring "positive integral": Ensure the final answer is an integer and is positive, and that it is the least such value.

Summary

To find the least positive integral value of $\alpha$ for which the angle between the given vectors is acute, we used the condition that the dot product of the vectors must be positive. We calculated the dot product as $\alpha^2 - 4\alpha - 4$. Setting this greater than zero, we solved the quadratic inequality $\alpha^2 - 4\alpha - 4 > 0$ to find that $\alpha < 2 - 2\sqrt{2}$ or $\alpha > 2 + 2\sqrt{2}$. Approximating the roots, we found the conditions to be $\alpha < -0.828$ or $\alpha > 4.828$. Considering only positive integral values of $\alpha$, the least value satisfying $\alpha > 4.828$ is $5$.

The final answer is $\boxed{5}$.