Question

The least positive integral value of \( \alpha \), for which the angle between the vectors \( \alpha \hat{i} - 2\hat{j} + 2\hat{k} \) and \( \alpha \hat{i} + 2\alpha \hat{j} - 2\hat{k} \) is acute, is ______.

Answer 1

Correct Answer: 5

Step 1. Condition for Vectors to be Acute: For the angle between two vectors to be acute, their dot product must be positive:
\(\vec{u} \cdot \vec{v} > 0\)
  Given vectors:
\(\vec{u} = \alpha i - 2j + 2k \quad \text{and} \quad \vec{v} = \alpha i + 2j - 2k\)
  We aim to find conditions on \( \alpha \) such that the dot product is positive.

Step 2. Calculate the Dot Product \( \vec{u} \cdot \vec{v} \): The dot product of two vectors \( \vec{u} \) and \( \vec{v} \) is given by:

  \(\vec{u} \cdot \vec{v} = (\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2)\)
  Compute each term:
  - The term \( \alpha \cdot \alpha \) gives: \( \alpha^2 \)
  - The term \( (-2) \cdot (2\alpha) \) gives: \( -4\alpha \)
  - The term \( (2) \cdot (-2) \) gives: \( -4 \)
  
  Combining these terms, we have:
\(\vec{u} \cdot \vec{v} = \alpha^2 - 4\alpha - 4\)

Step 3. Set Up the Inequality: For the angle between the vectors to be acute:
  \(\vec{u} \cdot \vec{v} > 0 \Rightarrow \alpha^2 - 4\alpha - 4 > 0\)
  This is a quadratic inequality. We can find the roots of the corresponding equation:
  \(\alpha^2 - 4\alpha - 4 = 0\)

Step 4. Solve the Quadratic Equation: Use the quadratic formula:
  \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  Here, \( a = 1 \), \( b = -4 \), and \( c = -4 \). Substituting these values:
  
  \(\alpha = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\)

  Simplifying:

 \(\alpha = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm 4\sqrt{2}}{2}\)
 
\(\alpha = 2 \pm 2\sqrt{2}\)

 Determine the Solution to the Inequality: The roots of the equation are:
 \(\alpha = 2 + 2\sqrt{2} \quad \text{and} \quad \alpha = 2 - 2\sqrt{2}\)
  The quadratic \( \alpha^2 - 4\alpha - 4 > 0 \) is positive outside the interval between these roots. Therefore:

\(\alpha < 2 - 2\sqrt{2} \quad \text{or} \quad \alpha > 2 + 2\sqrt{2}\)

 Since we are looking for the least positive integral value of \( \alpha \), we need to find the smallest integer greater than \( 2 + 2\sqrt{2} \).

 Approximate the value of \( 2 + 2\sqrt{2} \):

 \(\sqrt{2} \approx 1.414\)

 \(2 + 2\sqrt{2} \approx 2 + 2 \cdot 1.414 \approx 2 + 2.828 = 4.828\)

The smallest integer greater than 4.828 is 5.

Conclusion: The least positive integral value of \( \alpha \) that makes the angle between \( \vec{u} \) and \( \vec{v} \) acute is:
\(\alpha = 5\)