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

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:
$\alpha = 5$