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

To determine the smallest positive integer value for \( \alpha \) such that the angle between vectors \(\vec{A} = \alpha \hat{i} - 2\hat{j} + 2\hat{k}\) and \(\vec{B} = \alpha \hat{i} + 2\alpha \hat{j} - 2\hat{k}\) is acute, proceed as follows:

1. Recognize that an acute angle implies a positive dot product: \(\vec{A} \cdot \vec{B} > 0\).

2. Compute the dot product \(\vec{A} \cdot \vec{B}\):

\(\vec{A} \cdot \vec{B} = (\alpha \hat{i} - 2\hat{j} + 2\hat{k}) \cdot (\alpha \hat{i} + 2\alpha \hat{j} - 2\hat{k})\)

This calculation yields:

\(\alpha^2 + (-2)(2\alpha) + (2)(-2)\)

Simplifying the expression results in:

\(\alpha^2 - 4\alpha - 4\)

Further simplification leads to:

\(\alpha^2 - 2\alpha - 8\)

3. Establish the inequality for an acute angle:

\(\alpha^2 - 2\alpha - 8 > 0\)

4. Solve the quadratic inequality. First, find the roots of \(\alpha^2 - 2\alpha - 8 = 0\):

Using the quadratic formula \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-2, c=-8\):

\(\alpha = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2}\)

This yields:

\(\alpha = \frac{2 \pm 6}{2}\)

The roots are \(\alpha = 4\) and \(\alpha = -2\).

5. Determine the intervals where the inequality \(\alpha^2 - 2\alpha - 8 > 0\) holds. The critical values \(-2\) and \(4\) divide the number line into \((-\infty, -2)\), \((-2, 4)\), and \((4, \infty)\). Test a value from each interval; for example, testing \(\alpha = 5\) from \((4, \infty)\):

\(5^2 - 2 \cdot 5 - 8 = 25 - 10 - 8 = 7\). Since \(7>0\), the inequality is satisfied for \(\alpha>4\).

The smallest integer value of \( \alpha \) greater than 4 is 5.

6. Confirm that \(\alpha = 5\) satisfies the condition \(\alpha>4\). The smallest positive integral value is indeed 5.