The values of x for which the angle between the vectors `veca = xhati - 3hatj - hatk` and `vecb = 2xhati + xhatj - hatk` is acute and the angle between b and y-axis lies between `pi/2` and `pi` are:

Options

A. `xgt0`
B. `xlt0`
C. `xgt1only`
D. `xlt-1` only

Answer

To solve the problem, we need to find the values of \( x \) for which the angle between the vectors \( \vec{a} = x \hat{i} - 3 \hat{j} - \hat{k} \) and \( \vec{b} = 2x \hat{i} + x \hat{j} - \hat{k} \) is acute, and the angle between \( \vec{b} \) and the y-axis lies between \( \frac{\pi}{2} \) and \( \pi \).

### Step 1: Understand the conditions for the angles
1. The angle \( \theta \) between vectors \( \vec{a} \) and \( \vec{b} \) is acute if \( \cos \theta > 0 \).
2. The angle \( \alpha \) between vector \( \vec{b} \) and the y-axis is between \( \frac{\pi}{2} \) and \( \pi \) if \( \cos \alpha < 0 \).

### Step 2: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
The dot product \( \vec{a} \cdot \vec{b} \) is given by:
\[ \vec{a} \cdot \vec{b} = (x \hat{i} - 3 \hat{j} - \hat{k}) \cdot (2x \hat{i} + x \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ = x(2x) + (-3)(x) + (-1)(-1) = 2x^2 - 3x + 1 \]

### Step 3: Set the condition for an acute angle
For the angle to be acute:
\[ 2x^2 - 3x + 1 \geq 0 \]

### Step 4: Solve the quadratic inequality
To solve the inequality \( 2x^2 - 3x + 1 \geq 0 \), we first find the roots of the equation \( 2x^2 - 3x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
Thus, the roots are:
\[ x = 1 \quad \text{and} \quad x = \frac{1}{2} \]

### Step 5: Test intervals for the inequality
The critical points are \( x = \frac{1}{2} \) and \( x = 1 \). We test the intervals:
1. \( (-\infty, \frac{1}{2}) \)
2. \( \left(\frac{1}{2}, 1\right) \)
3. \( (1, \infty) \)

- For \( x < \frac{1}{2} \), choose \( x = 0 \):
\[ 2(0)^2 - 3(0) + 1 = 1 \geq 0 \quad \text{(True)} \]
- For \( \frac{1}{2} < x < 1 \), choose \( x = 0.75 \):
\[ 2(0.75)^2 - 3(0.75) + 1 = 1.125 - 2.25 + 1 = -0.125 \quad \text{(False)} \]
- For \( x > 1 \), choose \( x = 2 \):
\[ 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3 \geq 0 \quad \text{(True)} \]

Thus, the solution for the inequality is:
\[ x \in (-\infty, \frac{1}{2}] \cup [1, \infty) \]

### Step 6: Condition for \( \vec{b} \) and the y-axis
The angle \( \alpha \) between \( \vec{b} \) and the y-axis is given by:
\[ \cos \alpha = \frac{\vec{b} \cdot \hat{j}}{|\vec{b}|} \]
Calculating \( \vec{b} \cdot \hat{j} \):
\[ \vec{b} \cdot \hat{j} = x \]
For \( \cos \alpha < 0 \):
\[ x < 0 \]

### Step 7: Find the intersection of the conditions
We need to find the intersection of:
1. \( x \in (-\infty, \frac{1}{2}] \cup [1, \infty) \)
2. \( x < 0 \)

The intersection is:
\[ x \in (-\infty, 0) \]

### Final Answer
The values of \( x \) for which the conditions are satisfied are:
\[ \boxed{(-\infty, 0)} \]