The value of x for which the angle between ` veca = 2x^(2) hati + 4x hatj =hatk +hatk and vecb = 7hati -2hatj =x hatk` , is obtuse and the angle between ` vecb` and the z-axis is acute and less than `pi//6`, are

Options

A. `altxlt1//2`
B. `1//2ltxlt15`
C. `xlt1//2orxlt0`
D. none of these

Explanation

To solve the problem, we need to find the value of \( x \) such that:

1. The angle between the vectors \( \vec{a} \) and \( \vec{b} \) is obtuse.
2. The angle between \( \vec{b} \) and the z-axis is acute and less than \( \frac{\pi}{6} \).

### Step 1: Define the vectors

Given:
\[ \vec{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k} \]
\[ \vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k} \]

### Step 2: Condition for obtuse angle between \( \vec{a} \) and \( \vec{b} \)

The angle between two vectors is obtuse if their dot product is negative:
\[ \vec{a} \cdot \vec{b} < 0 \]

Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (2x^2)(7) + (4x)(-2) + (1)(x) \]
\[ = 14x^2 - 8x + x = 14x^2 - 7x \]

Setting the inequality:
\[ 14x^2 - 7x < 0 \]

Factoring out:
\[ 7x(2x - 1) < 0 \]

### Step 3: Solve the inequality

The critical points are \( x = 0 \) and \( x = \frac{1}{2} \). Testing intervals:

- For \( x < 0 \): \( 7x(2x - 1) > 0 \)
- For \( 0 < x < \frac{1}{2} \): \( 7x(2x - 1) < 0 \) (satisfied)
- For \( x > \frac{1}{2} \): \( 7x(2x - 1) > 0 \)

Thus, the solution for this condition is:
\[ 0 < x < \frac{1}{2} \]

### Step 4: Condition for angle between \( \vec{b} \) and the z-axis

The angle between \( \vec{b} \) and the z-axis is acute and less than \( \frac{\pi}{6} \). The cosine of the angle should be greater than \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \).

Using the formula:
\[ \cos \theta = \frac{\vec{b} \cdot \hat{k}}{|\vec{b}|} \]

Calculating \( \vec{b} \cdot \hat{k} \):
\[ \vec{b} \cdot \hat{k} = x \]

Calculating \( |\vec{b}| \):
\[ |\vec{b}| = \sqrt{7^2 + (-2)^2 + x^2} = \sqrt{49 + 4 + x^2} = \sqrt{53 + x^2} \]

Setting the inequality:
\[ \frac{x}{\sqrt{53 + x^2}} > \frac{\sqrt{3}}{2} \]

### Step 5: Solve the inequality

Squaring both sides:
\[ \frac{x^2}{53 + x^2} > \frac{3}{4} \]

Cross-multiplying:
\[ 4x^2 > 3(53 + x^2) \]
\[ 4x^2 > 159 + 3x^2 \]
\[ x^2 > 159 \]

Thus, we have:
\[ x > \sqrt{159} \quad \text{or} \quad x < -\sqrt{159} \]

### Step 6: Combine conditions

We have two conditions:
1. \( 0 < x < \frac{1}{2} \)
2. \( x > \sqrt{159} \) or \( x < -\sqrt{159} \)

Since \( \sqrt{159} \) is approximately \( 12.6 \), there are no values of \( x \) that can satisfy both conditions simultaneously.

### Conclusion

Thus, the value of \( x \) does not exist.