If `theta` is the angle between two vectors ` vec a\ a n d\ vec b ,\ t h e n\ vec adot vec bgeq0` only when `0

Options

A. `0lt theta lt (pi)/(2)`
B. `0 le theta le (pi)/(2)`
C. `0 lt theta lt pi`
D. `0 le theta le pi`

Explanation

To solve the problem, we need to determine the conditions under which the dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is greater than or equal to zero. The dot product is given by the formula:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) \]

where \(\theta\) is the angle between the two vectors, and \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes of the vectors \(\vec{a}\) and \(\vec{b}\).

### Step-by-Step Solution:

1. **Start with the given condition**:
We are given that:
\[ \vec{a} \cdot \vec{b} \geq 0 \]

2. **Substitute the dot product formula**:
Using the formula for the dot product, we can rewrite the condition as:
\[ |\vec{a}| |\vec{b}| \cos(\theta) \geq 0 \]

3. **Analyze the inequality**:
Since the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\) are always non-negative (they are lengths), we can focus on the cosine term:
\[ \cos(\theta) \geq 0 \]

4. **Determine when cosine is non-negative**:
The cosine function is non-negative in the following intervals:
- From \(0\) to \(\frac{\pi}{2}\) (first quadrant)
- At \(\theta = 0\) and \(\theta = \frac{\pi}{2}\), \(\cos(\theta) = 0\)

5. **Conclusion about the angle \(\theta\)**:
Therefore, the condition \(\cos(\theta) \geq 0\) is satisfied when:
\[ 0 \leq \theta \leq \frac{\pi}{2} \]

6. **Check the options**:
Looking at the options provided:
- a. \(0 < \theta < \frac{\pi}{2}\)
- b. \(0 \leq \theta < \frac{\pi}{2}\)
- c. \(0 < \theta < \pi\)
- d. \(0 \leq \theta \leq \pi\)

The correct option that matches our conclusion is:
\[ d. \quad 0 \leq \theta \leq \frac{\pi}{2} \]

### Final Answer:
The correct option is:
**d. \(0 \leq \theta \leq \frac{\pi}{2}\)**