A vector `vecA` is along the positive z-axis and its vector product with another vector `vecB` is zero, then vector `vecB` could be :

Options

A. `hati+hatj`
B. `4hati`
C. `hati+hatk`
D. `-7hatk`

Answer

To solve the problem, we need to analyze the conditions given in the question.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have a vector \(\vec{A}\) that is along the positive z-axis. This can be represented as:
\[ \vec{A} = A \hat{k} \]
where \(A\) is the magnitude of the vector and \(\hat{k}\) is the unit vector in the z-direction.

2. **Vector Product Condition**:
- We are told that the vector product (cross product) of \(\vec{A}\) and another vector \(\vec{B}\) is zero:
\[ \vec{A} \times \vec{B} = 0 \]
- The cross product of two vectors is zero if the vectors are parallel or if one of the vectors is the zero vector.

3. **Determining the Direction of \(\vec{B}\)**:
- Since \(\vec{A}\) is along the z-axis, for \(\vec{A} \times \vec{B} = 0\), vector \(\vec{B}\) must also be parallel to \(\vec{A}\). This means \(\vec{B}\) can be expressed as:
\[ \vec{B} = B \hat{k} \]
where \(B\) is some scalar multiple.

4. **Analyzing the Given Options**:
- We need to check which of the given options can be expressed as a multiple of \(\hat{k}\):
1. \( \hat{i} + \hat{j} \) (not parallel to \(\hat{k}\))
2. \( 4 \hat{i} \) (not parallel to \(\hat{k}\))
3. \( \hat{i} + \hat{k} \) (not purely along \(\hat{k}\))
4. \( -7 \hat{k} \) (this is parallel to \(\hat{k}\))

5. **Conclusion**:
- The only vector from the options that is parallel to \(\vec{A}\) (which is along the z-axis) is:
\[ \vec{B} = -7 \hat{k} \]
- Therefore, the answer is:
\[ \vec{B} = -7 \hat{k} \]

### Final Answer:
\(\vec{B} = -7 \hat{k}\)