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 determine which vector \( \vec{B} \) can produce a zero vector when crossed with the vector \( \vec{A} \), which is along the positive z-axis.

### Step-by-Step Solution:

1. **Define the Vector \( \vec{A} \)**:
Since \( \vec{A} \) is along the positive z-axis, we can represent it as:
\[ \vec{A} = n \hat{k} \]
where \( n \) is a scalar (magnitude of the vector) and \( \hat{k} \) is the unit vector in the z-direction.

2. **Understanding the Cross Product**:
The cross product \( \vec{A} \times \vec{B} \) is zero if and only if \( \vec{A} \) and \( \vec{B} \) are parallel or if either vector is zero. This means that \( \vec{B} \) must also have no component in the z-direction.

3. **Evaluate Each Option**:
We will evaluate each option to see if it satisfies the condition of being parallel to \( \vec{A} \) or having no z-component.

- **Option 1: \( \vec{B} = \hat{i} + \hat{j} \)**:
\[ \vec{B} = 1 \hat{i} + 1 \hat{j} + 0 \hat{k} \]
Calculate \( \vec{A} \times \vec{B} \):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & n \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - n) - \hat{j}(0 - n) + \hat{k}(0 - 0) = -n \hat{i} + n \hat{j} \]
This is not zero, so option 1 is incorrect.

- **Option 2: \( \vec{B} = 4 \hat{i} \)**:
\[ \vec{B} = 4 \hat{i} + 0 \hat{j} + 0 \hat{k} \]
Calculate \( \vec{A} \times \vec{B} \):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & n \\ 4 & 0 & 0 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - 4n) + \hat{k}(0 - 0) = 4n \hat{j} \]
This is not zero, so option 2 is incorrect.

- **Option 3: \( \vec{B} = \hat{i} + \hat{k} \)**:
\[ \vec{B} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k} \]
Calculate \( \vec{A} \times \vec{B} \):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & n \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - n) + \hat{k}(0 - 0) = n \hat{j} \]
This is not zero, so option 3 is incorrect.

- **Option 4: \( \vec{B} = -7 \hat{k} \)**:
\[ \vec{B} = 0 \hat{i} + 0 \hat{j} - 7 \hat{k} \]
Calculate \( \vec{A} \times \vec{B} \):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & n \\ 0 & 0 & -7 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 0) = 0 \]
This is zero, so option 4 is correct.

### Conclusion:
The vector \( \vec{B} \) that satisfies the condition \( \vec{A} \times \vec{B} = 0 \) is:
\[ \vec{B} = -7 \hat{k} \]