A vector `vecA` points vertically upward and `vecB` points towards north. The vector product `vecAxxvecB` is

Options

A. along west
B. along east
C. zero
D. vertically downward

Answer

To solve the problem of finding the vector product \(\vec{A} \times \vec{B}\), we can follow these steps:

### Step 1: Define the Vectors
- Let \(\vec{A}\) be the vector pointing vertically upward. In Cartesian coordinates, this can be represented as:
\[ \vec{A} = k \hat{k} \]
where \(\hat{k}\) is the unit vector in the z-direction (upward).

- Let \(\vec{B}\) be the vector pointing towards the north. In Cartesian coordinates, this can be represented as:
\[ \vec{B} = j \hat{j} \]
where \(\hat{j}\) is the unit vector in the y-direction (north).

### Step 2: Calculate the Cross Product
The vector product (cross product) \(\vec{A} \times \vec{B}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} \]

### Step 3: Evaluate the Determinant
Calculating the determinant, we have:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} \]
This simplifies to:
\[ \vec{A} \times \vec{B} = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = 0 \hat{i} + 0 \hat{j} + 1 \hat{i} = -\hat{i} \]

### Step 4: Interpret the Result
The result \(-\hat{i}\) indicates that the direction of the vector product \(\vec{A} \times \vec{B}\) is in the negative x-direction, which corresponds to the west direction.

### Conclusion
Thus, the vector product \(\vec{A} \times \vec{B}\) points towards the west.