A vector `vec(A)` points verically upward and `vec(B)` points towards north. The vector product `vec(A) xx vec(B)` is

Options

A. Zero
B. Along west
C. Along east
D. Vertically downward

Explanation

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

### Step 1: Identify the Directions of the Vectors
- Vector \( \vec{A} \) points vertically upward, which we can represent as the positive z-axis. Therefore, we can denote \( \vec{A} = k \hat{z} \), where \( k \) is a positive constant.
- Vector \( \vec{B} \) points towards the north, which we can represent as the positive y-axis. Thus, we can denote \( \vec{B} = j \hat{y} \), where \( j \) is a constant.

### Step 2: Write the Vector Product
The vector product \( \vec{A} \times \vec{B} \) can be calculated using the right-hand rule and the properties of cross products.

### Step 3: Apply the Cross Product Formula
Using the formula for the cross product:
\[ \vec{A} \times \vec{B} = (k \hat{z}) \times (j \hat{y}) \]
We can use the determinant form of the cross product:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & k \\ 0 & j & 0 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Calculating the determinant, we have:
\[ \vec{A} \times \vec{B} = \hat{i}(0 \cdot 0 - j \cdot k) - \hat{j}(0 \cdot 0 - 0 \cdot k) + \hat{k}(0 \cdot j - 0 \cdot 0) \]
This simplifies to:
\[ \vec{A} \times \vec{B} = -jk \hat{i} \]

### Step 5: Determine the Direction
Since \( -jk \hat{i} \) indicates that the result is in the negative x-direction, we can conclude that the vector product \( \vec{A} \times \vec{B} \) points towards the west.

### Final Answer
Thus, the vector product \( \vec{A} \times \vec{B} \) points towards the west, which can be represented as:
\[ \vec{A} \times \vec{B} = -\hat{i} \]