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

Options

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

Answer

To solve the problem, we need to determine the cross product of the two vectors, \(\vec{A}\) and \(\vec{B}\). Here's a step-by-step solution:

### Step 1: Define the Vectors
- The vector \(\vec{A}\) points vertically upward. In Cartesian coordinates, this can be represented as:
\[ \vec{A} = 0 \hat{i} + 0 \hat{j} + A \hat{k} \]
where \(A\) is the magnitude of the vector in the z-direction (upward).

- The vector \(\vec{B}\) points towards the north, which corresponds to the positive y-direction. Thus, it can be represented as:
\[ \vec{B} = 0 \hat{i} + B \hat{j} + 0 \hat{k} \]
where \(B\) is the magnitude of the vector in the y-direction (north).

### Step 2: Calculate the Cross Product
The 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 the vectors:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & A \\ 0 & B & 0 \end{vmatrix} \]

### Step 3: Expand the Determinant
Calculating the determinant gives:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 0 & A \\ B & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & A \\ 0 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 0 \\ 0 & B \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} 0 & A \\ B & 0 \end{vmatrix} = (0)(0) - (A)(B) = -AB \]

The second determinant is zero:
\[ \begin{vmatrix} 0 & A \\ 0 & 0 \end{vmatrix} = 0 \]

The third determinant is also zero:
\[ \begin{vmatrix} 0 & 0 \\ 0 & B \end{vmatrix} = 0 \]

Putting it all together:
\[ \vec{A} \times \vec{B} = -AB \hat{i} + 0 \hat{j} + 0 \hat{k} = -AB \hat{i} \]

### Step 4: Interpret the Result
The result \(-AB \hat{i}\) indicates that the direction of the vector \(\vec{A} \times \vec{B}\) is towards the west (negative x-direction).

### Final Answer
Thus, the vector produced by \(\vec{A} \times \vec{B}\) is directed towards the west.

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