A vector `vecF_(1)` is along the positive `X`-axis. its vectors product with another vector `vecF_(2)` is ? where `vecF_(2)` is along Y-axis

Explanation

To solve the problem of finding the vector product (cross product) of two vectors, `vecF_(1)` and `vecF_(2)`, where `vecF_(1)` is along the positive X-axis and `vecF_(2)` is along the Y-axis, we can follow these steps:

### Step 1: Define the Vectors
We start by defining the two vectors in a three-dimensional coordinate system:
- Let `vecF_(1) = F1 * i`, where `F1` is the magnitude of `vecF_(1)` and `i` is the unit vector in the X-direction.
- Let `vecF_(2) = F2 * j`, where `F2` is the magnitude of `vecF_(2)` and `j` is the unit vector in the Y-direction.

### Step 2: Write the Cross Product Formula
The cross product of two vectors `vecA` and `vecB` can be calculated using the determinant of a matrix:
\[ vecA \times vecB = \begin{vmatrix} i & j & k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \]
For our vectors:
- `vecF_(1) = F1 * i` implies `A_x = F1`, `A_y = 0`, `A_z = 0`.
- `vecF_(2) = F2 * j` implies `B_x = 0`, `B_y = F2`, `B_z = 0`.

### Step 3: Set Up the Determinant
Now we can set up the determinant for the cross product:
\[ vecF_(1) \times vecF_(2) = \begin{vmatrix} i & j & k \\ F1 & 0 & 0 \\ 0 & F2 & 0 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Calculating the determinant, we have:
\[ vecF_(1) \times vecF_(2) = i(0 \cdot 0 - 0 \cdot F2) - j(F1 \cdot 0 - 0 \cdot 0) + k(F1 \cdot F2 - 0 \cdot 0) \]
This simplifies to:
\[ vecF_(1) \times vecF_(2) = k(F1 \cdot F2) \]

### Step 5: Determine the Direction
The result shows that the vector product `vecF_(1) \times vecF_(2)` is along the positive Z-axis, represented by the unit vector `k`.

### Final Result
Thus, the vector product of `vecF_(1)` and `vecF_(2)` is:
\[ vecF_(1) \times vecF_(2) = F1 \cdot F2 \cdot k \]
where `k` indicates the direction along the positive Z-axis.