Explain why the answer is -k and why A = -i + 3j and b= i -2j find a x b

### Step 1: Understanding the Cross Product
Given two vectors \( \mathbf{A} = -\mathbf{i} + 3\mathbf{j} \) and \( \mathbf{B} = \mathbf{i} - 2\mathbf{j} \), you are asked to find \( \mathbf{A} \times \mathbf{B} \) (the cross product of the two vectors).

To calculate the cross product, we use the following formula:

\[
\mathbf{A} \times \mathbf{B} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{matrix} \right|
\]

Here, \( A_x \), \( A_y \), and \( A_z \) are the components of vector \( \mathbf{A} \), and \( B_x \), \( B_y \), and \( B_z \) are the components of vector \( \mathbf{B} \). Let's first break down the components of the two vectors:

- \( \mathbf{A} = -\mathbf{i} + 3\mathbf{j} \), so \( A_x = -1 \), \( A_y = 3 \), and \( A_z = 0 \) (since there is no \( \mathbf{k} \) component).
- \( \mathbf{B} = \mathbf{i} - 2\mathbf{j} \), so \( B_x = 1 \), \( B_y = -2 \), and \( B_z = 0 \) (since there is no \( \mathbf{k} \) component).

### Step 2: Applying the Determinant Formula

Now substitute these values into the determinant:

\[
\mathbf{A} \times \mathbf{B} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 0 \\ 1 & -2 & 0 \end{matrix} \right|
\]

We expand this determinant as follows:

\[
\mathbf{A} \times \mathbf{B} = \mathbf{i} \left( \begin{vmatrix} 3 & 0 \\ -2 & 0 \end{vmatrix} \right) - \mathbf{j} \left( \begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} \right) + \mathbf{k} \left( \begin{vmatrix} -1 & 3 \\ 1 & -2 \end{vmatrix} \right)
\]

### Step 3: Calculating the Determinants

Now, compute each of the 2x2 determinants:

1. For \( \mathbf{i} \):

\[
\begin{vmatrix} 3 & 0 \\ -2 & 0 \end{vmatrix} = (3)(0) - (0)(-2) = 0
\]

2. For \( \mathbf{j} \):

\[
\begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} = (-1)(0) - (0)(1) = 0
\]

3. For \( \mathbf{k} \):

\[
\begin{vmatrix} -1 & 3 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (3)(1) = 2 - 3 = -1
\]

### Step 4: Final Cross Product

Substitute the values of the determinants back into the equation:

\[
\mathbf{A} \times \mathbf{B} = 0\mathbf{i} - 0\mathbf{j} + (-1)\mathbf{k}
\]

Thus, the cross product is:

\[
\mathbf{A} \times \mathbf{B} = -\mathbf{k}
\]

### Step 5: Explanation of the Result
The result of the cross product is \( -\mathbf{k} \), which means the resultant vector points in the negative \( \mathbf{k} \)-direction. The magnitude of the vector is 1, and it lies perpendicular to the plane formed by \( \mathbf{A} \) and \( \mathbf{B} \).

### Conclusion
The answer is \( -\mathbf{k} \) because the cross product of two vectors in the plane defined by \( \mathbf{i} \) and \( \mathbf{j} \) results in a vector perpendicular to that plane, which is along the \( \mathbf{k} \)-axis.

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Would you like more details or have any further questions on this? 

Here are 5 related questions to think about:
1. What is the geometric meaning of the cross product?
2. How does the direction of the cross product relate to the right-hand rule?
3. How does changing the signs of the components of \( \mathbf{A} \) or \( \mathbf{B} \) affect the result?
4. What would happen if the vectors \( \mathbf{A} \) and \( \mathbf{B} \) were parallel?
5. How is the magnitude of the cross product related to the angle between the two vectors?

**Tip:** The magnitude of the cross product can be found using the formula \( |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \), where \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{B} \).

Explore how to calculate the cross product of vectors A = -i + 3j and B = i - 2j with a detailed step-by-step explanation. Learn key vector algebra concepts like the cross product formula, the right-hand rule, and determinant calculations. Suitable for Grades 11-12.

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