Find `(2A)xx(-3B)`, if `A =2hati - hatj` and `B =(hatj+hatk)`

Explanation

To solve the problem of finding \( (2A) \times (-3B) \) given \( A = 2\hat{i} - \hat{j} \) and \( B = \hat{j} + \hat{k} \), we will follow these steps:

### Step 1: Calculate \( 2A \)
Given \( A = 2\hat{i} - \hat{j} \):
\[ 2A = 2(2\hat{i} - \hat{j}) = 4\hat{i} - 2\hat{j} \]

### Step 2: Calculate \( -3B \)
Given \( B = \hat{j} + \hat{k} \):
\[ -3B = -3(\hat{j} + \hat{k}) = -3\hat{j} - 3\hat{k} \]

### Step 3: Set up the cross product \( (2A) \times (-3B) \)
Now we need to compute:
\[ (2A) \times (-3B) = (4\hat{i} - 2\hat{j}) \times (-3\hat{j} - 3\hat{k}) \]

### Step 4: Use the determinant method for the cross product
We can express the vectors in a determinant form:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 0 \\ 0 & -3 & -3 \end{vmatrix} \]

### Step 5: Calculate the determinant
Using the determinant formula:
\[ = \hat{i} \begin{vmatrix} -2 & 0 \\ -3 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 0 \\ 0 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & -2 \\ 0 & -3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \hat{i} \):
\[ = \hat{i}((-2)(-3) - (0)(-3)) = \hat{i}(6) \]
2. For \( \hat{j} \):
\[ = -\hat{j}((4)(-3) - (0)(0)) = -\hat{j}(-12) = 12\hat{j} \]
3. For \( \hat{k} \):
\[ = \hat{k}((4)(-3) - (-2)(0)) = \hat{k}(-12) \]

### Step 6: Combine the results
Putting it all together, we have:
\[ (2A) \times (-3B) = 6\hat{i} + 12\hat{j} - 12\hat{k} \]

### Final Answer
Thus, the result of \( (2A) \times (-3B) \) is:
\[ 6\hat{i} + 12\hat{j} - 12\hat{k} \]