`vecA=(hati-2hatj+6hatk)` and `vecB=(hati-2hatj+hatk)`, find the cross product between `vecA` and `vecB`.

Options

A. `5hati+10hatj`
B. `5hati-10hatj`
C. `10hati+5hatj`
D. `10hati-5hatj`

Answer

To find the cross product of vectors \(\vec{A}\) and \(\vec{B}\), we can use the determinant method. Let's break it down step by step.

### Step 1: Write down the vectors
Given:
\[ \vec{A} = \hat{i} - 2\hat{j} + 6\hat{k} \]
\[ \vec{B} = \hat{i} - 2\hat{j} + \hat{k} \]

### Step 2: Set up the determinant
The cross product \(\vec{A} \times \vec{B}\) can be calculated using the following determinant:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 6 \\ 1 & -2 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we can expand it as follows:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -2 & 6 \\ -2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 6 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 1 & -2 \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinants
1. For \(\hat{i}\):
\[ \begin{vmatrix} -2 & 6 \\ -2 & 1 \end{vmatrix} = (-2)(1) - (6)(-2) = -2 + 12 = 10 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 1 & 6 \\ 1 & 1 \end{vmatrix} = (1)(1) - (6)(1) = 1 - 6 = -5 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & -2 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (-2)(1) = -2 + 2 = 0 \]

### Step 5: Substitute back into the expression
Now substituting back into the expression for the cross product:
\[ \vec{A} \times \vec{B} = 10\hat{i} - (-5)\hat{j} + 0\hat{k} \]
This simplifies to:
\[ \vec{A} \times \vec{B} = 10\hat{i} + 5\hat{j} + 0\hat{k} \]
or simply:
\[ \vec{A} \times \vec{B} = 10\hat{i} + 5\hat{j} \]

### Final Answer
The cross product \(\vec{A} \times \vec{B}\) is:
\[ \boxed{10\hat{i} + 5\hat{j}} \]