Find `a times b` if `a=i-j+k" and "b=i+j-3k`.

Answer

To find the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), we will use the determinant method. The vectors are given as:

\[ \mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} \quad \text{and} \quad \mathbf{b} = \mathbf{i} + \mathbf{j} - 3\mathbf{k} \]

### Step 1: Set up the determinant

We will set up a determinant with the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row, the components of vector \( \mathbf{a} \) in the second row, and the components of vector \( \mathbf{b} \) in the third row.

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -3 \end{vmatrix} \]

### Step 2: Calculate the determinant

To calculate the determinant, we will expand it using the first row:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 1 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

1. For the first determinant:
\[ \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (1)(1) = 3 - 1 = 2 \]

2. For the second determinant:
\[ \begin{vmatrix} 1 & 1 \\ 1 & -3 \end{vmatrix} = (1)(-3) - (1)(1) = -3 - 1 = -4 \]

3. For the third determinant:
\[ \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \]

### Step 4: Substitute back into the expression

Now substituting these values back into the expression for \( \mathbf{a} \times \mathbf{b} \):

\[ \mathbf{a} \times \mathbf{b} = \mathbf{i}(2) - \mathbf{j}(-4) + \mathbf{k}(2) \]

This simplifies to:

\[ \mathbf{a} \times \mathbf{b} = 2\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \]

### Final Result

Thus, the cross product \( \mathbf{a} \times \mathbf{b} \) is:

\[ \mathbf{a} \times \mathbf{b} = 2\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \]