The vector `[(i-j+k) times (2i-3j-k)] times [(-3i+j+k) times (2j+k)]` is given by

Options

A. `3i+5j-3k`
B. `-5(3i-5j-3k)`
C. `5(3i+5j-3k)`
D. `(15i-25j+15k)`

Answer

To solve the problem, we need to compute the vector expression given by:

\[ [(\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} - 3\mathbf{j} - \mathbf{k})] \times [(-3\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{j} + \mathbf{k})] \]

### Step 1: Compute the first cross product \((\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} - 3\mathbf{j} - \mathbf{k})\)

We set up the determinant for the cross product:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 2 & -3 & -1 \end{vmatrix} \]

Calculating this determinant:

\[ = \mathbf{i} \begin{vmatrix} -1 & 1 \\ -3 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ 2 & -3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 1 \\ -3 & -1 \end{vmatrix} = (-1)(-1) - (1)(-3) = 1 + 3 = 4\)
2. \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3\)
3. \(\begin{vmatrix} 1 & -1 \\ 2 & -3 \end{vmatrix} = (1)(-3) - (-1)(2) = -3 + 2 = -1\)

Substituting back, we have:

\[ = 4\mathbf{i} - (-3)\mathbf{j} - 1\mathbf{k} = 4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k} \]

### Step 2: Compute the second cross product \((-3\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{j} + \mathbf{k})\)

Set up the determinant for this cross product:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & 1 \\ 0 & 2 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = \mathbf{i} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -3 & 1 \\ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -3 & 1 \\ 0 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (1)(2) = 1 - 2 = -1\)
2. \(\begin{vmatrix} -3 & 1 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (1)(0) = -3\)
3. \(\begin{vmatrix} -3 & 1 \\ 0 & 2 \end{vmatrix} = (-3)(2) - (1)(0) = -6\)

Substituting back, we have:

\[ = -1\mathbf{i} - (-3)\mathbf{j} - 6\mathbf{k} = -1\mathbf{i} + 3\mathbf{j} - 6\mathbf{k} \]

### Step 3: Compute the final cross product

Now we need to find:

\[ (4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k}) \times (-1\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \]

Set up the determinant:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & -1 \\ -1 & 3 & -6 \end{vmatrix} \]

Calculating this determinant:

\[ = \mathbf{i} \begin{vmatrix} 3 & -1 \\ 3 & -6 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 4 & -1 \\ -1 & -6 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 3 & -1 \\ 3 & -6 \end{vmatrix} = (3)(-6) - (-1)(3) = -18 + 3 = -15\)
2. \(\begin{vmatrix} 4 & -1 \\ -1 & -6 \end{vmatrix} = (4)(-6) - (-1)(-1) = -24 - 1 = -25\)
3. \(\begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} = (4)(3) - (3)(-1) = 12 + 3 = 15\)

Substituting back, we have:

\[ = -15\mathbf{i} + 25\mathbf{j} + 15\mathbf{k} \]

### Final Answer

Thus, the final vector is:

\[ -15\mathbf{i} + 25\mathbf{j} + 15\mathbf{k} \]