Given the vectors `vec u=2 hat i-hat j-hat k and vec v=hat i-hat j+2hat k and vec w=hat i-hat k` If the volume of the parallelopiped having `- cvec u,vec v and c vec w` as concurrent edges, is 8 then `c` can be equal to

Options

A. a) `pm2`
B. b) `4`
C. c) `8`
D. d) cannot be determine

Answer

To solve the problem step by step, we will follow the outlined procedure to find the value of \( c \) such that the volume of the parallelepiped formed by the vectors \( -c\vec{u}, \vec{v}, \) and \( c\vec{w} \) equals 8.

### Step 1: Define the vectors
We are given the vectors:
\[ \vec{u} = 2\hat{i} - \hat{j} - \hat{k} \]
\[ \vec{v} = \hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{w} = \hat{i} - \hat{k} \]

### Step 2: Express the modified vectors
We need to express the vectors \( -c\vec{u}, \vec{v}, \) and \( c\vec{w} \):
\[ -c\vec{u} = -c(2\hat{i} - \hat{j} - \hat{k}) = -2c\hat{i} + c\hat{j} + c\hat{k} \]
\[ \vec{v} = \hat{i} - \hat{j} + 2\hat{k} \]
\[ c\vec{w} = c(\hat{i} - \hat{k}) = c\hat{i} - c\hat{k} \]

### Step 3: Set up the volume formula
The volume \( V \) of the parallelepiped formed by the vectors \( \vec{a}, \vec{b}, \vec{c} \) is given by the scalar triple product:
\[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]
In this case, we can write:
\[ V = |(-c\vec{u}) \cdot (\vec{v} \times (c\vec{w}))| \]
This can be simplified to:
\[ V = |(-c) \cdot \text{det}(-2c, 1, c; 1, -1, 0; 1, 2, -1)| \]

### Step 4: Calculate the determinant
We need to calculate the determinant:
\[ \text{det} \begin{pmatrix} -2c & c & c \\ 1 & -1 & 2 \\ 1 & 0 & -1 \end{pmatrix} \]

### Step 5: Simplify the determinant
We can factor out \( c \) from the first row and the third row:
\[ = c^2 \cdot \text{det} \begin{pmatrix} -2 & 1 & 1 \\ 1 & -1 & 2 \\ 0 & 1 & -1 \end{pmatrix} \]

Now we calculate the determinant:
\[ = -2 \cdot \text{det} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} + 1 \cdot \text{det} \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} + 1 \cdot \text{det} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \]

Calculating these:
1. \(\text{det} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} = (-1)(-1) - (2)(1) = 1 - 2 = -1\)
2. \(\text{det} \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} = (1)(-1) - (2)(0) = -1\)
3. \(\text{det} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = (1)(1) - (-1)(0) = 1\)

So, the determinant becomes:
\[ = -2(-1) + 1(-1) + 1(1) = 2 - 1 + 1 = 2 \]

### Step 6: Set the volume equal to 8
Now we have:
\[ V = |c^2 \cdot 2| = 8 \]
This simplifies to:
\[ 2c^2 = 8 \implies c^2 = 4 \implies c = \pm 2 \]

### Conclusion
Thus, the possible values of \( c \) are:
\[ \boxed{\pm 2} \]