Find the volume of the parallelopiped, whose three coterminous edges are represented by the vectors `hati+hatj+hatk,hati-hatj+hatk,hati+2hatj-hatk`.

Answer

To find the volume of the parallelepiped formed by the vectors \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), and \(\vec{c} = \hat{i} + 2\hat{j} - \hat{k}\), we can use the scalar triple product formula, which is given by the determinant of a matrix formed by these vectors.

### Step 1: Write the vectors in component form
The vectors can be represented as:
- \(\vec{a} = (1, 1, 1)\)
- \(\vec{b} = (1, -1, 1)\)
- \(\vec{c} = (1, 2, -1)\)

### Step 2: Set up the determinant
The volume \(V\) of the parallelepiped is given by the absolute value of the determinant of the matrix formed by these vectors:

\[ V = \left| \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} \right| \]

### Step 3: Calculate the determinant
To calculate the determinant, we can use the formula for a \(3 \times 3\) determinant:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this to our matrix:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 1((-1)(-1) - (1)(2)) - 1((1)(-1) - (1)(1)) + 1((1)(2) - (1)(1)) \]

Calculating each term:

1. First term: \(1(1 - 2) = 1(-1) = -1\)
2. Second term: \(-1(-1 - 1) = -1(-2) = 2\)
3. Third term: \(1(2 - 1) = 1(1) = 1\)

Now, combining these results:

\[ -1 + 2 + 1 = 2 \]

### Step 4: Take the absolute value
The volume of the parallelepiped is:

\[ V = |2| = 2 \]

### Final Answer
The volume of the parallelepiped is \(2\) cubic units.

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