If `a=hati+hatj+hatk and b=hati-hatj`, then vectors `((a*hati)hati+(a*hatj)hatj+(a*hatk)hatk),{(b*hati)hati+(bhatj)hatj+(b*hatk)hatk} and( hati+hatj-2hatk)`

Options

A. are mutually perpendicular
B. are coplanar
C. form a parallepiped of volume 6 units
D. form a parallelopiped of volume 3 units

Answer

To solve the problem, we need to find the scalar triple product of the three vectors defined in the question. Let's break down the solution step by step.

### Given:
1. **Vector a**: \( \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \)
2. **Vector b**: \( \mathbf{b} = \hat{i} - \hat{j} \)

### Step 1: Calculate the first vector \( \mathbf{x} \)
The first vector is defined as:
\[ \mathbf{x} = (\mathbf{a} \cdot \hat{i}) \hat{i} + (\mathbf{a} \cdot \hat{j}) \hat{j} + (\mathbf{a} \cdot \hat{k}) \hat{k} \]

Calculating the dot products:
- \( \mathbf{a} \cdot \hat{i} = 1 \)
- \( \mathbf{a} \cdot \hat{j} = 1 \)
- \( \mathbf{a} \cdot \hat{k} = 1 \)

Thus, we have:
\[ \mathbf{x} = 1 \hat{i} + 1 \hat{j} + 1 \hat{k} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate the second vector \( \mathbf{y} \)
The second vector is defined as:
\[ \mathbf{y} = (\mathbf{b} \cdot \hat{i}) \hat{i} + (\mathbf{b} \cdot \hat{j}) \hat{j} + (\mathbf{b} \cdot \hat{k}) \hat{k} \]

Calculating the dot products:
- \( \mathbf{b} \cdot \hat{i} = 1 \)
- \( \mathbf{b} \cdot \hat{j} = -1 \)
- \( \mathbf{b} \cdot \hat{k} = 0 \)

Thus, we have:
\[ \mathbf{y} = 1 \hat{i} - 1 \hat{j} + 0 \hat{k} = \hat{i} - \hat{j} \]

### Step 3: Define the third vector \( \mathbf{z} \)
The third vector is given as:
\[ \mathbf{z} = \hat{i} + \hat{j} - 2 \hat{k} \]

### Step 4: Calculate the scalar triple product \( \mathbf{x} \cdot (\mathbf{y} \times \mathbf{z}) \)
The scalar triple product can be calculated using the determinant of the matrix formed by the components of the vectors:
\[ \text{Volume} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -2 \end{vmatrix} \]

### Step 5: Calculate the determinant
Expanding the determinant:
\[ = 1 \begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (0)(1) = 2 \)

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

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

Putting it all together:
\[ = 1(2) - 1(-2) + 1(2) = 2 + 2 + 2 = 6 \]

### Final Result
The volume of the parallel pipette formed by the vectors is \( 6 \) cubic units.

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