If `a = i+j+k and b=i-j` then the vectors
`(a.i)i+(a.j)j+(a.k)k, `
`(b.i)+(b.j)j+(b.k)k, and i+j-2k`

Options

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

Answer

To solve the problem, we need to find the vectors \( R_1 \), \( R_2 \), and \( R_3 \) based on the given vectors \( \mathbf{a} \) and \( \mathbf{b} \), and then determine their relationships (mutually perpendicular, coplanar, etc.).

### Step 1: Define the vectors
Given:
\[ \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = \mathbf{i} - \mathbf{j} \]

### Step 2: Calculate \( R_1 \)
The first vector \( R_1 \) is defined as:
\[ R_1 = (a_i)\mathbf{i} + (a_j)\mathbf{j} + (a_k)\mathbf{k} \]
Where \( a_i, a_j, a_k \) are the components of vector \( \mathbf{a} \).

From \( \mathbf{a} \):
- \( a_i = 1 \)
- \( a_j = 1 \)
- \( a_k = 1 \)

Thus,
\[ R_1 = 1\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]

### Step 3: Calculate \( R_2 \)
The second vector \( R_2 \) is defined as:
\[ R_2 = (b_i)\mathbf{i} + (b_j)\mathbf{j} + (b_k)\mathbf{k} \]
Where \( b_i, b_j, b_k \) are the components of vector \( \mathbf{b} \).

From \( \mathbf{b} \):
- \( b_i = 1 \)
- \( b_j = -1 \)
- \( b_k = 0 \)

Thus,
\[ R_2 = 1\mathbf{i} - 1\mathbf{j} + 0\mathbf{k} = \mathbf{i} - \mathbf{j} \]

### Step 4: Define \( R_3 \)
The third vector \( R_3 \) is given as:
\[ R_3 = \mathbf{i} + \mathbf{j} - 2\mathbf{k} \]

### Step 5: Check for mutual perpendicularity
To check if the vectors \( R_1 \), \( R_2 \), and \( R_3 \) are mutually perpendicular, we need to calculate the dot products:

1. **Calculate \( R_1 \cdot R_2 \)**:
\[ R_1 \cdot R_2 = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot (\mathbf{i} - \mathbf{j}) = 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot 0 = 1 - 1 + 0 = 0 \]

2. **Calculate \( R_1 \cdot R_3 \)**:
\[ R_1 \cdot R_3 = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot (-2) = 1 + 1 - 2 = 0 \]

3. **Calculate \( R_2 \cdot R_3 \)**:
\[ R_2 \cdot R_3 = (\mathbf{i} - \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 1 \cdot 1 + (-1) \cdot 1 + 0 \cdot (-2) = 1 - 1 + 0 = 0 \]

Since all dot products are zero, the vectors \( R_1 \), \( R_2 \), and \( R_3 \) are mutually perpendicular.

### Step 6: Check for coplanarity
Vectors are coplanar if the scalar triple product is zero. Since \( R_1 \), \( R_2 \), and \( R_3 \) are mutually perpendicular, they cannot be coplanar.

### Step 7: Calculate the volume of the parallelepiped
The volume \( V \) of the parallelepiped formed by the vectors can be calculated using the scalar triple product:
\[ V = |R_1 \cdot (R_2 \times R_3)| \]

First, calculate \( R_2 \times R_3 \):
\[ R_2 \times R_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 0 \\ 1 & 1 & -2 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}((-1)(-2) - (0)(1)) - \mathbf{j}(1 \cdot (-2) - 0 \cdot 1) + \mathbf{k}(1 \cdot 1 - (-1) \cdot 1) \]
\[ = \mathbf{i}(2) - \mathbf{j}(-2) + \mathbf{k}(1 + 1) \]
\[ = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \]

Now calculate \( R_1 \cdot (R_2 \times R_3) \):
\[ R_1 \cdot (R_2 \times R_3) = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot (2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2 + 2 + 2 = 6 \]

Thus, the volume of the parallelepiped is \( 6 \).

### Conclusion
The vectors \( R_1 \), \( R_2 \), and \( R_3 \) are mutually perpendicular and the volume of the parallelepiped formed by them is \( 6 \).