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

Options

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

Answer

To solve the problem step by step, we will first define the vectors \( \mathbf{a} \) and \( \mathbf{b} \), then calculate the required vectors, and finally check if they are mutually perpendicular and if they are coplanar.

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

### Step 2: Calculate the First Vector
We need to calculate:
\[ \mathbf{A} = (\mathbf{a} \cdot \hat{i}) \hat{i} + (\mathbf{a} \cdot \hat{j}) \hat{j} + (\mathbf{a} \cdot \hat{k}) \hat{k} \]

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

Thus,
\[ \mathbf{A} = 1 \cdot \hat{i} + 1 \cdot \hat{j} + 1 \cdot \hat{k} = \hat{i} + \hat{j} + \hat{k} \]

### Step 3: Calculate the Second Vector
Now we calculate:
\[ \mathbf{B} = (\mathbf{b} \cdot \hat{i}) \hat{i} + (\mathbf{b} \cdot \hat{j}) \hat{j} + (\mathbf{b} \cdot \hat{k}) \hat{k} \]

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

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

### Step 4: Define the Third Vector
The third vector is given as:
\[ \mathbf{C} = \hat{i} + \hat{j} - 2\hat{k} \]

### Step 5: Check for Mutual Perpendicularity
To check if the vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are mutually perpendicular, we need to compute the dot products.

1. Calculate \( \mathbf{A} \cdot \mathbf{B} \):
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j}) = 1 - 1 + 0 = 0 \]

2. Calculate \( \mathbf{B} \cdot \mathbf{C} \):
\[ \mathbf{B} \cdot \mathbf{C} = (\hat{i} - \hat{j}) \cdot (\hat{i} + \hat{j} - 2\hat{k}) = 1 - 1 + 0 = 0 \]

3. Calculate \( \mathbf{A} \cdot \mathbf{C} \):
\[ \mathbf{A} \cdot \mathbf{C} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} - 2\hat{k}) = 1 + 1 - 2 = 0 \]

Since all dot products are zero, \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are mutually perpendicular.

### Step 6: Check for Coplanarity
To check if the vectors are coplanar, we can calculate the scalar triple product:
\[ \text{Scalar Triple Product} = \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \]

We can use the determinant method:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ = 1 \cdot \begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 0 \\ 1 & -2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating each 2x2 determinant:
1. \( (-1)(-2) - (0)(1) = 2 \)
2. \( (1)(-2) - (0)(1) = -2 \)
3. \( (1)(1) - (-1)(1) = 2 \)

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

Since the scalar triple product is non-zero (6), the vectors are not coplanar.

### Conclusion
- The vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are mutually perpendicular.
- They are not coplanar.