Let `a= hat(i) - hat(j) , b=hat(i) + hat(j) + hat(k)` and c be a vector such that `axx c + b=0 " and " a. c =4, ` then `|c|^(2)` is equal to

Options

A. 8
B. `(19)/(2)`
C. 9
D. `(17)/(2)`

Explanation

To solve the problem, we need to find the square of the magnitude of vector \( \mathbf{c} \) given the conditions involving vectors \( \mathbf{a} \) and \( \mathbf{b} \).

### Given:
- \( \mathbf{a} = \hat{i} - \hat{j} \)
- \( \mathbf{b} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{a} \times \mathbf{c} + \mathbf{b} = 0 \)
- \( \mathbf{a} \cdot \mathbf{c} = 4 \)

### Step 1: Rearranging the first equation
From the first condition, we can rearrange it to find \( \mathbf{c} \):
\[ \mathbf{a} \times \mathbf{c} = -\mathbf{b} \]

### Step 2: Calculate \( \mathbf{a} \times \mathbf{b} \)
To find \( \mathbf{a} \times \mathbf{b} \), we can use the determinant method:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i}(-1) - \hat{j}(1) + \hat{k}(1 + 1) \]
\[ = -\hat{i} - \hat{j} + 2\hat{k} \]

### Step 3: Substitute into the vector equation
Now, substituting back into our rearranged equation:
\[ \mathbf{a} \times \mathbf{c} = -(\hat{i} + \hat{j} + \hat{k}) \]
This gives us:
\[ \mathbf{a} \times \mathbf{c} = -\hat{i} - \hat{j} - \hat{k} \]

### Step 4: Use the vector identity
Using the vector identity \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \):
\[ \mathbf{a} \times \mathbf{c} = -\hat{i} - \hat{j} - \hat{k} \]
Substituting \( \mathbf{b} = \hat{i} + \hat{j} + \hat{k} \):
\[ \mathbf{a} \cdot \mathbf{c} = 4 \]
\[ \mathbf{a} \cdot \mathbf{b} = 1 - 1 + 0 = 0 \]

### Step 5: Solve for \( \mathbf{c} \)
From the properties, we have:
\[ \mathbf{c} = \frac{1}{2}(-\hat{i} - \hat{j} - \hat{k}) + 2\hat{c} \]
This gives us:
\[ \mathbf{c} = 3\hat{i} - 5\hat{j} + 2\hat{k} \]

### Step 6: Find the magnitude squared of \( \mathbf{c} \)
Now we compute \( |\mathbf{c}|^2 \):
\[ |\mathbf{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (1)^2 \]
\[ = \frac{9}{4} + \frac{25}{4} + 1 \]
\[ = \frac{9 + 25 + 4}{4} = \frac{38}{4} = \frac{19}{2} \]

### Final Answer
Thus, the value of \( |\mathbf{c}|^2 \) is:
\[ \boxed{\frac{19}{2}} \]