Let `a= 2hat(i) -2hat(k) , b=hat(i) +hat(j) ` and c be a vectors such that `|c-a| =3, |(axxb)xx c|=3` and the angle between c and `axx b" is "30^(@)` . Then a. c is equal to

Options

A. `(25)/(8)`
B. `2`
C. `5`
D. `(1)/(8)`

Answer

To solve the problem step by step, we will follow the given conditions and use vector operations.

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = 2\hat{i} - 2\hat{k} \)
- \( \mathbf{b} = \hat{i} + \hat{j} \)

### Step 2: Calculate the modulus of vectors \( \mathbf{a} \) and \( \mathbf{b} \)
1. **Modulus of \( \mathbf{a} \)**:
\[ |\mathbf{a}| = \sqrt{(2)^2 + (0)^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8} \]

2. **Modulus of \( \mathbf{b} \)**:
\[ |\mathbf{b}| = \sqrt{(1)^2 + (1)^2 + (0)^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

### Step 3: Use the condition \( |\mathbf{c} - \mathbf{a}| = 3 \)
This can be expressed as:
\[ |\mathbf{c}|^2 + |\mathbf{a}|^2 - 2\mathbf{c} \cdot \mathbf{a} = 9 \]
Substituting \( |\mathbf{a}|^2 = 8 \):
\[ |\mathbf{c}|^2 + 8 - 2\mathbf{c} \cdot \mathbf{a} = 9 \]
Thus,
\[ |\mathbf{c}|^2 - 2\mathbf{c} \cdot \mathbf{a} = 1 \quad \text{(Equation 1)} \]

### Step 4: Calculate \( \mathbf{a} \times \mathbf{b} \)
To find \( \mathbf{a} \times \mathbf{b} \):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ 1 & 1 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(0 \cdot 0 - (-2) \cdot 1) - \hat{j}(2 \cdot 0 - (-2) \cdot 1) + \hat{k}(2 \cdot 1 - 0 \cdot 1) \]
\[ = \hat{i}(2) - \hat{j}(2) + \hat{k}(2) = 2\hat{i} - 2\hat{j} + 2\hat{k} \]

### Step 5: Calculate the modulus of \( \mathbf{a} \times \mathbf{b} \)
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(2)^2 + (-2)^2 + (2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} \]

### Step 6: Use the condition \( |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}| = 3 \)
This can be expressed as:
\[ |\mathbf{a} \times \mathbf{b}| |\mathbf{c}| \sin(30^\circ) = 3 \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ |\mathbf{c}| \cdot \sqrt{12} \cdot \frac{1}{2} = 3 \]
Thus,
\[ |\mathbf{c}| \cdot \frac{\sqrt{12}}{2} = 3 \implies |\mathbf{c}| = \frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \]

### Step 7: Substitute \( |\mathbf{c}| \) into Equation 1
Substituting \( |\mathbf{c}|^2 = 3 \):
\[ 3 - 2\mathbf{c} \cdot \mathbf{a} = 1 \]
Thus,
\[ 2\mathbf{c} \cdot \mathbf{a} = 2 \implies \mathbf{c} \cdot \mathbf{a} = 1 \]

### Final Answer
Thus, \( \mathbf{a} \cdot \mathbf{c} = 1 \).