If `a=(1)/(7)(2hat(i)+3hat(j)+6hat(k)): b=(1)/(7)(6hat(i)+2hat(j)-3hat(k)): c=c_1hat(i)+c_2hat(j)+c_2hat(k)` and matrix `A=[[(2)/(7), (3)/(7), (6)/(7)], [(6)/(7), (2)/(7), -(3)/(7)], [c_1, c_2, c_3]] and AT^(T)=I`, then c

Options

A. (a)`(3hat(i)+6hat(j)+2hat(k))/(7)`
B. (b)`(3hat(i)-6hat(j)+2hat(k))/(7)`
C. (c)`(-3hat(i)+6hat(j)-2hat(k))/(7)`
D. (d)`-(3hat(i)+6hat(j)+2hat(k))/(7)`

Answer

To solve the problem, we need to find the vector \( c \) given the vectors \( a \) and \( b \), and the matrix \( A \). The condition \( A A^T = I \) implies that the rows of \( A \) are orthonormal vectors.

### Step 1: Write down the vectors \( a \) and \( b \)

Given:
\[ a = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
\[ b = \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) \]

### Step 2: Write down the matrix \( A \)

The matrix \( A \) is given by:
\[ A = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\ \frac{6}{7} & \frac{2}{7} & -\frac{3}{7} \\ c_1 & c_2 & c_3 \end{bmatrix} \]

### Step 3: Calculate \( A A^T \)

We need to compute \( A A^T \) and set it equal to the identity matrix \( I \).

The transpose of \( A \) is:
\[ A^T = \begin{bmatrix} \frac{2}{7} & \frac{6}{7} & c_1 \\ \frac{3}{7} & \frac{2}{7} & c_2 \\ \frac{6}{7} & -\frac{3}{7} & c_3 \end{bmatrix} \]

Now, we compute \( A A^T \):
\[ A A^T = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\ \frac{6}{7} & \frac{2}{7} & -\frac{3}{7} \\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix} \frac{2}{7} & \frac{6}{7} & c_1 \\ \frac{3}{7} & \frac{2}{7} & c_2 \\ \frac{6}{7} & -\frac{3}{7} & c_3 \end{bmatrix} \]

### Step 4: Calculate the elements of \( A A^T \)

1. **First row, first column**:
\[ \left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2 = \frac{4}{49} + \frac{9}{49} + \frac{36}{49} = \frac{49}{49} = 1 \]

2. **First row, second column**:
\[ \frac{2}{7} \cdot \frac{6}{7} + \frac{3}{7} \cdot \frac{2}{7} + \frac{6}{7} \cdot -\frac{3}{7} = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = 0 \]

3. **First row, third column**:
\[ \frac{2}{7}c_1 + \frac{3}{7}c_2 + \frac{6}{7}c_3 = 0 \]

4. **Second row, first column**:
\[ \frac{6}{7} \cdot \frac{2}{7} + \frac{2}{7} \cdot \frac{3}{7} + -\frac{3}{7} \cdot \frac{6}{7} = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = 0 \]

5. **Second row, second column**:
\[ \left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2 = \frac{36}{49} + \frac{4}{49} + \frac{9}{49} = \frac{49}{49} = 1 \]

6. **Second row, third column**:
\[ \frac{6}{7}c_1 + \frac{2}{7}c_2 - \frac{3}{7}c_3 = 0 \]

7. **Third row, first column**:
\[ c_1 \cdot \frac{2}{7} + c_2 \cdot \frac{6}{7} + c_3 \cdot c_1 = 0 \]

8. **Third row, second column**:
\[ c_1 \cdot \frac{3}{7} + c_2 \cdot \frac{2}{7} + c_3 \cdot c_2 = 0 \]

9. **Third row, third column**:
\[ c_1^2 + c_2^2 + c_3^2 = 1 \]

### Step 5: Solve the equations

From the equations derived, we have:
1. \( \frac{2}{7}c_1 + \frac{3}{7}c_2 + \frac{6}{7}c_3 = 0 \)
2. \( \frac{6}{7}c_1 + \frac{2}{7}c_2 - \frac{3}{7}c_3 = 0 \)
3. \( c_1^2 + c_2^2 + c_3^2 = 1 \)

By solving these equations, we can express \( c_1, c_2, c_3 \) in terms of each other and find their values.

### Step 6: Substitute values and simplify

After solving the equations, we find:
\[ c = \frac{1}{7}(-3\hat{i} + 6\hat{j} - 2\hat{k}) \]

### Final Answer

Thus, the vector \( c \) is:
\[ c = \frac{1}{7}(-3\hat{i} + 6\hat{j} - 2\hat{k}) \]