The vectors `vec(a)= x hat(i) + z hat(k), vec(b) = hat(k), vec(c )` are such that they form a right handed system. What is `vec(c )` equal to ?

Options

A. `hat(j)`
B. `z hat(i) - x hat(k)`
C. `y hat(i) - xhat(j)`
D. `x hat(i) - y hat(j)`

Answer

To find the vector \(\vec{c}\) such that the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) form a right-handed system, we can use the cross product.

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Given vectors are:
\[ \vec{a} = x \hat{i} + 0 \hat{j} + z \hat{k} \]
\[ \vec{b} = 0 \hat{i} + 0 \hat{j} + 1 \hat{k} \]
- We need to find \(\vec{c}\).

2. **Use the Right-Hand Rule**:
- For a right-handed system, we have:
\[ \vec{c} = \vec{b} \times \vec{a} \]

3. **Set Up the Cross Product**:
- The cross product can be calculated using the determinant of a matrix:
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ x & 0 & z \end{vmatrix} \]

4. **Calculate the Determinant**:
- Expanding the determinant along the first row:
\[ \vec{c} = \hat{i} \begin{vmatrix} 0 & 1 \\ 0 & z \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 0 \\ x & 0 \end{vmatrix} \]

5. **Calculate Each Minor**:
- The first minor:
\[ \begin{vmatrix} 0 & 1 \\ 0 & z \end{vmatrix} = 0 \cdot z - 1 \cdot 0 = 0 \]
- The second minor:
\[ \begin{vmatrix} 0 & 1 \\ x & z \end{vmatrix} = 0 \cdot z - 1 \cdot x = -x \]
- The third minor:
\[ \begin{vmatrix} 0 & 0 \\ x & 0 \end{vmatrix} = 0 \cdot 0 - 0 \cdot x = 0 \]

6. **Combine the Results**:
- Putting it all together:
\[ \vec{c} = \hat{i}(0) - \hat{j}(-x) + \hat{k}(0) = x \hat{j} \]

7. **Final Result**:
- Thus, the vector \(\vec{c}\) is:
\[ \vec{c} = x \hat{j} \]