`hata` and `hatb` are two mutually perpendicular unit vectors. If the vectors `xhata+xhatb+z(hataxxhatb),hata+(hataxxhatb)` and `zhata+zhatb+y(hataxxhatb)` lie in a plane, then `z` is

Options

A. A.M is `x` and `y`
B. G.M. of `x` and `y`
C. H.M. of `x` and `y`
D. equal to zero

Explanation

To solve the problem, we need to determine the value of \( z \) given that the vectors \( \mathbf{x}\hat{a} + \mathbf{x}\hat{b} + z(\hat{a} \times \hat{b}) \), \( \hat{a} + \hat{a} + \hat{b} \), and \( z\hat{a} + z\hat{b} + y(\hat{a} \times \hat{b}) \) lie in the same plane.

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \( \hat{a} \) and \( \hat{b} \) be two mutually perpendicular unit vectors. This means \( \hat{a} \cdot \hat{b} = 0 \) and \( |\hat{a}| = |\hat{b}| = 1 \).
- The cross product \( \hat{a} \times \hat{b} \) will give another unit vector that is perpendicular to both \( \hat{a} \) and \( \hat{b} \).

2. **Identifying the Vectors**:
- The first vector is \( \mathbf{v_1} = \mathbf{x}\hat{a} + \mathbf{x}\hat{b} + z(\hat{a} \times \hat{b}) \).
- The second vector is \( \mathbf{v_2} = 2\hat{a} + \hat{b} \).
- The third vector is \( \mathbf{v_3} = z\hat{a} + z\hat{b} + y(\hat{a} \times \hat{b}) \).

3. **Condition for Coplanarity**:
- For the vectors \( \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \) to lie in the same plane, the scalar triple product must be zero:
\[ \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) = 0 \]

4. **Calculating the Cross Product**:
- We need to compute \( \mathbf{v_2} \times \mathbf{v_3} \). This involves using the determinant of a matrix formed by the components of the vectors.

5. **Setting Up the Determinant**:
- The coefficients of \( \hat{a}, \hat{b}, \hat{a} \times \hat{b} \) in \( \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \) can be expressed as:
- For \( \mathbf{v_1} \): Coefficients are \( (x, x, z) \)
- For \( \mathbf{v_2} \): Coefficients are \( (2, 1, 0) \)
- For \( \mathbf{v_3} \): Coefficients are \( (z, z, y) \)

6. **Forming the Determinant**:
- The determinant can be set up as follows:
\[ \begin{vmatrix} x & x & z \\ 2 & 1 & 0 \\ z & z & y \end{vmatrix} = 0 \]

7. **Calculating the Determinant**:
- Expanding the determinant:
\[ x(1 \cdot y - 0 \cdot z) - x(2 \cdot y - 0 \cdot z) + z(2z - 1x) = 0 \]
- Simplifying gives:
\[ xy - 2xy + 2z^2 - xz = 0 \]
- Rearranging leads to:
\[ 2z^2 - xz - xy = 0 \]

8. **Solving the Quadratic Equation**:
- This is a quadratic equation in \( z \):
\[ 2z^2 - xz - xy = 0 \]
- Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{x \pm \sqrt{x^2 + 8xy}}{4} \]

9. **Finding the Value of \( z \)**:
- The geometric mean of \( x \) and \( y \) is \( z = \sqrt{xy} \).

### Final Answer:
Thus, \( z \) is the geometric mean of \( x \) and \( y \):
\[ z = \sqrt{xy} \]