The vectors `hat(i)- 2x hat(j) - 3y hat(k) and hat(i) + 3x hat(j) + 2y hat(k)` are orthogonal to each other. Then, the locus of the point (x, y) is:

Options

A. Hyperbola
B. Ellipse
C. Parabola
D. Circle

Answer

To solve the problem, we need to find the locus of the point (x, y) such that the given vectors are orthogonal. Let's break it down step by step.

### Step 1: Identify the vectors
The two vectors given are:
1. \( \mathbf{A} = \hat{i} - 2x \hat{j} - 3y \hat{k} \)
2. \( \mathbf{B} = \hat{i} + 3x \hat{j} + 2y \hat{k} \)

### Step 2: Use the condition for orthogonality
Vectors are orthogonal if their dot product is zero. Therefore, we need to calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \) and set it equal to zero.

### Step 3: Calculate the dot product
The dot product of \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = (1)(1) + (-2x)(3x) + (-3y)(2y) \]
Calculating each term:
- The first term: \( 1 \cdot 1 = 1 \)
- The second term: \( -2x \cdot 3x = -6x^2 \)
- The third term: \( -3y \cdot 2y = -6y^2 \)

Putting it all together:
\[ \mathbf{A} \cdot \mathbf{B} = 1 - 6x^2 - 6y^2 \]

### Step 4: Set the dot product to zero
Now, we set the dot product equal to zero:
\[ 1 - 6x^2 - 6y^2 = 0 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 6x^2 + 6y^2 = 1 \]

### Step 6: Simplify the equation
Dividing the entire equation by 6:
\[ x^2 + y^2 = \frac{1}{6} \]

### Step 7: Identify the locus
The equation \( x^2 + y^2 = \frac{1}{6} \) represents a circle with a radius of \( \sqrt{\frac{1}{6}} \).

### Conclusion
Thus, the locus of the point (x, y) is a circle.