If the vectors `hat(i) - 2x hat(j) + 3 y hat(k) ` and `hat(i) +2x hat(j) - y hat(k)` are perpendicular, then the locus of (x,y) is

Options

A. a circle
B. on ellipse
C. a hyperbola
D. none of these

Answer

To solve the problem, we need to find the locus of points \((x, y)\) such that the given vectors are perpendicular. Here’s the step-by-step solution:

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

### Step 2: Use the condition for perpendicular vectors
Two vectors are perpendicular 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 \(\mathbf{A} \cdot \mathbf{B}\) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (1)(1) + (-2x)(2x) + (3y)(-y) \]
\[ = 1 - 4x^2 - 3y^2 \]

### Step 4: Set the dot product to zero
Since the vectors are perpendicular, we set the dot product equal to zero:
\[ 1 - 4x^2 - 3y^2 = 0 \]

### Step 5: Rearrange the equation
Rearranging the equation gives:
\[ 4x^2 + 3y^2 = 1 \]

### Step 6: Identify the locus
The equation \(4x^2 + 3y^2 = 1\) represents an ellipse in the Cartesian plane.

### Conclusion
Thus, the locus of the points \((x, y)\) such that the vectors are perpendicular is an ellipse.

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