If the vectors `hat(i)-x hat(j)-y hat(k) and hat(i)+x hat(j)+y hat(k)` are orthogonal to each other, then what is the locus of the point (x, y) ?

Options

A. a parabola
B. an ellipse
C. a circle
D. a straight line

Explanation

To solve the problem, we need to find the locus of the point \((x, y)\) such that the vectors \(\hat{i} - x\hat{j} - y\hat{k}\) and \(\hat{i} + x\hat{j} + y\hat{k}\) are orthogonal to each other.

### Step-by-Step Solution:

1. **Identify the Vectors:**
The two vectors given are:
\[ \mathbf{A} = \hat{i} - x\hat{j} - y\hat{k} \]
\[ \mathbf{B} = \hat{i} + x\hat{j} + y\hat{k} \]

2. **Use the Orthogonality Condition:**
Two vectors are orthogonal if their dot product is zero. Therefore, we need to calculate:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]

3. **Calculate the Dot Product:**
The dot product \(\mathbf{A} \cdot \mathbf{B}\) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} - x\hat{j} - y\hat{k}) \cdot (\hat{i} + x\hat{j} + y\hat{k}) \]
Using the properties of the dot product:
\[ = \hat{i} \cdot \hat{i} + \hat{i} \cdot (x\hat{j}) + \hat{i} \cdot (y\hat{k}) - x\hat{j} \cdot \hat{i} - x\hat{j} \cdot (x\hat{j}) - x\hat{j} \cdot (y\hat{k}) - y\hat{k} \cdot \hat{i} - y\hat{k} \cdot (x\hat{j}) - y\hat{k} \cdot (y\hat{k}) \]
Simplifying this, we know that \(\hat{i} \cdot \hat{i} = 1\) and all other dot products between different unit vectors are zero:
\[ = 1 - x^2 - y^2 \]

4. **Set the Dot Product to Zero:**
To find the condition for orthogonality:
\[ 1 - x^2 - y^2 = 0 \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ x^2 + y^2 = 1 \]

6. **Interpret the Result:**
The equation \(x^2 + y^2 = 1\) represents a circle centered at the origin with a radius of 1.

### Final Answer:
The locus of the point \((x, y)\) is a circle given by the equation:
\[ x^2 + y^2 = 1 \]