The vectors `hat(i)- x hat(j)- y k and hat(i) + x hat(j) + y hat(k)` are orthogonal to reach 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

Answer

To find the locus of the point (x, y) given 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, we can follow these steps:

### Step 1: Define the vectors
Let:
- \(\mathbf{A} = \hat{i} - x \hat{j} - y \hat{k}\)
- \(\mathbf{B} = \hat{i} + x \hat{j} + y \hat{k}\)

### Step 2: Use the condition for orthogonality
Two vectors are orthogonal if their dot product is zero. Therefore, we need to find:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]

### Step 3: Calculate the dot product
Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} - x \hat{j} - y \hat{k}) \cdot (\hat{i} + x \hat{j} + y \hat{k}) \]
Using the properties of the dot product:
\[ = \hat{i} \cdot \hat{i} + \hat{i} \cdot (x \hat{j}) + \hat{i} \cdot (y \hat{k}) - x \hat{j} \cdot \hat{i} - x \hat{j} \cdot (x \hat{j}) - x \hat{j} \cdot (y \hat{k}) - y \hat{k} \cdot \hat{i} - y \hat{k} \cdot (x \hat{j}) - y \hat{k} \cdot (y \hat{k}) \]
Since the dot products of different unit vectors are zero and the dot product of a unit vector with itself is one, we simplify:
\[ = 1 - x^2 - y^2 \]

### Step 4: Set the dot product equal to zero
Setting the dot product equal to zero gives:
\[ 1 - x^2 - y^2 = 0 \]

### Step 5: Rearranging the equation
Rearranging the equation, we get:
\[ x^2 + y^2 = 1 \]

### Step 6: Identify the locus
The equation \(x^2 + y^2 = 1\) represents a circle centered at the origin (0, 0) with a radius of 1.

### Conclusion
Thus, the locus of the point (x, y) is a circle with radius 1 centered at the origin.

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