What is the vector equally inclined to the vectors `hat(i)+3hat(j) and 3hat(i)+hat(j)`?

Options

A. `hat(i)+hat(j)`
B. `2hat(i)-hat(j)`
C. `2hat(i)+hat(j)`
D. None of these

Explanation

To find the vector that is equally inclined to the vectors \(\hat{i} + 3\hat{j}\) and \(3\hat{i} + \hat{j}\), we can follow these steps:

### Step 1: Define the Vectors
Let:
- Vector \(A = \hat{i} + 3\hat{j}\)
- Vector \(B = 3\hat{i} + \hat{j}\)

### Step 2: Calculate the Magnitudes of the Vectors
The magnitude of vector \(A\) is calculated as follows:
\[ |A| = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

The magnitude of vector \(B\) is calculated as follows:
\[ |B| = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 3: Find the Cosine of the Angles
We need to find a vector \(C\) that is equally inclined to both \(A\) and \(B\). Let’s denote vector \(C\) as \(C = x\hat{i} + y\hat{j}\).

To find the angle between vectors \(A\) and \(C\), we use the dot product:
\[ \cos(\theta_1) = \frac{A \cdot C}{|A| |C|} = \frac{(1)(x) + (3)(y)}{\sqrt{10} \sqrt{x^2 + y^2}} \]

To find the angle between vectors \(B\) and \(C\):
\[ \cos(\theta_2) = \frac{B \cdot C}{|B| |C|} = \frac{(3)(x) + (1)(y)}{\sqrt{10} \sqrt{x^2 + y^2}} \]

### Step 4: Set the Cosines Equal
Since the angles are equal, we set the cosines equal:
\[ \frac{(1)(x) + (3)(y)}{\sqrt{10} \sqrt{x^2 + y^2}} = \frac{(3)(x) + (1)(y)}{\sqrt{10} \sqrt{x^2 + y^2}} \]

### Step 5: Simplify the Equation
We can eliminate the common terms:
\[ 1x + 3y = 3x + 1y \]
Rearranging gives:
\[ 3y - 1y = 3x - 1x \implies 2y = 2x \implies y = x \]

### Step 6: Find the Vector
Since \(y = x\), we can express vector \(C\) as:
\[ C = x\hat{i} + x\hat{j} = x(\hat{i} + \hat{j}) \]
Thus, any vector of the form \(k(\hat{i} + \hat{j})\) where \(k\) is a scalar will be equally inclined to both \(A\) and \(B\).

### Conclusion
The vector that is equally inclined to the vectors \(\hat{i} + 3\hat{j}\) and \(3\hat{i} + \hat{j}\) is of the form \(k(\hat{i} + \hat{j})\).