A vector of magnitude 4 which is equally inclined to the vectors `hati +hatj,hatj +hatk and hatk +hati`, is

Options

A. `4/sqrt3 (hati -hatj -hatk)`
B. ` 4/sqrt3 (hati +hatj -hatk)`
C. ` 4/sqrt3 (hati +hatj +hatk)`
D. `4/sqrt3 (hati +hatj -hatk)`

Explanation

To solve the problem of finding a vector of magnitude 4 that is equally inclined to the vectors \( \hat{i} + \hat{j} \), \( \hat{j} + \hat{k} \), and \( \hat{k} + \hat{i} \), we will follow these steps:

### Step 1: Define the Vector
Let the vector be represented as:
\[ \mathbf{r} = x \hat{i} + y \hat{j} + z \hat{k} \]
Given that the magnitude of this vector is 4, we can express this as:
\[ |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2} = 4 \]
Squaring both sides gives:
\[ x^2 + y^2 + z^2 = 16 \quad \text{(Equation 1)} \]

### Step 2: Equally Inclined Condition
The vector \( \mathbf{r} \) is equally inclined to the three vectors \( \hat{i} + \hat{j} \), \( \hat{j} + \hat{k} \), and \( \hat{k} + \hat{i} \). This means that the angles between \( \mathbf{r} \) and each of these vectors are the same.

Using the property of the dot product, we can express the cosine of the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) as:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]
Since the angles are equal, we can set up the following equations:
\[ \frac{\mathbf{r} \cdot (\hat{i} + \hat{j})}{|\mathbf{r}| |\hat{i} + \hat{j}|} = \frac{\mathbf{r} \cdot (\hat{j} + \hat{k})}{|\mathbf{r}| |\hat{j} + \hat{k}|} = \frac{\mathbf{r} \cdot (\hat{k} + \hat{i})}{|\mathbf{r}| |\hat{k} + \hat{i}|} \]

### Step 3: Calculate Dot Products
Now we calculate the dot products:
1. For \( \hat{i} + \hat{j} \):
\[ \mathbf{r} \cdot (\hat{i} + \hat{j}) = x + y \]
The magnitude \( |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

2. For \( \hat{j} + \hat{k} \):
\[ \mathbf{r} \cdot (\hat{j} + \hat{k}) = y + z \]
The magnitude \( |\hat{j} + \hat{k}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

3. For \( \hat{k} + \hat{i} \):
\[ \mathbf{r} \cdot (\hat{k} + \hat{i}) = z + x \]
The magnitude \( |\hat{k} + \hat{i}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

### Step 4: Set Up the Equations
From the equal angles condition, we have:
\[ \frac{x + y}{4 \sqrt{2}} = \frac{y + z}{4 \sqrt{2}} = \frac{z + x}{4 \sqrt{2}} \]
This simplifies to:
\[ x + y = y + z \quad \text{(Equation 2)} \]
\[ y + z = z + x \quad \text{(Equation 3)} \]
\[ z + x = x + y \quad \text{(Equation 4)} \]

### Step 5: Solve the System of Equations
From Equation 2:
\[ x + y = y + z \implies x = z \]
From Equation 3:
\[ y + z = z + x \implies y = x \]
From Equation 4:
\[ z + x = x + y \implies z = y \]
Thus, we find that:
\[ x = y = z \]

### Step 6: Substitute Back
Substituting \( x = y = z \) into Equation 1:
\[ x^2 + x^2 + x^2 = 16 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}} \]
Since \( x = y = z \), we have:
\[ \mathbf{r} = \pm \frac{4}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]

### Final Answer
The vector of magnitude 4 that is equally inclined to the given vectors is:
\[ \mathbf{r} = \pm \frac{4}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]