`a =hati+hatj-hatk,b= hati-2hatj+hatk,c=hati-hatj-hatk`, then a vector in plane of a and b whose projection on c is of magnitude `((1)/(sqrt(3)))` is given by :

Options

A. `2hati-3hatj+2hatk`
B. `4hati-7hatj+4hatk`
C. `4hati-2hatj+2hatk`
D. `4hati+13hatj-10hatk`

Answer

To solve the problem step by step, we will find a vector \( \mathbf{U} \) in the plane of vectors \( \mathbf{A} \) and \( \mathbf{B} \) whose projection on vector \( \mathbf{C} \) has a specific magnitude.

### Step 1: Define the vectors
Given:
\[ \mathbf{A} = \hat{i} + \hat{j} - \hat{k} \]
\[ \mathbf{B} = \hat{i} - 2\hat{j} + \hat{k} \]
\[ \mathbf{C} = \hat{i} - \hat{j} - \hat{k} \]

### Step 2: Express vector \( \mathbf{U} \)
Since \( \mathbf{U} \) is in the plane of \( \mathbf{A} \) and \( \mathbf{B} \), we can express it as:
\[ \mathbf{U} = \mu \mathbf{A} + \lambda \mathbf{B} \]
Substituting the values of \( \mathbf{A} \) and \( \mathbf{B} \):
\[ \mathbf{U} = \mu (\hat{i} + \hat{j} - \hat{k}) + \lambda (\hat{i} - 2\hat{j} + \hat{k}) \]
This simplifies to:
\[ \mathbf{U} = (\mu + \lambda) \hat{i} + (\mu - 2\lambda) \hat{j} + (\mu + \lambda) \hat{k} \]

### Step 3: Find the projection of \( \mathbf{U} \) on \( \mathbf{C} \)
The projection of \( \mathbf{U} \) on \( \mathbf{C} \) is given by:
\[ \text{Projection of } \mathbf{U} \text{ on } \mathbf{C} = \frac{\mathbf{U} \cdot \mathbf{C}}{|\mathbf{C}|} \]
We need to find \( |\mathbf{C}| \):
\[ |\mathbf{C}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]
Thus, the projection becomes:
\[ \frac{\mathbf{U} \cdot \mathbf{C}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]
This implies:
\[ \mathbf{U} \cdot \mathbf{C} = 1 \]

### Step 4: Calculate \( \mathbf{U} \cdot \mathbf{C} \)
Calculating the dot product:
\[ \mathbf{U} \cdot \mathbf{C} = (\mu + \lambda) \cdot 1 + (\mu - 2\lambda) \cdot (-1) + (\mu + \lambda) \cdot (-1) \]
This simplifies to:
\[ \mathbf{U} \cdot \mathbf{C} = \mu + \lambda - \mu + 2\lambda - \mu - \lambda = -\mu + 2\lambda \]
Setting this equal to 1:
\[ -\mu + 2\lambda = 1 \]

### Step 5: Solve for \( \mu \) and \( \lambda \)
Rearranging gives:
\[ \mu = 2\lambda - 1 \]

### Step 6: Substitute \( \mu \) into \( \mathbf{U} \)
Substituting \( \mu \) back into \( \mathbf{U} \):
\[ \mathbf{U} = (2\lambda - 1 + \lambda) \hat{i} + (2\lambda - 1 - 2\lambda) \hat{j} + (2\lambda - 1 + \lambda) \hat{k} \]
This simplifies to:
\[ \mathbf{U} = (3\lambda - 1)\hat{i} + (-1)\hat{j} + (3\lambda - 1)\hat{k} \]

### Step 7: Set conditions for \( \mathbf{U} \)
We can set conditions for \( \mathbf{U} \) to find specific values for \( \lambda \). For example, we can set \( 3\lambda - 1 = 2 \) or \( 3\lambda - 1 = 4 \).

### Step 8: Solve for \( \lambda \)
1. If \( 3\lambda - 1 = 2 \):
\[ 3\lambda = 3 \implies \lambda = 1 \]
Substituting back gives:
\[ \mathbf{U} = (3(1) - 1)\hat{i} + (-1)\hat{j} + (3(1) - 1)\hat{k} = 2\hat{i} - \hat{j} + 2\hat{k} \]

2. If \( 3\lambda - 1 = 4 \):
\[ 3\lambda = 5 \implies \lambda = \frac{5}{3} \]
Substituting back gives:
\[ \mathbf{U} = (3(\frac{5}{3}) - 1)\hat{i} + (-1)\hat{j} + (3(\frac{5}{3}) - 1)\hat{k} = 4\hat{i} - \hat{j} + 4\hat{k} \]

### Final Result
Thus, the vector \( \mathbf{U} \) can be either:
\[ \mathbf{U} = 2\hat{i} - \hat{j} + 2\hat{k} \quad \text{or} \quad \mathbf{U} = 4\hat{i} - \hat{j} + 4\hat{k} \]