`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)/(sqrt3)` is given by :

Options

A. `2hati - 3hatj + 2hatk`
B. `4hati -7hatj +4hatk`
C. `4hati -2hatj +2hatk`
D. none

Answer

To solve the problem step by step, we need to find a vector \( \mathbf{R} \) that lies in the plane of vectors \( \mathbf{a} \) and \( \mathbf{b} \) and has a projection on vector \( \mathbf{c} \) of magnitude \( \frac{1}{\sqrt{3}} \).

### 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 \( \mathbf{R} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \)
Since \( \mathbf{R} \) lies in the plane of \( \mathbf{a} \) and \( \mathbf{b} \), we can express it as:
\[ \mathbf{R} = \lambda \mathbf{a} + \mu \mathbf{b} \]
Substituting the vectors:
\[ \mathbf{R} = \lambda (\hat{i} + \hat{j} - \hat{k}) + \mu (\hat{i} - 2\hat{j} + \hat{k}) \]
This simplifies to:
\[ \mathbf{R} = (\lambda + \mu) \hat{i} + (\lambda - 2\mu) \hat{j} + (-\lambda + \mu) \hat{k} \]

### Step 3: Calculate the Projection of \( \mathbf{R} \) on \( \mathbf{c} \)
The projection of \( \mathbf{R} \) on \( \mathbf{c} \) is given by:
\[ \text{Projection} = \frac{\mathbf{R} \cdot \mathbf{c}}{|\mathbf{c}|} \]
First, we need to find \( |\mathbf{c}| \):
\[ |\mathbf{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]

Now, we compute \( \mathbf{R} \cdot \mathbf{c} \):
\[ \mathbf{R} \cdot \mathbf{c} = (\lambda + \mu)(1) + (\lambda - 2\mu)(-1) + (-\lambda + \mu)(-1) \]
This simplifies to:
\[ \mathbf{R} \cdot \mathbf{c} = \lambda + \mu - \lambda + 2\mu + \lambda - \mu = 2\mu + \lambda \]

### Step 4: Set Up the Equation for the Projection Magnitude
We know the magnitude of the projection is \( \frac{1}{\sqrt{3}} \):
\[ \frac{2\mu + \lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Multiplying both sides by \( \sqrt{3} \):
\[ 2\mu + \lambda = 1 \]

### Step 5: Solve for \( \lambda \) in terms of \( \mu \)
From the equation \( 2\mu + \lambda = 1 \):
\[ \lambda = 1 - 2\mu \]

### Step 6: Substitute \( \lambda \) back into \( \mathbf{R} \)
Substituting \( \lambda \) into \( \mathbf{R} \):
\[ \mathbf{R} = (1 - 2\mu + \mu) \hat{i} + (1 - 2\mu - 2\mu) \hat{j} + (-(1 - 2\mu) + \mu) \hat{k} \]
This simplifies to:
\[ \mathbf{R} = (1 - \mu) \hat{i} + (1 - 4\mu) \hat{j} + (3\mu - 1) \hat{k} \]

### Step 7: Choose a Value for \( \mu \)
Let’s choose \( \mu = 0 \):
\[ \mathbf{R} = 1 \hat{i} + 1 \hat{j} - 1 \hat{k} \]
This gives:
\[ \mathbf{R} = \hat{i} + \hat{j} - \hat{k} \]

### Step 8: Verify the Projection
Now we check if this \( \mathbf{R} \) satisfies the projection condition:
\[ \mathbf{R} \cdot \mathbf{c} = 1 - 1 + 1 = 1 \]
The magnitude of the projection:
\[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Thus, the projection condition is satisfied.

### Final Answer
The vector \( \mathbf{R} \) in the plane of \( \mathbf{a} \) and \( \mathbf{b} \) whose projection on \( \mathbf{c} \) has a magnitude of \( \frac{1}{\sqrt{3}} \) is:
\[ \mathbf{R} = \hat{i} + \hat{j} - \hat{k} \]