Let `veca=hati + hatj +hatk,vecb=hati- hatj + hatk and vecc= hati-hatj - hatk` be three vectors. A vectors `vecv` in the plane of `veca and vecb` , whose projection on `vecc is 1/sqrt3` is given by

Options

A. `hati-3hatj + 3hatk`
B. `-3hati-3hatj +hatk `
C. `3hati -hatj + 3hatk`
D. `hati+ 3hatj -3hatk`

Explanation

To solve the problem, we need to find a vector \( \vec{v} \) in the plane of vectors \( \vec{a} \) and \( \vec{b} \) such that its projection onto vector \( \vec{c} \) is \( \frac{1}{\sqrt{3}} \).

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{c} = \hat{i} - \hat{j} - \hat{k} \]

### Step 2: Express \( \vec{v} \) in terms of \( \vec{a} \) and \( \vec{b} \)
Since \( \vec{v} \) is in the plane of \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{v} \) as a linear combination of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{v} = \mu \vec{a} + \lambda \vec{b} \]
Substituting the values of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{v} = \mu (\hat{i} + \hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \]
\[ \vec{v} = (\mu + \lambda) \hat{i} + (\mu - \lambda) \hat{j} + (\mu + \lambda) \hat{k} \]

### Step 3: Calculate the projection of \( \vec{v} \) onto \( \vec{c} \)
The projection of \( \vec{v} \) onto \( \vec{c} \) is given by:
\[ \text{Projection} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} \]
We need this projection to equal \( \frac{1}{\sqrt{3}} \).

### Step 4: Calculate \( |\vec{c}| \)
First, we calculate the magnitude of \( \vec{c} \):
\[ |\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]

### Step 5: Set up the equation
Now, we set up the equation:
\[ \vec{v} \cdot \vec{c} = 1 \]
Calculating \( \vec{v} \cdot \vec{c} \):
\[ \vec{v} \cdot \vec{c} = [(\mu + \lambda) \hat{i} + (\mu - \lambda) \hat{j} + (\mu + \lambda) \hat{k}] \cdot [\hat{i} - \hat{j} - \hat{k}] \]
\[ = (\mu + \lambda) \cdot 1 + (\mu - \lambda) \cdot (-1) + (\mu + \lambda) \cdot (-1) \]
\[ = (\mu + \lambda) - (\mu - \lambda) - (\mu + \lambda) \]
\[ = \mu + \lambda - \mu + \lambda - \mu - \lambda = -\mu + \lambda \]
Thus, we have:
\[ -\mu + \lambda = 1 \]

### Step 6: Solve for \( \mu \) and \( \lambda \)
From the equation \( -\mu + \lambda = 1 \), we can express \( \lambda \):
\[ \lambda = \mu + 1 \]

### Step 7: Substitute back into \( \vec{v} \)
Substituting \( \lambda \) back into the expression for \( \vec{v} \):
\[ \vec{v} = \mu \vec{a} + (\mu + 1) \vec{b} \]
\[ = \mu (\hat{i} + \hat{j} + \hat{k}) + (\mu + 1)(\hat{i} - \hat{j} + \hat{k}) \]
\[ = \mu \hat{i} + \mu \hat{j} + \mu \hat{k} + (\mu + 1) \hat{i} - (\mu + 1) \hat{j} + (\mu + 1) \hat{k} \]
\[ = (2\mu + 1) \hat{i} + (-1) \hat{j} + (2\mu + 1) \hat{k} \]

### Step 8: Identify the correct option
Now we can see that the coefficients of \( \hat{i} \) and \( \hat{k} \) are the same, and the coefficient of \( \hat{j} \) is different. We need to check the options provided.

After checking the options, we find that:
\[ \vec{v} = 3 \hat{i} - \hat{j} + 3 \hat{k} \]
is the correct vector.

### Final Answer:
\[ \vec{v} = 3 \hat{i} - \hat{j} + 3 \hat{k} \]