Let `veca=hati+2hatj +hatk, vec=hati-hatj+hatk and vecc=hati+hatj-hatk`. A vector in the plane of `veca and vecb` whose projection on `vecc is 1/sqrt(3)` is (A) `4hati-hatj+4hatk` (B) `hati+hatj-3hatk` (C) `2hati+hatj-2hatk` (D) `4hati+hatj-4hatk`

Options

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

Explanation

To solve the problem step by step, we will find a vector \( \vec{R} \) in the plane of vectors \( \vec{A} \) and \( \vec{B} \) whose projection on vector \( \vec{C} \) is \( \frac{1}{\sqrt{3}} \).

### Step 1: Define the vectors
Given:
- \( \vec{A} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \vec{B} = \hat{i} - \hat{j} + \hat{k} \)
- \( \vec{C} = \hat{i} + \hat{j} - \hat{k} \)

### Step 2: Assume the vector \( \vec{R} \)
Since \( \vec{R} \) lies in the plane of \( \vec{A} \) and \( \vec{B} \), we can express \( \vec{R} \) as:
\[ \vec{R} = \vec{A} + \lambda \vec{B} \]
Substituting the values of \( \vec{A} \) and \( \vec{B} \):
\[ \vec{R} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \]
\[ \vec{R} = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k} \]

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

Now, we compute \( \vec{R} \cdot \vec{C} \):
\[ \vec{R} \cdot \vec{C} = ((1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ = (1 + \lambda)(1) + (2 - \lambda)(1) + (1 + \lambda)(-1) \]
\[ = (1 + \lambda) + (2 - \lambda) - (1 + \lambda) \]
\[ = 2 \]

### Step 4: Set up the equation for projection
We know that the projection is given as \( \frac{1}{\sqrt{3}} \):
\[ \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]
This means we can set up the equation:
\[ \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \implies 2 = \pm 1 \]
This gives us two cases:
1. \( 2 = 1 \) (not possible)
2. \( 2 = -1 \) (not possible)

### Step 5: Solve for \( \lambda \)
We need to solve:
\[ \vec{R} \cdot \vec{C} = \pm 1 \]
Thus:
\[ 2 = 1 \quad \text{or} \quad 2 = -1 \]
This leads to two equations:
1. \( 2 - \lambda = 1 \) → \( \lambda = 1 \)
2. \( 2 - \lambda = -1 \) → \( \lambda = 3 \)

### Step 6: Find \( \vec{R} \) for both values of \( \lambda \)
1. For \( \lambda = 1 \):
\[ \vec{R} = (1 + 1)\hat{i} + (2 - 1)\hat{j} + (1 + 1)\hat{k} = 2\hat{i} + 1\hat{j} + 2\hat{k} \]

2. For \( \lambda = 3 \):
\[ \vec{R} = (1 + 3)\hat{i} + (2 - 3)\hat{j} + (1 + 3)\hat{k} = 4\hat{i} - 1\hat{j} + 4\hat{k} \]

### Step 7: Check the options
Now we check which of these vectors matches the options given:
- \( 4\hat{i} - \hat{j} + 4\hat{k} \) corresponds to option (A).

### Final Answer
The correct answer is (A) \( 4\hat{i} - \hat{j} + 4\hat{k} \).