Let `a=hat(i)+2hat(j)+hat(k), b=hat(i)-hat(j)+hat(k), c=hat(i)+hat(j)-hat(k)`. A vector coplanar to a and b has a projection along c of magnitude `(1)/(sqrt(3))`, then the vector is

Options

A. `4hat(i)-hat(j)+4hat(k)`
B. `4hat(i)+hat(j)-4hat(k)`
C. `2hat(i)+hat(j)+hat(k)`
D. None of these

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and elaborate on each step.

### Step 1: Define the vectors
We have three vectors defined as:
- \( \mathbf{a} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \)
- \( \mathbf{c} = \hat{i} + \hat{j} - \hat{k} \)

### Step 2: Express the coplanar vector
Let \( \mathbf{R} \) be the vector that is coplanar to \( \mathbf{a} \) and \( \mathbf{b} \). We can express \( \mathbf{R} \) as a linear combination of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{R} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \lambda \) is a scalar.

### Step 3: Substitute the vectors
Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \) into the equation for \( \mathbf{R} \):
\[ \mathbf{R} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \]
This simplifies to:
\[ \mathbf{R} = (1 + \lambda) \hat{i} + (2 - \lambda) \hat{j} + (1 + \lambda) \hat{k} \]

### Step 4: Find the projection of \( \mathbf{R} \) along \( \mathbf{c} \)
The projection of \( \mathbf{R} \) along \( \mathbf{c} \) is given by the formula:
\[ \text{Projection of } \mathbf{R} \text{ on } \mathbf{c} = \frac{\mathbf{R} \cdot \mathbf{c}}{|\mathbf{c}|} \]
We know that this projection has a magnitude of \( \frac{1}{\sqrt{3}} \).

### Step 5: Calculate \( |\mathbf{c}| \)
First, we calculate the magnitude of \( \mathbf{c} \):
\[ |\mathbf{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]

### Step 6: Set up the equation for the projection
Thus, we have:
\[ \frac{\mathbf{R} \cdot \mathbf{c}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]
This implies:
\[ \mathbf{R} \cdot \mathbf{c} = 1 \]

### Step 7: Calculate \( \mathbf{R} \cdot \mathbf{c} \)
Now we compute the dot product \( \mathbf{R} \cdot \mathbf{c} \):
\[ \mathbf{R} \cdot \mathbf{c} = ((1 + \lambda) \hat{i} + (2 - \lambda) \hat{j} + (1 + \lambda) \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ = (1 + \lambda)(1) + (2 - \lambda)(1) + (1 + \lambda)(-1) \]
\[ = (1 + \lambda) + (2 - \lambda) - (1 + \lambda) \]
\[ = 2 \]

### Step 8: Set up the equation
Now we set the equation:
\[ 2 = 1 \]
This leads to:
\[ 1 + \lambda - 1 - \lambda = 0 \]
This simplifies to:
\[ 1 + \lambda + 2 - \lambda - 1 - \lambda = 0 \]
\[ 1 = 2 - \lambda \]
Thus:
\[ \lambda = 1 \]

### Step 9: Find \( \mathbf{R} \) for \( \lambda = 1 \)
Substituting \( \lambda = 1 \) back into the equation for \( \mathbf{R} \):
\[ \mathbf{R} = (1 + 1) \hat{i} + (2 - 1) \hat{j} + (1 + 1) \hat{k} = 2\hat{i} + 1\hat{j} + 2\hat{k} \]

### Step 10: Find \( \mathbf{R} \) for \( \lambda = 3 \)
Now substituting \( \lambda = 3 \):
\[ \mathbf{R} = (1 + 3) \hat{i} + (2 - 3) \hat{j} + (1 + 3) \hat{k} = 4\hat{i} - 1\hat{j} + 4\hat{k} \]

### Final Answer
Thus, the vectors that are coplanar to \( \mathbf{a} \) and \( \mathbf{b} \) with the specified projection along \( \mathbf{c} \) are:
1. \( \mathbf{R_1} = 2\hat{i} + \hat{j} + 2\hat{k} \)
2. \( \mathbf{R_2} = 4\hat{i} - \hat{j} + 4\hat{k} \)