A vector of magnitude `sqrt2` coplanar with the vectors `veca=hati+hatj+2hatk and vecb = hati + hatj + hatk,` and perpendicular to the vector `vecc = hati + hatj +hatk` is

Options

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

Explanation

To find a vector \( \vec{d} \) of magnitude \( \sqrt{2} \) that is coplanar with the vectors \( \vec{a} \) and \( \vec{b} \), and perpendicular to the vector \( \vec{c} \), we can follow these steps:

### Step 1: Define the vectors
Given:
- \( \vec{a} = \hat{i} + \hat{j} + 2\hat{k} \)
- \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Find the cross product of \( \vec{a} \) and \( \vec{b} \)
To find a vector that is coplanar with \( \vec{a} \) and \( \vec{b} \), we can use the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 1 - 2 \cdot 1) - \hat{j} (1 \cdot 1 - 2 \cdot 1) + \hat{k} (1 \cdot 1 - 1 \cdot 1) \]
\[ = \hat{i} (1 - 2) - \hat{j} (1 - 2) + \hat{k} (0) \]
\[ = -\hat{i} + \hat{j} \]

### Step 3: Find a vector perpendicular to \( \vec{c} \)
To find a vector \( \vec{d} \) that is perpendicular to \( \vec{c} \), we can express \( \vec{d} \) as:
\[ \vec{d} = k(-\hat{i} + \hat{j}) \quad \text{for some scalar } k \]

### Step 4: Ensure \( \vec{d} \) has a magnitude of \( \sqrt{2} \)
The magnitude of \( \vec{d} \) is given by:
\[ |\vec{d}| = |k| \sqrt{(-1)^2 + 1^2} = |k| \sqrt{2} \]
Setting this equal to \( \sqrt{2} \):
\[ |k| \sqrt{2} = \sqrt{2} \]
Thus, \( |k| = 1 \), which gives us \( k = 1 \) or \( k = -1 \).

### Step 5: Write the final vectors
This gives us two possible vectors:
1. \( \vec{d} = -\hat{i} + \hat{j} \)
2. \( \vec{d} = \hat{i} - \hat{j} \)

Both vectors are coplanar with \( \vec{a} \) and \( \vec{b} \) and perpendicular to \( \vec{c} \).

### Conclusion
The vector \( \vec{d} \) of magnitude \( \sqrt{2} \) that is coplanar with \( \vec{a} \) and \( \vec{b} \), and perpendicular to \( \vec{c} \) can be either:
- \( \vec{d} = -\hat{i} + \hat{j} \) or
- \( \vec{d} = \hat{i} - \hat{j} \)