Vectors perpendicular to`hati-hatj-hatk` and in the plane of `hati+hatj+hatk and -hati+hatj+hatk` are (A) `hati+hatk` (B) `2hati+hatj+hatk` (C) `3hati+2hatj+hatk` (D) `-4hati-2hatj-2hatk`

Options

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

Explanation

To solve the problem of finding vectors that are perpendicular to the vector \(\hat{i} - \hat{j} - \hat{k}\) and lie in the plane defined by the vectors \(\hat{i} + \hat{j} + \hat{k}\) and \(-\hat{i} + \hat{j} + \hat{k}\), we will follow these steps:

### Step 1: Define the vectors
Let:
- \(\alpha = \hat{i} - \hat{j} - \hat{k}\)
- \(\beta = \hat{i} + \hat{j} + \hat{k}\)
- \(\gamma = -\hat{i} + \hat{j} + \hat{k}\)

We are looking for a vector \(\mathbf{v} = x \hat{i} + y \hat{j} + z \hat{k}\) that is perpendicular to \(\alpha\) and lies in the plane formed by \(\beta\) and \(\gamma\).

### Step 2: Condition for coplanarity
The vectors \(\alpha\), \(\beta\), and \(\gamma\) are coplanar if the determinant of the matrix formed by their coefficients is zero:

\[ \begin{vmatrix} 1 & -1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand along the first row:

\[ = 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ = 1(1 \cdot 1 - 1 \cdot 1) + 1(1 \cdot 1 - (-1) \cdot 1) + 1(1 \cdot 1 - (-1) \cdot 1) \]
\[ = 0 + 2 + 2 = 4 \neq 0 \]

This indicates that the vectors are not coplanar. However, we need to find the condition for \(\mathbf{v}\) to be in the plane of \(\beta\) and \(\gamma\).

### Step 4: Find the condition for \(\mathbf{v}\)
For \(\mathbf{v}\) to lie in the plane of \(\beta\) and \(\gamma\), we can express \(\mathbf{v}\) as a linear combination of \(\beta\) and \(\gamma\):

\[ \mathbf{v} = a(\hat{i} + \hat{j} + \hat{k}) + b(-\hat{i} + \hat{j} + \hat{k}) \]

This gives us:

\[ \mathbf{v} = (a - b)\hat{i} + (a + b)\hat{j} + (a + b)\hat{k} \]

### Step 5: Perpendicularity condition
The vector \(\mathbf{v}\) must also satisfy the condition of being perpendicular to \(\alpha\):

\[ \mathbf{v} \cdot \alpha = 0 \]

Calculating the dot product:

\[ (x, y, z) \cdot (1, -1, -1) = x - y - z = 0 \]

### Step 6: Solve the equations
From the perpendicularity condition, we have:

\[ x = y + z \]

Substituting \(x = y + z\) into the expression for \(\mathbf{v}\):

\[ \mathbf{v} = (y + z - b)\hat{i} + (y + b)\hat{j} + (y + b)\hat{k} \]

### Step 7: Check the options
Now we check the options given in the question to see which ones satisfy both conditions:

1. **Option A:** \(\hat{i} + \hat{k}\) → Not satisfying.
2. **Option B:** \(2\hat{i} + \hat{j} + \hat{k}\) → Satisfies.
3. **Option C:** \(3\hat{i} + 2\hat{j} + \hat{k}\) → Not satisfying.
4. **Option D:** \(-4\hat{i} - 2\hat{j} - 2\hat{k}\) → Satisfies.

Thus, the correct options are B and D.