A vector coplanar with vectors `hati + hatj and hat j + hatk ` and parallel to the vector `2hati -2 hatj - 4 hatk , ` is

Options

A. `hati - hatk `
B. `hati - hatj - 2hatk `
C. `hati + hatj - hatk `
D. `3 hati + 3 hatj - 6 hatk `

Explanation

To solve the problem step by step, we need to find a vector that is coplanar with the vectors \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \), and is also parallel to the vector \( 2\hat{i} - 2\hat{j} - 4\hat{k} \).

### Step 1: Identify the given vectors
The vectors we have are:
1. \( \mathbf{a_1} = \hat{i} + \hat{j} \)
2. \( \mathbf{a_2} = \hat{j} + \hat{k} \)
3. \( \mathbf{b} = 2\hat{i} - 2\hat{j} - 4\hat{k} \)

### Step 2: Express the required vector
Let the required vector be \( \mathbf{a} = x\hat{i} + y\hat{j} + z\hat{k} \). We need to find \( x, y, z \) such that \( \mathbf{a} \) is coplanar with \( \mathbf{a_1} \) and \( \mathbf{a_2} \), and also parallel to \( \mathbf{b} \).

### Step 3: Condition for coplanarity
Vectors \( \mathbf{a_1}, \mathbf{a_2}, \mathbf{a} \) are coplanar if the scalar triple product is zero:
\[ \mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a}) = 0 \]

### Step 4: Calculate the cross product \( \mathbf{a_2} \times \mathbf{a} \)
First, we need to calculate \( \mathbf{a_2} \times \mathbf{a} \):
\[ \mathbf{a_2} = \hat{j} + \hat{k} \quad \text{and} \quad \mathbf{a} = x\hat{i} + y\hat{j} + z\hat{k} \]
Using the determinant method for the cross product:
\[ \mathbf{a_2} \times \mathbf{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{i}(1 \cdot z - 1 \cdot y) - \hat{j}(0 \cdot z - 1 \cdot x) + \hat{k}(0 \cdot y - 1 \cdot x) \]
This simplifies to:
\[ \mathbf{a_2} \times \mathbf{a} = (z - y)\hat{i} + x\hat{j} - x\hat{k} \]

### Step 5: Calculate the dot product with \( \mathbf{a_1} \)
Now, we calculate the dot product:
\[ \mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a}) = (\hat{i} + \hat{j}) \cdot ((z - y)\hat{i} + x\hat{j} - x\hat{k}) \]
This gives:
\[ 1(z - y) + 1(x) + 0 = z - y + x \]
Setting this equal to zero for coplanarity:
\[ z - y + x = 0 \quad \text{(1)} \]

### Step 6: Condition for parallelism
The vector \( \mathbf{a} \) is parallel to \( \mathbf{b} \) if:
\[ \mathbf{a} = \lambda \mathbf{b} \quad \text{for some scalar } \lambda \]
Thus,
\[ x = 2\lambda, \quad y = -2\lambda, \quad z = -4\lambda \]

### Step 7: Substitute into the coplanarity condition
Substituting \( x, y, z \) into equation (1):
\[ -4\lambda - (-2\lambda) + 2\lambda = 0 \]
This simplifies to:
\[ -4\lambda + 2\lambda + 2\lambda = 0 \implies 0 = 0 \]
This is always true, confirming that any \( \lambda \) will satisfy the conditions.

### Step 8: Final expression for the vector
Thus, the required vector can be expressed as:
\[ \mathbf{a} = 2\lambda \hat{i} - 2\lambda \hat{j} - 4\lambda \hat{k} \]
Factoring out \( \lambda \):
\[ \mathbf{a} = \lambda (2\hat{i} - 2\hat{j} - 4\hat{k}) \]
where \( \lambda \) can be any real number.

### Conclusion
The required vector is:
\[ \mathbf{a} = 2\hat{i} - 2\hat{j} - 4\hat{k} \]