What is the value of `lambda ` for which the vectors `hat(i)-hat(j)+hat(k), 2 hat(i)+hat(j)-hat(k) and lambda hat(i)-hat(j)+lambda hat(k)` are co-planar?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the value of \( \lambda \) for which the vectors \( \hat{i} - \hat{j} + \hat{k} \), \( 2\hat{i} + \hat{j} - \hat{k} \), and \( \lambda \hat{i} - \hat{j} + \lambda \hat{k} \) are coplanar, we can use the determinant method. The vectors are coplanar if the determinant of the matrix formed by these vectors is equal to zero.

### Step-by-step Solution:

1. **Write the vectors in component form**:
- The first vector \( \vec{A} = \hat{i} - \hat{j} + \hat{k} \) can be represented as \( (1, -1, 1) \).
- The second vector \( \vec{B} = 2\hat{i} + \hat{j} - \hat{k} \) can be represented as \( (2, 1, -1) \).
- The third vector \( \vec{C} = \lambda \hat{i} - \hat{j} + \lambda \hat{k} \) can be represented as \( (\lambda, -1, \lambda) \).

2. **Set up the determinant**:
We need to form a matrix with these vectors as rows:
\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} \]

3. **Calculate the determinant**:
The determinant can be calculated using the formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Here, substituting the values:
\[ D = 1 \cdot \left(1 \cdot \lambda - (-1) \cdot (-1)\right) - (-1) \cdot \left(2 \cdot \lambda - (-1) \cdot (-1)\right) + 1 \cdot \left(2 \cdot (-1) - 1 \cdot \lambda\right) \]
Simplifying this gives:
\[ D = 1 \cdot (\lambda - 1) + (2\lambda - 1) + (-2 - \lambda) \]
\[ D = \lambda - 1 + 2\lambda - 1 - 2 - \lambda \]
\[ D = (1 + 2 - 1 - 2) + \lambda = 2\lambda - 4 \]

4. **Set the determinant to zero**:
For the vectors to be coplanar, we set the determinant equal to zero:
\[ 2\lambda - 4 = 0 \]

5. **Solve for \( \lambda \)**:
\[ 2\lambda = 4 \implies \lambda = 2 \]

### Final Answer:
The value of \( \lambda \) for which the vectors are coplanar is \( \lambda = 2 \).