What is the value of `lamda` for which the vectors `hat(i) - hat(j) + hat(k), 2hat(i) + hat(j)- hat(k) and lamda hat(i) - hat(j)+ lamda (k)` are coplanar

Options

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

Answer

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 follow these steps:

### Step 1: Write down the vectors
The vectors are:
1. \( \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \)
2. \( \mathbf{B} = 2\hat{i} + \hat{j} - \hat{k} \)
3. \( \mathbf{C} = \lambda \hat{i} - \hat{j} + \lambda \hat{k} \)

### Step 2: Form the matrix of coefficients
We can represent these vectors in a matrix form, where each row corresponds to the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \):

\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the value of \( \lambda \) for which these vectors are coplanar, we need to set the determinant of this matrix to zero:

\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} = 0 \]

### Step 4: Expand the determinant
Using the cofactor expansion along the first row:

\[ = 1 \cdot \begin{vmatrix} 1 & -1 \\ -1 & \lambda \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & -1 \\ \lambda & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ \lambda & -1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ -1 & \lambda \end{vmatrix} = 1 \cdot \lambda - (-1) \cdot (-1) = \lambda - 1 \)

2. \( \begin{vmatrix} 2 & -1 \\ \lambda & \lambda \end{vmatrix} = 2 \cdot \lambda - (-1) \cdot \lambda = 2\lambda + \lambda = 3\lambda \)

3. \( \begin{vmatrix} 2 & 1 \\ \lambda & -1 \end{vmatrix} = 2 \cdot (-1) - 1 \cdot \lambda = -2 - \lambda \)

Putting it all together, we have:

\[ (\lambda - 1) + 3\lambda + (-2 - \lambda) = 0 \]

### Step 5: Simplify the equation
Combine like terms:

\[ \lambda - 1 + 3\lambda - 2 - \lambda = 0 \]
\[ (1 + 3 - 1)\lambda - 3 = 0 \]
\[ 3\lambda - 3 = 0 \]

### Step 6: Solve for \( \lambda \)
Now, isolate \( \lambda \):

\[ 3\lambda = 3 \implies \lambda = 1 \]

Thus, the value of \( \lambda \) for which the vectors are coplanar is \( \lambda = 1 \).