If `veca=2hati+lambda hatj+3hatk, vecb=3hati+3hatj+5hatk, vec c=lambda hati+2hatj+2hatk` are linearly dependent vectors, then the number of possible values of `lambda` is :
i) 0
ii)1
iii)2
iv)More than 2

Options

A. 0
B. 1
C. 2
D. More than 2

Answer

To determine the number of possible values of \( \lambda \) such that the vectors \( \vec{a} = 2\hat{i} + \lambda \hat{j} + 3\hat{k} \), \( \vec{b} = 3\hat{i} + 3\hat{j} + 5\hat{k} \), and \( \vec{c} = \lambda \hat{i} + 2\hat{j} + 2\hat{k} \) are linearly dependent, we can follow these steps:

### Step 1: Set up the matrix
We can represent the vectors as rows of a matrix \( M \):
\[ M = \begin{bmatrix} 2 & \lambda & 3 \\ 3 & 3 & 5 \\ \lambda & 2 & 2 \end{bmatrix} \]

### Step 2: Calculate the determinant
For the vectors to be linearly dependent, the determinant of the matrix \( M \) must be zero:
\[ \text{det}(M) = 0 \]

Calculating the determinant:
\[ \text{det}(M) = 2 \begin{vmatrix} 3 & 5 \\ 2 & 2 \end{vmatrix} - \lambda \begin{vmatrix} 3 & 5 \\ \lambda & 2 \end{vmatrix} + 3 \begin{vmatrix} 3 & 3 \\ \lambda & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 3 & 5 \\ 2 & 2 \end{vmatrix} = (3 \cdot 2) - (5 \cdot 2) = 6 - 10 = -4 \)
2. \( \begin{vmatrix} 3 & 5 \\ \lambda & 2 \end{vmatrix} = (3 \cdot 2) - (5 \cdot \lambda) = 6 - 5\lambda \)
3. \( \begin{vmatrix} 3 & 3 \\ \lambda & 2 \end{vmatrix} = (3 \cdot 2) - (3 \cdot \lambda) = 6 - 3\lambda \)

Substituting these into the determinant:
\[ \text{det}(M) = 2(-4) - \lambda(6 - 5\lambda) + 3(6 - 3\lambda) \]
\[ = -8 - \lambda(6 - 5\lambda) + 18 - 9\lambda \]
\[ = -8 + 18 - 6\lambda + 5\lambda^2 - 9\lambda \]
\[ = 10 + 5\lambda^2 - 15\lambda \]

### Step 3: Set the determinant to zero
Now, we set the determinant to zero:
\[ 5\lambda^2 - 15\lambda + 10 = 0 \]

### Step 4: Simplify the equation
Dividing the entire equation by 5:
\[ \lambda^2 - 3\lambda + 2 = 0 \]

### Step 5: Factor the quadratic equation
Factoring the quadratic:
\[ (\lambda - 2)(\lambda - 1) = 0 \]

### Step 6: Solve for \( \lambda \)
Setting each factor to zero gives us:
\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]
\[ \lambda - 1 = 0 \quad \Rightarrow \quad \lambda = 1 \]

### Conclusion
The possible values of \( \lambda \) are \( 1 \) and \( 2 \). Therefore, the number of possible values of \( \lambda \) is **2**.