The vectors `lambdahati + hatj + 2hatk, hati + lambdahatj +hatk, 2hati - hatj + 2hatk` are coplanar, if:

Options

A. `lambda =2`
B. `lambda=0`
C. `lambda=1`
D. `lambda =-1`

Answer

To determine the value of \( \lambda \) for which the vectors \( \mathbf{a} = \lambda \hat{i} + \hat{j} + 2\hat{k} \), \( \mathbf{b} = \hat{i} + \lambda \hat{j} + \hat{k} \), and \( \mathbf{c} = 2\hat{i} - \hat{j} + 2\hat{k} \) are coplanar, we will use the condition that the scalar triple product of the vectors must be zero.

### Step-by-step Solution:

1. **Write the Vectors**:
\[ \mathbf{a} = \lambda \hat{i} + \hat{j} + 2\hat{k} \]
\[ \mathbf{b} = \hat{i} + \lambda \hat{j} + \hat{k} \]
\[ \mathbf{c} = 2\hat{i} - \hat{j} + 2\hat{k} \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) must equal zero for the vectors to be coplanar. This can be calculated using the determinant of a matrix formed by the components of the vectors:
\[ \begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & 1 \\ 2 & -1 & 2 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We will calculate the determinant using the formula for a 3x3 matrix:
\[ D = \lambda \begin{vmatrix} \lambda & 1 \\ -1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} + 2 \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} \]

- Calculate the first determinant:
\[ \begin{vmatrix} \lambda & 1 \\ -1 & 2 \end{vmatrix} = \lambda \cdot 2 - 1 \cdot (-1) = 2\lambda + 1 \]

- Calculate the second determinant:
\[ \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 2 = 0 \]

- Calculate the third determinant:
\[ \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) - \lambda \cdot 2 = -1 - 2\lambda \]

Putting it all together:
\[ D = \lambda(2\lambda + 1) - 1(0) + 2(-1 - 2\lambda) \]
\[ D = \lambda(2\lambda + 1) - 2 - 4\lambda \]
\[ D = 2\lambda^2 + \lambda - 2 - 4\lambda \]
\[ D = 2\lambda^2 - 3\lambda - 2 \]

4. **Set the Determinant to Zero**:
\[ 2\lambda^2 - 3\lambda - 2 = 0 \]

5. **Solve the Quadratic Equation**:
We can use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = -2 \):
\[ \lambda = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ \lambda = \frac{3 \pm \sqrt{9 + 16}}{4} \]
\[ \lambda = \frac{3 \pm 5}{4} \]
This gives us two solutions:
\[ \lambda = \frac{8}{4} = 2 \quad \text{and} \quad \lambda = \frac{-2}{4} = -\frac{1}{2} \]

### Final Answer:
The vectors are coplanar if \( \lambda = 2 \) or \( \lambda = -\frac{1}{2} \).