It is given that the vectors`vec(a)=(2hat(i)-2hat(k)), vec(b)=hat(i)+(lambda +1)hat(j) and vec(c)=(4hat(j)+2hat(k))` are coplanar. Then, the value of `lambda` is

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. 2
D. 1

Explanation

To find the value of \( \lambda \) such that the vectors \( \vec{a} = 2\hat{i} - 2\hat{k} \), \( \vec{b} = \hat{i} + (\lambda + 1)\hat{j} \), and \( \vec{c} = 4\hat{j} + 2\hat{k} \) are coplanar, we need to use the condition that the scalar triple product of the vectors is zero.

### Step 1: Write the vectors in component form
We have:
- \( \vec{a} = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} \)
- \( \vec{b} = \begin{pmatrix} 1 \\ \lambda + 1 \\ 0 \end{pmatrix} \)
- \( \vec{c} = \begin{pmatrix} 0 \\ 4 \\ 2 \end{pmatrix} \)

### Step 2: Set up the scalar triple product
The scalar triple product can be calculated using the determinant of a matrix formed by the vectors:
\[ \text{Scalar Triple Product} = \begin{vmatrix} 2 & 0 & -2 \\ 1 & \lambda + 1 & 0 \\ 0 & 4 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
We will calculate the determinant using the formula for a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, \( d, e, f \) are the elements of the second row, and \( g, h, i \) are the elements of the third row.

Substituting the values:
\[ \text{Det} = 2 \begin{vmatrix} \lambda + 1 & 0 \\ 4 & 2 \end{vmatrix} - 0 + (-2) \begin{vmatrix} 1 & \lambda + 1 \\ 0 & 4 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} \lambda + 1 & 0 \\ 4 & 2 \end{vmatrix} = (\lambda + 1) \cdot 2 - 0 \cdot 4 = 2(\lambda + 1) \)
2. \( \begin{vmatrix} 1 & \lambda + 1 \\ 0 & 4 \end{vmatrix} = 1 \cdot 4 - 0 \cdot (\lambda + 1) = 4 \)

Thus, substituting back:
\[ \text{Det} = 2(2(\lambda + 1)) - 2(4) = 4(\lambda + 1) - 8 \]

### Step 4: Set the determinant to zero for coplanarity
For the vectors to be coplanar, we set the determinant to zero:
\[ 4(\lambda + 1) - 8 = 0 \]

### Step 5: Solve for \( \lambda \)
\[ 4(\lambda + 1) = 8 \]
\[ \lambda + 1 = 2 \]
\[ \lambda = 1 \]

### Conclusion
The value of \( \lambda \) is \( 1 \).