For what value of `lambda` are the vectors `lambda hat(i)+(1 + lambda) hat(j) + (1 + 2 lambda) hat( k) and (1-lambda) hat(i)+lambda hat(j) + 2 hat(k)` perpendicular ?

Options

A. `-1//3`
B. `1//3`
C. `2//3`
D. 1

Explanation

To determine the value of \( \lambda \) for which the vectors

\[ \mathbf{A} = \lambda \hat{i} + (1 + \lambda) \hat{j} + (1 + 2\lambda) \hat{k} \]

and

\[ \mathbf{B} = (1 - \lambda) \hat{i} + \lambda \hat{j} + 2 \hat{k} \]

are perpendicular, we need to use the property that two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Write the dot product formula**:
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \]
where \( A_x, A_y, A_z \) are the components of vector \( \mathbf{A} \) and \( B_x, B_y, B_z \) are the components of vector \( \mathbf{B} \).

2. **Identify the components**:
From the vectors, we have:
- \( A_x = \lambda \), \( A_y = 1 + \lambda \), \( A_z = 1 + 2\lambda \)
- \( B_x = 1 - \lambda \), \( B_y = \lambda \), \( B_z = 2 \)

3. **Calculate the dot product**:
Now, substituting the components into the dot product formula:
\[ \mathbf{A} \cdot \mathbf{B} = (\lambda)(1 - \lambda) + (1 + \lambda)(\lambda) + (1 + 2\lambda)(2) \]

4. **Expand the expression**:
Expanding each term:
\[ \mathbf{A} \cdot \mathbf{B} = \lambda(1 - \lambda) + (1 + \lambda)\lambda + 2(1 + 2\lambda) \]
\[ = \lambda - \lambda^2 + \lambda + \lambda^2 + 2 + 4\lambda \]

5. **Combine like terms**:
Combining the terms:
\[ = \lambda - \lambda^2 + \lambda + \lambda^2 + 2 + 4\lambda = 6\lambda + 2 \]

6. **Set the dot product to zero**:
For the vectors to be perpendicular, we set the dot product equal to zero:
\[ 6\lambda + 2 = 0 \]

7. **Solve for \( \lambda \)**:
Rearranging gives:
\[ 6\lambda = -2 \]
\[ \lambda = -\frac{2}{6} = -\frac{1}{3} \]

### Conclusion:
The value of \( \lambda \) for which the vectors are perpendicular is

\[ \lambda = -\frac{1}{3}. \]