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

Options

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

Answer

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 1: Write the dot product of the two vectors
The dot product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:

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

### Step 2: Calculate the dot product
Using the distributive property of the dot product, we calculate:

\[ \mathbf{A} \cdot \mathbf{B} = \lambda(1 - \lambda) + (1 + \lambda)\lambda + (1 + 2\lambda)(2) \]

Now, let's simplify each term:

1. \( \lambda(1 - \lambda) = \lambda - \lambda^2 \)
2. \( (1 + \lambda)\lambda = \lambda + \lambda^2 \)
3. \( (1 + 2\lambda)(2) = 2 + 4\lambda \)

Combining these, we have:

\[ \mathbf{A} \cdot \mathbf{B} = (\lambda - \lambda^2) + (\lambda + \lambda^2) + (2 + 4\lambda) \]

### Step 3: Combine like terms
Now, let's combine the terms:

\[ \mathbf{A} \cdot \mathbf{B} = \lambda - \lambda^2 + \lambda + \lambda^2 + 2 + 4\lambda \]

This simplifies to:

\[ \mathbf{A} \cdot \mathbf{B} = (1 + 4 + 1)\lambda + 2 = 6\lambda + 2 \]

### Step 4: Set the dot product equal to zero
Since the vectors are perpendicular, we set the dot product equal to zero:

\[ 6\lambda + 2 = 0 \]

### Step 5: Solve for \( \lambda \)
Now, we solve for \( \lambda \):

\[ 6\lambda = -2 \]
\[ \lambda = -\frac{2}{6} = -\frac{1}{3} \]

### Conclusion
Thus, the value of \( \lambda \) for which the vectors are perpendicular is

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

---