The vectors `2hati-lamdahatj+3lamdahatk and (1+lamda)hati-2lamdahatj+hatk` include an acute angle for (A) all values of `lambda`(B) `lamda lt -2 `and `lamdagt-1/2` (C) `lambda=-1/2` (D) `lamdaepsilon [-2,-1/2]`

Options

A. All real m
B. `m lt -2 " or " m gt -(1)/(2)`
C. `m=-(1)/(2)`
D. `m in [-2,-(1)/(2)]`

Answer

To solve the problem, we need to determine for which values of \( \lambda \) the vectors

\[ \mathbf{A} = 2\hat{i} - \lambda \hat{j} + 3\lambda \hat{k} \]

and

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

include an acute angle. The angle between two vectors is acute if the cosine of the angle is positive, which means that the dot product of the two vectors must be greater than zero.

### Step-by-step Solution:

1. **Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)**:
\[ \mathbf{A} \cdot \mathbf{B} = (2\hat{i} - \lambda \hat{j} + 3\lambda \hat{k}) \cdot ((1 + \lambda)\hat{i} - 2\lambda \hat{j} + \hat{k}) \]
Using the distributive property of the dot product:
\[ = 2(1 + \lambda) + (-\lambda)(-2\lambda) + (3\lambda)(1) \]
\[ = 2 + 2\lambda + 2\lambda^2 + 3\lambda \]
\[ = 2\lambda^2 + 5\lambda + 2 \]

2. **Set up the inequality for an acute angle**:
We need:
\[ 2\lambda^2 + 5\lambda + 2 > 0 \]

3. **Factor the quadratic expression**:
To factor \( 2\lambda^2 + 5\lambda + 2 \), we can use the quadratic formula or factorization:
\[ 2\lambda^2 + 5\lambda + 2 = (2\lambda + 1)(\lambda + 2) \]

4. **Determine the critical points**:
The critical points from the factors are:
\[ 2\lambda + 1 = 0 \implies \lambda = -\frac{1}{2} \]
\[ \lambda + 2 = 0 \implies \lambda = -2 \]

5. **Test intervals around the critical points**:
We will test the sign of \( (2\lambda + 1)(\lambda + 2) \) in the intervals:
- \( (-\infty, -2) \)
- \( (-2, -\frac{1}{2}) \)
- \( (-\frac{1}{2}, \infty) \)

- For \( \lambda < -2 \) (e.g., \( \lambda = -3 \)):
\[ (2(-3) + 1)(-3 + 2) = (-6 + 1)(-1) = 5 > 0 \]
- For \( -2 < \lambda < -\frac{1}{2} \) (e.g., \( \lambda = -1 \)):
\[ (2(-1) + 1)(-1 + 2) = (-2 + 1)(1) = -1 < 0 \]
- For \( \lambda > -\frac{1}{2} \) (e.g., \( \lambda = 0 \)):
\[ (2(0) + 1)(0 + 2) = (1)(2) = 2 > 0 \]

6. **Conclusion**:
The expression \( 2\lambda^2 + 5\lambda + 2 > 0 \) is satisfied for:
- \( \lambda < -2 \)
- \( \lambda > -\frac{1}{2} \)

Therefore, the values of \( \lambda \) for which the vectors include an acute angle are:
\[ \lambda \in (-\infty, -2) \cup (-\frac{1}{2}, \infty) \]

### Final Answer:
The correct option is (B) \( \lambda < -2 \) and \( \lambda > -\frac{1}{2} \).