The vectors `a = 2lambda^(2) i+4lambdaj+k` and `b=7i-2j+lambdak` make an obtuse angle whereas the angle between b and k is acute and less then `pi//6` domain of `lambda` is

Options

A. `0 lt lambda lt (1)/(2)`
B. `lambda gt sqrt(159)`
C. `-(1)/(2) lt lambda lt 0`
D. null set

Answer

To solve the problem, we need to analyze the conditions given for the vectors \( \mathbf{a} \) and \( \mathbf{b} \) in relation to the angle they form with each other and with the unit vector \( \mathbf{k} \).

### Given Vectors:
- \( \mathbf{a} = 2\lambda^2 \mathbf{i} + 4\lambda \mathbf{j} + \mathbf{k} \)
- \( \mathbf{b} = 7\mathbf{i} - 2\mathbf{j} + \lambda \mathbf{k} \)

### Step 1: Condition for Obtuse Angle between \( \mathbf{a} \) and \( \mathbf{b} \)
The angle between two vectors is obtuse if their dot product is less than zero. Therefore, we compute the dot product \( \mathbf{a} \cdot \mathbf{b} \):

\[ \mathbf{a} \cdot \mathbf{b} = (2\lambda^2)(7) + (4\lambda)(-2) + (1)(\lambda) \]

Calculating this gives:

\[ \mathbf{a} \cdot \mathbf{b} = 14\lambda^2 - 8\lambda + \lambda = 14\lambda^2 - 7\lambda \]

Setting the dot product less than zero for the obtuse angle condition:

\[ 14\lambda^2 - 7\lambda < 0 \]

Factoring out \( 7\lambda \):

\[ 7\lambda(2\lambda - 1) < 0 \]

### Step 2: Analyzing the Inequality
To find the values of \( \lambda \) that satisfy this inequality, we consider the critical points where \( 7\lambda = 0 \) or \( 2\lambda - 1 = 0 \):

1. \( \lambda = 0 \)
2. \( \lambda = \frac{1}{2} \)

Using a number line, we test intervals:
- For \( \lambda < 0 \): \( 7\lambda < 0 \) and \( 2\lambda - 1 < 0 \) (positive product)
- For \( 0 < \lambda < \frac{1}{2} \): \( 7\lambda > 0 \) and \( 2\lambda - 1 < 0 \) (negative product)
- For \( \lambda > \frac{1}{2} \): \( 7\lambda > 0 \) and \( 2\lambda - 1 > 0 \) (positive product)

Thus, the solution for this condition is:

\[ 0 < \lambda < \frac{1}{2} \]

### Step 3: Condition for Acute Angle between \( \mathbf{b} \) and \( \mathbf{k} \)
The angle between \( \mathbf{b} \) and \( \mathbf{k} \) is acute if their dot product is greater than zero:

\[ \mathbf{b} \cdot \mathbf{k} = 7(0) + (-2)(0) + \lambda(1) = \lambda > 0 \]

This condition is satisfied for:

\[ \lambda > 0 \]

### Step 4: Combining Conditions
From the two conditions, we have:
1. \( 0 < \lambda < \frac{1}{2} \)
2. \( \lambda > 0 \)

The combined domain of \( \lambda \) is:

\[ 0 < \lambda < \frac{1}{2} \]

### Step 5: Additional Condition
The angle between \( \mathbf{b} \) and \( \mathbf{k} \) is also stated to be less than \( \frac{\pi}{6} \). Therefore, we need to ensure that:

\[ \cos(\theta) > \frac{\sqrt{3}}{2} \]

Using the formula for cosine:

\[ \cos(\theta) = \frac{\mathbf{b} \cdot \mathbf{k}}{|\mathbf{b}| |\mathbf{k}|} \]

Calculating \( |\mathbf{b}| \):

\[ |\mathbf{b}| = \sqrt{7^2 + (-2)^2 + \lambda^2} = \sqrt{49 + 4 + \lambda^2} = \sqrt{53 + \lambda^2} \]

Thus,

\[ \cos(\theta) = \frac{\lambda}{\sqrt{53 + \lambda^2}} > \frac{\sqrt{3}}{2} \]

### Step 6: Solve the Inequality
Squaring both sides gives:

\[ \frac{\lambda^2}{53 + \lambda^2} > \frac{3}{4} \]

Cross-multiplying and simplifying leads to:

\[ 4\lambda^2 > 3(53 + \lambda^2) \implies 4\lambda^2 > 159 + 3\lambda^2 \implies \lambda^2 > 159 \implies \lambda > \sqrt{159} \]

### Step 7: Final Domain
Now we have two conditions:
1. \( 0 < \lambda < \frac{1}{2} \)
2. \( \lambda > \sqrt{159} \)

Since \( \sqrt{159} \) is approximately \( 12.6 \), there is no overlap between these two intervals. Therefore, the final domain of \( \lambda \) is:

\[ \text{Null set} \]

### Conclusion
The domain of \( \lambda \) is empty, meaning there are no values of \( \lambda \) that satisfy both conditions.