If the vectors `a = (2, log_(3)x,lambda)` and `b = (-3,lambdalog_(3) x, log_(3)x)` are inclined at an acute angle, then

Options

A. `lambda =0`
B. `lambda gt 0`
C. `lambda gt 0`
D. none of these

Answer

To solve the problem, we need to analyze the vectors \( \mathbf{a} \) and \( \mathbf{b} \) given by:

\[ \mathbf{a} = (2, \log_3 x, \lambda) \]
\[ \mathbf{b} = (-3, \lambda \log_3 x, \log_3 x) \]

We know that the vectors are inclined at an acute angle, which means that their dot product must be positive. The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:

\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

### Step 1: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)

Substituting the components of \( \mathbf{a} \) and \( \mathbf{b} \):

\[ \mathbf{a} \cdot \mathbf{b} = (2)(-3) + (\log_3 x)(\lambda \log_3 x) + (\lambda)(\log_3 x) \]

Calculating each term:

\[ = -6 + \lambda (\log_3 x)^2 + \lambda (\log_3 x) \]

### Step 2: Set the dot product greater than zero

Since the vectors are inclined at an acute angle, we set the dot product greater than zero:

\[ -6 + \lambda (\log_3 x)^2 + \lambda (\log_3 x) > 0 \]

### Step 3: Rearranging the inequality

Rearranging gives:

\[ \lambda (\log_3 x)^2 + \lambda (\log_3 x) > 6 \]

Factoring out \( \lambda \):

\[ \lambda \left( (\log_3 x)^2 + \log_3 x \right) > 6 \]

### Step 4: Analyze the expression

For the inequality to hold, we need to consider the sign of \( \lambda \) and the expression \( (\log_3 x)^2 + \log_3 x \).

1. **If \( \lambda > 0 \)**, then \( (\log_3 x)^2 + \log_3 x \) must also be positive for the inequality to hold.
2. **If \( \lambda < 0 \)**, then \( (\log_3 x)^2 + \log_3 x \) must be negative, which is not possible since \( \lambda \) would make the left side negative.

### Step 5: Finding conditions for \( x \)

The expression \( (\log_3 x)^2 + \log_3 x \) can be rewritten as:

\[ t^2 + t \quad \text{where } t = \log_3 x \]

This is a quadratic equation in \( t \). The roots can be found using the quadratic formula:

\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 0}}{2 \cdot 1} = \frac{-1 \pm 1}{2} \]

The roots are \( t = 0 \) and \( t = -1 \). The expression is positive for \( t < -1 \) or \( t > 0 \).

### Step 6: Conclusion about \( x \)

Thus, we have:

- \( \log_3 x > 0 \) implies \( x > 3 \).
- \( \log_3 x < -1 \) implies \( x < \frac{1}{3} \).

### Final Result

The vectors \( \mathbf{a} \) and \( \mathbf{b} \) are inclined at an acute angle if \( x > 3 \) or \( x < \frac{1}{3} \) and \( \lambda > 0 \).