If the vectors `veca = ( c log_(2) x ) hati - 6hatj + 3hatk ` and `vecb=(log_(2)x )hati + 2hatj + (2clog_(2)x)hatk` make an obtuse angle for any ` x = ( 0 , oo) ` then c belongs to

Options

A. ` ( - oo, 0)`
B. ` ( - oo , -4//3)`
C. ` ( -4//3,0)`
D. ` (-4//3, oo)`

Explanation

To solve the problem, we need to determine the values of \( c \) such that the vectors \( \vec{a} \) and \( \vec{b} \) make an obtuse angle for any \( x \) in the interval \( (0, \infty) \).

### Step 1: Write down the vectors
Given:
\[ \vec{a} = (c \log_2 x) \hat{i} - 6 \hat{j} + 3 \hat{k} \]
\[ \vec{b} = (\log_2 x) \hat{i} + 2 \hat{j} + (2c \log_2 x) \hat{k} \]

### Step 2: Use the dot product condition for obtuse angles
The angle between two vectors is obtuse if their dot product is negative:
\[ \vec{a} \cdot \vec{b} < 0 \]

### Step 3: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
Using the formula for the dot product:
\[ \vec{a} \cdot \vec{b} = (c \log_2 x)(\log_2 x) + (-6)(2) + (3)(2c \log_2 x) \]
This simplifies to:
\[ \vec{a} \cdot \vec{b} = c (\log_2 x)^2 - 12 + 6c \log_2 x \]
Combining terms, we have:
\[ \vec{a} \cdot \vec{b} = c (\log_2 x)^2 + 6c \log_2 x - 12 \]

### Step 4: Set up the inequality
We need:
\[ c (\log_2 x)^2 + 6c \log_2 x - 12 < 0 \]

### Step 5: Substitute \( y = \log_2 x \)
Since \( x \) is in \( (0, \infty) \), \( y \) can take any real value. Thus, we rewrite the inequality:
\[ c y^2 + 6c y - 12 < 0 \]

### Step 6: Analyze the quadratic inequality
This is a quadratic inequality in \( y \). For the quadratic \( ay^2 + by + c < 0 \) to hold for all \( y \), the following conditions must be satisfied:
1. \( a < 0 \) (which means \( c < 0 \))
2. The discriminant must be less than zero.

### Step 7: Calculate the discriminant
The discriminant \( D \) of the quadratic \( c y^2 + 6c y - 12 \) is given by:
\[ D = (6c)^2 - 4(c)(-12) = 36c^2 + 48c \]
We require:
\[ 36c^2 + 48c < 0 \]

### Step 8: Factor the discriminant
Factoring gives:
\[ 12c(3c + 4) < 0 \]
This inequality holds when:
1. \( c < 0 \) and \( 3c + 4 > 0 \) (which gives \( c > -\frac{4}{3} \))
2. \( c > 0 \) and \( 3c + 4 < 0 \) (not possible since \( c > 0 \))

### Step 9: Combine the conditions
Thus, we have:
\[ -\frac{4}{3} < c < 0 \]

### Conclusion
The values of \( c \) that make the angle between the vectors obtuse for any \( x \in (0, \infty) \) are:
\[ c \in \left(-\frac{4}{3}, 0\right) \]