The set of all `alpha`, for which the vector `vec a = alpha t hat i + 6 hat j – 3 hat k` and `vec b = t hat i – 2 hat j – 2 alpha t hat k` are inclined at a obtuse angle for all `t in R`, is

Options

A. `[0, 1)`
B. `(- 2, 0]`
C. `(-4/3 , 1)`
D. `(-4/3 , 0 ]`

Answer

To find the set of all values of \( \alpha \) for which the vectors \( \vec{a} = \alpha \hat{i} + 6 \hat{j} - 3 \hat{k} \) and \( \vec{b} = \hat{i} - 2 \hat{j} - 2\alpha \hat{k} \) are inclined at an obtuse angle for all \( t \in \mathbb{R} \), we need to analyze the dot product of the two vectors.

### Step 1: Calculate the dot product of the vectors
The dot product \( \vec{a} \cdot \vec{b} \) is given by:
\[ \vec{a} \cdot \vec{b} = (\alpha \hat{i} + 6 \hat{j} - 3 \hat{k}) \cdot (\hat{i} - 2 \hat{j} - 2\alpha \hat{k}) \]
Calculating this, we have:
\[ \vec{a} \cdot \vec{b} = \alpha \cdot 1 + 6 \cdot (-2) + (-3) \cdot (-2\alpha) \]
\[ = \alpha - 12 + 6\alpha \]
\[ = 7\alpha - 12 \]

### Step 2: Set the condition for obtuse angle
For the vectors to be inclined at an obtuse angle, the dot product must be less than zero:
\[ 7\alpha - 12 < 0 \]

### Step 3: Solve the inequality
Rearranging the inequality gives:
\[ 7\alpha < 12 \]
\[ \alpha < \frac{12}{7} \]

### Step 4: Determine the range of \( \alpha \)
Next, we need to ensure that this condition holds for all \( t \in \mathbb{R} \). Since the expression \( 7\alpha - 12 \) is linear in \( \alpha \), we also need to check if there are any restrictions on \( \alpha \) that would make the quadratic formed by the dot product non-negative for some \( t \).

### Step 5: Analyze the quadratic expression
The expression \( \vec{a} \cdot \vec{b} \) can be viewed as a quadratic in \( t \):
\[ \alpha t^2 + 6\alpha t - 12 < 0 \]
This is a quadratic equation in \( t \) with coefficients:
- \( A = \alpha \)
- \( B = 6\alpha \)
- \( C = -12 \)

For this quadratic to be negative for all \( t \), the discriminant must be less than zero:
\[ D = B^2 - 4AC < 0 \]
Calculating the discriminant:
\[ D = (6\alpha)^2 - 4(\alpha)(-12) = 36\alpha^2 + 48\alpha \]
Setting the discriminant less than zero:
\[ 36\alpha^2 + 48\alpha < 0 \]
Factoring out \( 12\alpha \):
\[ 12\alpha(3\alpha + 4) < 0 \]

### Step 6: Solve the inequality
The critical points are \( \alpha = 0 \) and \( \alpha = -\frac{4}{3} \). We analyze the intervals:
1. \( \alpha < -\frac{4}{3} \)
2. \( -\frac{4}{3} < \alpha < 0 \)
3. \( \alpha > 0 \)

Testing these intervals:
- For \( \alpha < -\frac{4}{3} \), the product is positive.
- For \( -\frac{4}{3} < \alpha < 0 \), the product is negative.
- For \( \alpha > 0 \), the product is positive.

Thus, the solution for \( \alpha \) that satisfies both conditions is:
\[ -\frac{4}{3} < \alpha < 0 \]

### Final Answer
The set of all \( \alpha \) for which the vectors are inclined at an obtuse angle for all \( t \in \mathbb{R} \) is:
\[ \alpha \in \left(-\frac{4}{3}, 0\right) \]