Key Concepts and Formulas

• Dot Product and Angle Between Vectors: The dot product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between them.
• Obtuse Angle Condition: An obtuse angle $\theta$ lies in the interval $(\frac{\pi}{2}, \pi)$. For such angles, $\cos \theta < 0$.
• Quadratic Inequality for All Real Numbers: For a quadratic function $f(t) = At^2 + Bt + C$ to be strictly negative ($f(t) < 0$) for all $t \in \mathbb{R}$, two conditions must be met:
  • If $A \neq 0$, the parabola must open downwards ($A < 0$) and have no real roots (discriminant $D = B^2 - 4AC < 0$).
  • If $A = 0$, the function reduces to a linear or constant function ($Bt + C$). If $B=0$, it's a constant ($C$), which must be negative. If $B \neq 0$, it's a line, which cannot be strictly negative for all $t$.

Step-by-Step Solution

Step 1: Understand the Condition for an Obtuse Angle We are given that the vectors $\vec{a}$ and $\vec{b}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$. The angle $\theta$ between two vectors is obtuse if $\frac{\pi}{2} < \theta < \pi$. In this range, $\cos \theta < 0$. Using the dot product formula, $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. Since $|\vec{a}| > 0$ and $|\vec{b}| > 0$ (we will verify this shortly), the condition $\cos \theta < 0$ is equivalent to $\vec{a} \cdot \vec{b} < 0$.

Step 2: Verify that the Vectors are Non-Zero for all $t \in \mathbb{R}$ Let's check if $\vec{a}$ or $\vec{b}$ can be the zero vector for any $t$. For $\vec{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k}$: $|\vec{a}|^2 = (\alpha t)^2 + 6^2 + (-3)^2 = \alpha^2 t^2 + 36 + 9 = \alpha^2 t^2 + 45$. Since $\alpha^2 t^2 \ge 0$, $|\vec{a}|^2 \ge 45$. Thus, $\vec{a}$ is never the zero vector.

For $\vec{b} = t \hat{i} - 2 \hat{j} - 2 \alpha t \hat{k}$: $|\vec{b}|^2 = t^2 + (-2)^2 + (-2 \alpha t)^2 = t^2 + 4 + 4 \alpha^2 t^2 = (1+4\alpha^2)t^2 + 4$. Since $(1+4\alpha^2)t^2 \ge 0$, $|\vec{b}|^2 \ge 4$. Thus, $\vec{b}$ is never the zero vector. Since both vectors are always non-zero, the condition $\vec{a} \cdot \vec{b} < 0$ is indeed the correct one.

Step 3: Calculate the Dot Product $\vec{a} \cdot \vec{b}$ Given $\vec{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 \alpha t \hat{k}$. The dot product is: $\vec{a} \cdot \vec{b} = (\alpha t)(t) + (6)(-2) + (-3)(-2 \alpha t)$ $\vec{a} \cdot \vec{b} = \alpha t^2 - 12 + 6 \alpha t$ Rearranging the terms in descending order of $t$: $\vec{a} \cdot \vec{b} = \alpha t^2 + 6 \alpha t - 12$

Step 4: Set up the Inequality for an Obtuse Angle We require $\vec{a} \cdot \vec{b} < 0$ for all $t \in \mathbb{R}$. This translates to the inequality: $\alpha t^2 + 6 \alpha t - 12 < 0 \quad \text{for all } t \in \mathbb{R}$

Step 5: Analyze the Quadratic Inequality for All $t \in \mathbb{R}$ Let $P(t) = \alpha t^2 + 6 \alpha t - 12$. We need $P(t) < 0$ for all $t \in \mathbb{R}$. We consider two cases based on the coefficient of $t^2$, which is $\alpha$.

Case 1: $\alpha = 0$ If $\alpha = 0$, the inequality becomes: $(0) t^2 + 6(0) t - 12 < 0$ $-12 < 0$ This statement is true for all $t \in \mathbb{R}$. Therefore, $\alpha = 0$ is a valid solution.

Case 2: $\alpha \neq 0$ In this case, $P(t)$ is a quadratic function. For $P(t) < 0$ for all $t \in \mathbb{R}$, two conditions must be satisfied:

1. The parabola must open downwards: The coefficient of $t^2$ must be negative. $\alpha < 0$
2. The quadratic must have no real roots: The discriminant must be negative. The discriminant $D$ of $\alpha t^2 + 6 \alpha t - 12$ is given by $B^2 - 4AC$, where $A=\alpha$, $B=6\alpha$, and $C=-12$. $D = (6\alpha)^2 - 4(\alpha)(-12) < 0$ $36\alpha^2 + 48\alpha < 0$ To solve this inequality for $\alpha$, we factor out $12\alpha$: $12\alpha (3\alpha + 4) < 0$ The roots of $12\alpha(3\alpha + 4) = 0$ are $\alpha = 0$ and $3\alpha + 4 = 0 \Rightarrow \alpha = -\frac{4}{3}$. Since the expression $12\alpha(3\alpha + 4)$ is a quadratic in $\alpha$ with a positive leading coefficient ($36\alpha^2$), it is negative between its roots. Thus, the solution to this discriminant inequality is: $-\frac{4}{3} < \alpha < 0$

Now, we need to satisfy both conditions from Case 2: $\alpha < 0$ AND $-\frac{4}{3} < \alpha < 0$. The intersection of these two conditions is: $-\frac{4}{3} < \alpha < 0$

Step 6: Combine the Solutions from Both Cases From Case 1, we found $\alpha = 0$ is a solution. From Case 2, we found $\alpha \in \left(-\frac{4}{3}, 0\right)$ is a solution. Combining these two sets of solutions, the set of all possible values for $\alpha$ is the union: $\left(-\frac{4}{3}, 0\right) \cup \{0\}$ This union simplifies to: $\left(-\frac{4}{3}, 0\right]$

Common Mistakes & Tips

• Forgetting the $A=0$ Case: When analyzing a quadratic inequality required to hold for all $t$, always check the scenario where the leading coefficient ($A$) is zero. This can lead to a linear or constant inequality that might have valid solutions.
• Sign Errors in Discriminant: Be careful with signs when calculating and solving the discriminant inequality. A small error here can lead to an incorrect interval.
• Interpreting "Obtuse Angle": Ensure you correctly translate "obtuse angle" to $\cos \theta < 0$, which in turn means $\vec{a} \cdot \vec{b} < 0$ for non-zero vectors.

Summary To find the values of $\alpha$ for which the vectors are inclined at an obtuse angle for all $t$, we used the condition that their dot product must be negative. We calculated the dot product, which resulted in a quadratic expression in $t$ involving $\alpha$. We then analyzed this quadratic inequality, considering the case where the leading coefficient is zero and the case where it is non-zero. The conditions for the quadratic to be negative for all $t$ required the leading coefficient to be negative and the discriminant to be negative. Combining the results from all cases yielded the final range for $\alpha$.

The final answer is $\boxed{\left(-\frac{4}{3}, 0\right]}$.