Question

The set of all α, for which the vectors $\vec{a} = \alpha \hat{ti} + 6\hat{j} - 3\hat{k}$ and $\vec{b} = \hat{ti} - 2\hat{j} - 2\alpha t\hat{k}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$ is:

• A. [0,1)
• B. (-2,0]
• C. $\left[-\frac{4}{3}, 0\right]$
• D. $\left[-\frac{4}{3}, 1\right]$

Answer 1

Answer: (C)

$\left[-\frac{4}{3}, 0\right]$

Answer 2

The dot product of $\vec{a}$ and $\vec{b}$ is:
$\vec{a} \cdot \vec{b} = \alpha t + 6(-2) + (-3)(-2\alpha t) = \alpha t - 12 + 6\alpha t.$
$\vec{a} \cdot \vec{b} = (\alpha + 6\alpha)t - 12 = 7\alpha t - 12.$
For the angle to be obtuse:
\vec{a} \cdot \vec{b} < 0\.
This gives:
7\alpha t - 12 < 0 \implies t(7\alpha) - 12 < 0\.
For all $t \in \mathbb{R}$, this inequality holds only if:
\alpha < 0 \quad \text{and} \quad -12 < 0\.
To ensure obtuse angles:
-\frac{4}{3} < \alpha < 0\.
Final Answer: $(- \frac{4}{3}, 0)$.