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

The Correct Option is C

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

The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\(\vec{a} \cdot \vec{b} = (\alpha)(1) + (6)(-2) + (-3)(-2\alpha t)\)

Which simplifies to:

\(\vec{a} \cdot \vec{b} = \alpha - 12 + 6\alpha t\)

For the vectors to be inclined at an obtuse angle, the dot product must be negative:

\(\alpha - 12 + 6\alpha t < 0\)

We can rearrange this to:

\(\alpha(1 + 6t) < 12\)

This inequality should hold for all values of \(t \in \mathbb{R}\). Consider two cases for different values of \(t\):

• If \(1 + 6t > 0\), then \(\alpha < \frac{12}{1 + 6t}\).
• If \(1 + 6t < 0\), then \(\alpha > \frac{12}{1 + 6t}\).

For these conditions to hold for all values of \(t\), we consider boundary behavior:

• As \(t \to -\frac{1}{6}^{+}\), \(1 + 6t \to 0^{+}\), which makes \(\frac{12}{1 + 6t} \to +\infty\).
• As \(t \to -\frac{1}{6}^{-}\), \(1 + 6t \to 0^{-}\), which makes \(\frac{12}{1 + 6t} \to -\infty\).

Consequently, the entire range of \(( -\infty, 0 )\) is suitable for \(\alpha\). Hence, we only need to consider:

The set \(\left[-\frac{4}{3}, 0\right]\) because \(\alpha < 0\) satisfies the condition for all \(t\).

Thus, the correct answer is \(\left[-\frac{4}{3}, 0\right]\).

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: \((- \frac{4}{3}, 0)\).