The set of all values of \(t\) for which the angle between the vectors \(2t^2\hat{\imath}+4t\hat{\jmath}+\hat{k}\) and \(7\hat{\imath}-2\hat{\jmath}+t\hat{k}\) is obtuse is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle \left(0, \frac{1}{2}\right) \)
Let \(\vec{a} = 2t^2\hat{\imath}+4t\hat{\jmath}+\hat{k}\) and \(\vec{b} = 7\hat{\imath}-2\hat{\jmath}+t\hat{k}\). The angle is obtuse when \(\cos\theta \lt 0\), that is, when the scalar product is negative.
\(\vec{a}\cdot\vec{b} = 14t^2-8t+t\)
\( = 14t^2-7t = 7t(2t-1)\)
\(7t(2t-1) \lt 0\)
\( \displaystyle 0 \lt t \lt \frac{1}{2} \)
The angle can never be \(\pi\): that needs \(\vec{a}\) to point exactly opposite to \(\vec{b}\), but the x-components \(2t^2 \ge 0\) and \(7 \gt 0\) never have opposite signs. So every \(t\) in \(\left(0, \frac{1}{2}\right)\) gives an obtuse angle.
Why the other options are incorrect
Option (1) \( \displaystyle (-\infty, 0)\cup\left(\frac{1}{2}, \infty\right) \) — Incorrect. There \(\vec{a}\cdot\vec{b} \gt 0\), so the angle is acute.
Option (2) \( \displaystyle \left[0, \frac{1}{2}\right] \) — Incorrect. At \(t = 0\) and \(t = \frac{1}{2}\), \(\vec{a}\cdot\vec{b} = 0\) and the angle is a right angle, not obtuse.
Option (3) \( \displaystyle \left(-\frac{1}{2}, 0\right) \) — Incorrect. For negative \(t\) both \(7t\) and \(2t-1\) are negative, so \(\vec{a}\cdot\vec{b} \gt 0\).