If the angle between \(\vec{a}=2y^2\hat{\imath}+4y\hat{\jmath}+\hat{k}\) and \(\vec{b}=7\hat{\imath}-2\hat{\jmath}+y\hat{k}\) is obtuse, then:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(0 \lt y \lt \displaystyle\frac{1}{2}\)
Since \(\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta\), the angle is obtuse when the dot product is negative (and the vectors do not point in exactly opposite directions).
\(\vec a\cdot\vec b = (2y^2)(7) + (4y)(-2) + (1)(y)\)
\(= 14y^2 - 7y\)
\(= 7y(2y - 1)\)
The product of 7y and 2y − 1 is negative when the two factors have opposite signs, that is, between the roots 0 and 1/2:
\( \displaystyle 7y(2y - 1) \lt 0 \text{ when } 0 \lt y \lt \frac{1}{2} \)
The vectors are never in opposite directions: \(\vec a = k\vec b\) would need 1 = ky (from \(\hat{k}\)) and 4y = −2k (from \(\hat{\jmath}\)), so \(1 = -2y^2\), which is impossible.