If the angle between $\vec a=2y^2\hat i+4y\hat j+\hat k$ and $\vec b=7\hat i-2\hat j+y\hat k$ is obtuse, then:

$0<y<\frac{1}{2}$

The correct answer is Option (4) → $0<y<\frac{1}{2}$

for an angle to be obtuse, we must have,

$a.b=|a||b|\cos θ$

where,

$a.b<0$

$∴a.b=(2y^2)(7)+(4y)(-2)+(1)(y)$

$=14y^2-7y$

and,

$⇒14y^2-7y$

$7y(2y-1)<0$

Critical points are 0 and $\frac{1}{2}$

and,

for $0<y<\frac{1}{2}$

$7y$ is positive

$(2y-1)$ is negative

$∴(7y)(2y-1)<0\,∀\,0<y<\frac{1}{2}$