The values of x for which the angle between the vectors $\vec a = x\hat i-3\hat j-\hat k$ and $\vec b = 2x\hat i+x\hat j-\hat k$ is acute and the angle between the vectors b and the y-axis lies between $π/2$ and $π$, are

$<0$

The angle between $\vec a$ and $\vec b$ is acute. Therefore,

$\vec a.\vec b>0$

$⇒2x^2 - 3x+1>0$

$⇒(2x-1) (x-1)>0⇒ x<\frac{1}{2}$ or, $x>1$  ...(i)

The angle between b and the y-axis lies between $\frac{π}{2}$ and $π$.

$∴\vec b.\hat j<0⇒ x <0$   ...(ii)

From (i) and (ii), we obtain $x < 0$.