11. Let vectors $a$ and $b$ satisfy $|a|=1$, $|b|=2$, and the angle between $a$ and $b$ is $60^{\circ}$. If the angle between the vectors $7a+2tb$ and $ta+b$ is obtuse, then the range of the real number $t$ is $\qquad$
11. $\left(-7,-\frac{\sqrt{14}}{2}\right) \cup\left(-\frac{\sqrt{14}}{2},-\frac{1}{2}\right)$.

From $(7 a+2 t b) \cdot(t a+b)<0$, we get
$-7<t<-\frac{1}{2}$.
When $7 a+2 t b$ is collinear with $t a+b$, we have $\frac{7}{t}=\frac{2 t}{1}, t= \pm \frac{\sqrt{14}}{2}$.
When $t=-\frac{\sqrt{14}}{2}$, $7 a+2 t b$ and $t a+b$ are in opposite directions, so $t \neq-\frac{\sqrt{14}}{2}$.