I understand that for vectors $u$ and $v$, $cos\theta=\frac{u\cdot v}{|u||v|}$
This means that $u\cdot v= |u||v|\cos\theta$
So the dot product can only be negative when $\cos\theta<0$
$\cos\theta<0$ when $\frac{\pi}{2}<\theta<\frac{3\pi}{2}$
$\cos\theta>0$ when $0<\theta<\frac{\pi}{2}$
$\cos\theta=0$ when $\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$
My answers are as follows:
a) $a\cdot f$ positive since $\theta<\frac{\pi}{2}$
b) no clue how to do
c) $d \cdot b$ positive since $\theta<\frac{\pi}{2}$
d) $a \cdot b$ negative since $\theta>\frac{\pi}{2}$
e) $c \cdot e$ zero since $\theta=0$
Can someone please verify my answers for a, c, d, e? And how do I do part b)?
$$c\cdot e\neq0$$Since the dot product is $0$ if and only if the vectors are orthogonal, as $\cos(\pi/2)=0$. $c$ is in fact a $\pi$ turn away from $e$, so their dot product is as negative as it gets.
You were on the right track; for $b)$, make the same observation you made with $d)$: the turn between $f$ and $c$ is greater than $\pi/2$, so the dot product is negative.