Problem 6.11

6.11. a)

We know that

$\theta = \cos^{-1} \frac{\vec{A} \cdot \vec{B}}{(A)(B)}$

So, plugging in for $\vec{A}$ and $\vec{B}$, we find

$\theta = \cos^{-1} \frac{12 + 8}{\sqrt{(13)(32)}}$

or $\theta = 11.3{}^{\circ}$

b)

again plugging in to find $\theta$, we find

$\theta = \cos^{-1} \frac{-6 - 16}{\sqrt{(29)(20)}}$

or $\theta = 156{}^{\circ}$

c)

again plugging in to find $\theta$, we find

$\theta = \cos^{-1} \frac{-6 + 8}{\sqrt{(9)(25)}}$

or $\theta = 82.3{}^{\circ}$