Angle between vectors a and b if a + 3b is perpendicular to.... Determine the angle between vectors $a$ and $b( a \ne 0, b \ne0)$ if $a + 3b$ is perpendicular to $2a - b$ and $a + 7b$ is perpendicular to $2a + b$.
I've done this:
$(a + 3b) \cdot (2a - b) = 0$
$\dots 5a⋅b = 3b⋅b - 2a⋅a$
$(a + 7b) \cdot (2a + b) = 0$
$\cdots 15a⋅b = -2 a⋅a - 7b\cdot b$
which gives me:
$a⋅b = (-(b⋅b + a⋅a))/5$
$cos(\theta) = (-b - a)/5$
Now I'm not really sure how to proceed

You have $$5a\cdot b=3b\cdot b-2a\cdot a$$ $$15a\cdot b=-7b\cdot b-2a\cdot a$$so subtract the 2nd equation from the first to get $$-10a\cdot b=10b\cdot b\Rightarrow b\cdot b=-a\cdot b$$
Also $$35a\cdot b=21b\cdot b-14a\cdot a$$ $$45a\cdot b=-21b\cdot b-6a\cdot a\Rightarrow80a\cdot b=-20a\cdot a$$ so similarly $$a\cdot a=-4a\cdot b$$ 
Then finally 
$$\cos\theta=\frac{a\cdot b}{|a||b|}\\
\cos\theta=\frac{a\cdot b}{\sqrt{(a\cdot a) (b\cdot b)}}=\frac{a\cdot b}{\sqrt{(-4a\cdot b)(-a\cdot b)}}=\frac{1}{2}.$$
So $\theta=\pi/3$.
