You are given that $\left\langle a+3b,2a-3b\right\rangle=0$ and $\left\langle a-4b,a+2b \right\rangle=0$. Hence
$$\left\langle a+3b,2a-3b\right\rangle=2\left\langle a,a\right\rangle-9\left\langle b,b\right\rangle+3\left\langle a,b\right\rangle=0$$ and
$$\left\langle a-4b,a+2b \right\rangle=\left\langle a,a \right\rangle-8\left\langle b,b \right\rangle-2\left\langle a,b \right\rangle=0.$$(Here I assumed that you are working with a real inner-product space).
By subtracting the second equation twice from the first, we obtain
$$\left\langle b,b \right\rangle+\left\langle a,b \right\rangle=0.$$
Plugging the previous equation into the second yields
$$\|a\|^2=6\|b\|^2.$$
Hence $\frac{\|b\|}{\|a\|}=\sqrt{\frac{1}{6}}.$
Now you know that $\left\langle a,b \right\rangle=\cos(\theta)\|a\|\|b\|.$
Thus, after plugging the previous results in the third equation, we get
$$\cos(\theta)=-\frac{\|b\|}{\|a\|}=-\sqrt{\frac{1}{6}}.$$
There is nothing special here the dot product $AB\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line.
Note also that vector $AB$ is a direction vector for the line orthogonal to the given line in $B$ and passing through $A$.
Best Answer
Hint: Since the two vectors aren't null vectors, we have: $$\vec v\bullet\vec u=\|\vec v\|\|\vec u\|\cos\theta=0\iff \cos\theta=0$$ where $\theta$ is the angle between the two vectors.