<s>[INST] Find a vector that bisects the smaller of the two angles formed by vectors <3,4> and <5,-12>. The solution is <8,-1>. I tried finding the angle between the two vectors, but wasn't sure what to do next. 
 [/INST] The cosines of the angles have to be the same and positive (for the smaller angle). Therefore 
$$\frac{3 a + 4 b}{\sqrt{3^2 + 4^2} \cdot \sqrt {a^2 + b^2} } = \frac{5 a + (-12) b}{\sqrt{5^2 + (-12)^2} \cdot \sqrt {a^2 + b^2} }>0$$ or
$$\frac{3 a + 4 b}{5 } = \frac{5 a + (-12) b}{13 }>0$$
Check that $(a,b) = (8,-1)$ is a solution. Indeed:
$$\frac{3 \cdot 8 + 4 \cdot (-1)}{5}= 4$$ and
$$\frac{5 \cdot 8 + (-12) \cdot (-1)}{13}=4$$
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