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. 
 A: One way to do this, is to normalize the two vectors. If you then have two vectors $u,v$, such that $|u|=|v|$, then a vector which bisects them is $u+v$. This is because the diagonal of a rhombus bisects the angle it cuts (by similarity of triangles).
A: Let $u = \langle 3, 4\rangle$ and let $v = \langle 5, -12 \rangle.$
Let $w = \langle x, y \rangle$ be a vector that bisects the smaller angle between $u$ and $v$, and let that angle be $2\theta$ (so that $\theta$ is the angle between $u$ and $w$ and also the angle between $v$ and $w$.)  Then
$$ u \cdot w = \|u\| \|w\| \cos \theta,$$
$$ v \cdot w = \|v\| \|w\| \cos \theta.$$
Therefore
$$ \frac{u \cdot w} {v \cdot w} = \frac{\|u\| \|w\| \cos \theta}{\|v\| \|w\| \cos \theta}
 =  \frac{\|u\|}{\|v\|} = \frac{5}{13}.$$
But also
$$ \frac{u \cdot w} {v \cdot w} = \frac{3x + 4y}{5x - 12y}.$$
Therefore
$$\begin{eqnarray}\frac{3x + 4y}{5x - 12y} &=& \frac{5}{13},\\
39x + 52y &=& 25x - 60y, \\
14 x &=& -112y,\\
x &=& -8y,\\
w &=& \langle -8y, y \rangle.\end{eqnarray}$$
Now you just need to choose an appropriate value of $y.$
The vector is in the wrong direction (it bisects the larger angle between $u$ and $v$)
if you choose $y > 0,$ so let's try $y < 0.$ Specifically, $y = -1$ is a good choice,
resulting in the vector $ \langle 8, -1 \rangle. $
A: 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
$$\frac{5 \cdot 8 + (-12) \cdot (-1)}{13}=4$$
A: Let $u=<3,4>$ and $v=<5,-12>$
Adding the unit vectors of $u$ and $v$ will give you a bisecting vector, since the magnitudes of unit vectors are equivalent.
Picture of diagram demonstrating why this is true
Finding the unit vectors:
$\widehat{u} = \frac{u}{\rVert u \rVert}$
$\widehat{u} = (\frac{1}{\sqrt{3^{2}+4^{2}}})<3,4> = \frac{<3,4>}{\sqrt{25}} = <\frac{3}{5}, \frac{4}{5}>$
$\widehat{v} = (\frac{1}{\sqrt{5^{2}+(-12)^{2}}})<5,-12> = \frac{<5, -12>}{\sqrt{169}} = <\frac{5}{13}, \frac{-12}{13}>$
Then just add $\widehat{u}$ and $\widehat{v}$:
$<\frac{3}{5}+\frac{5}{13}, \frac{4}{5}-\frac{12}{13}> = <\frac{64}{65},\frac{-8}{65}>$
And you can make this result cleaner if you want (note that the above is still an acceptable and correct answer):
$(65)(<\frac{64}{65},\frac{-8}{65}>)=<64,-8>$
$\frac{<64,-8>}{8} = <8,-1>$
And there you have it! No angle calculations necessary.
A: One geometric solution: normalize the two vectors so that they have the same length and "connect the tips," drawing an isosceles triangle. The angle bisector will cross the midpoint of the opposite side of the triangle. Edit: Peter's way is a little cleaner.
Alternatively, if you want to explicitly get the angle, you can use the dot product: $u \cdot v = |u||v|\cos\theta$. Then you can halve the angle, and use the same formula again to find your bisector.
