Express the unit vector vec x through vec u and vec v…

Express the unit vector $\vec x$ through $\vec u$ and $\vec v$, given that $\vec x$ bisects the angle between $\vec a=\vec u+\vec v$ and $\vec b=2\vec u+\vec v$ and is coplanar with them, where $|\vec u|=2$, $|\vec v|=4$ and $\angle(\vec u,\vec v)=120^\circ$.

Step 1: Basic products

\[ \vec u^{\,2}=4,\qquad \vec v^{\,2}=16,\qquad \vec u\cdot\vec v=2\cdot4\cdot\left(-\tfrac12\right)=-4. \]

Step 2: Lengths of $\vec a$ and $\vec b$

\[ |\vec a|^2=4-8+16=12,\qquad |\vec a|=2\sqrt3, \]

\[ |\vec b|^2=4\cdot4+4(-4)+16=16,\qquad |\vec b|=4. \]

Step 3: Bisector direction

The sum of the two unit vectors is the diagonal of a rhombus, so it bisects the angle: \[\vec w=\frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}=\frac{\vec u+\vec v}{2\sqrt3}+\frac{2\vec u+\vec v}{4}=\frac{3+\sqrt3}{6}\,\vec u+\frac{3+2\sqrt3}{12}\,\vec v.\]

Step 4: Length of $\vec w$

Shorter than expanding the coefficients: with $\hat a=\vec a/|\vec a|$, $\hat b=\vec b/|\vec b|$, \[\vec a\cdot\vec b=(\vec u+\vec v)\cdot(2\vec u+\vec v)=8+3(-4)+16=12,\qquad \hat a\cdot\hat b=\frac{12}{2\sqrt3\cdot4}=\frac{\sqrt3}{2},\] so $\angle(\vec a,\vec b)=30^\circ$ and

\[ |\vec w|^2=1+2\cdot\frac{\sqrt3}{2}+1=2+\sqrt3,\qquad |\vec w|=\sqrt{2+\sqrt3}=\frac{\sqrt6+\sqrt2}{2}. \]

This agrees with the long computation in the source.

Step 5: The unit bisector

\[ \vec x=\frac{\vec w}{|\vec w|},\qquad \frac{1}{|\vec w|}=\frac{2}{\sqrt6+\sqrt2}=\frac{\sqrt6-\sqrt2}{2}, \]

\[ \vec x=\frac{\sqrt6-\sqrt2}{2}\left[\frac{3+\sqrt3}{6}\,\vec u+\frac{3+2\sqrt3}{12}\,\vec v\right]. \]

The coefficients simplify, because $(3+\sqrt3)(\sqrt6-\sqrt2)=2\sqrt6$ and $(3+2\sqrt3)(\sqrt6-\sqrt2)=\sqrt6+3\sqrt2$: \[\vec x=\frac{\sqrt6}{6}\,\vec u+\frac{\sqrt6+3\sqrt2}{24}\,\vec v\approx0.408\,\vec u+0.279\,\vec v\]

Step 6: Checks

Unit length:

\[ |\vec x|^2=\frac{6}{36}\cdot4+2\cdot\frac{\sqrt6(\sqrt6+3\sqrt2)}{144}(-4)+\frac{(\sqrt6+3\sqrt2)^2}{576}\cdot16=\frac23-\frac{6+6\sqrt3}{18}+\frac{24+6\sqrt{12}}{36}=1.\ \checkmark \]

Equal angles: $\vec w\cdot\hat a=1+\hat a\cdot\hat b=\vec w\cdot\hat b$, so $\cos\angle(\vec x,\vec a)=\cos\angle(\vec x,\vec b)$. Each half-angle is $15^\circ$.