5.6. Find the unit vector collinear with the vector directed along the bisector of angle $B A C$ of triangle $A B C$, if its vertices are given: $A(1 ; 1 ; 1), B(3 ; 0 ; 1), C(0 ; 3 ; 1)$.
5.6. Let's find the coordinates and magnitudes of vectors $\overline{A B}$ and $\overline{A C}$; we have $\overline{A B}(2 ; -1 ; 0), \overline{A C}(-1 ; 2 ; 0)$,

$$
|\overrightarrow{A B}|=\sqrt{2^{2}+(-1)^{2}+0^{2}}=\sqrt{5},|\overrightarrow{A C}|=\sqrt{(-1)^{2}+2^{2}+0^{2}}=\sqrt{5}
$$

Since $|\overline{A B}|=|\overline{A C}|$, $\overline{A D}=\overline{A B}+\overline{A C}$ is the diagonal of the rhombus $A B D C$ (Fig. 5.4), and therefore the bisector of angle BAC. We have $\overline{A D}=\overline{A B}+\overline{A C}=(2 ; -1 ; 0)+(-1 ; 2 ; 0)=(1 ; 1 ; 0)$ and $|\overline{A D}|=\sqrt{2}$. Let $\bar{e}$ be the unit vector in the same direction as vector $\overline{A D}$, i.e., $\bar{e}=\frac{\overline{A D}}{|\overrightarrow{A D}|}$. Then, using formula (5.18), we finally get $\bar{e}\left(\frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{2}} ; 0\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-600.jpg?height=317&width=448&top_left_y=1368&top_left_x=707)

Fig. 5.4

Answer: $\left(\frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{2}} ; 0\right)$.