The position vectors of the vertices $$A,\,B,\,C$$   of a triangle are $$\overrightarrow i - \overrightarrow j - 3\overrightarrow k ,\,2\overrightarrow i + \overrightarrow j - 2\overrightarrow k $$       and $$ - 5\overrightarrow i + 2\overrightarrow j - 6\overrightarrow k $$     respectively. The length of the bisector $$AD$$  of the angle $$BAC$$  where $$D$$ is on the line segment $$BC,$$  is :

Solution :
$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \left( {2\overrightarrow i + \overrightarrow j - 2\overrightarrow k } \right) - \left( {\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right) = \overrightarrow i + 2\overrightarrow j + \overrightarrow k \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = \left( { - 5\overrightarrow i + 2\overrightarrow j - 6\overrightarrow k } \right) - \left( {\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right) = - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k \cr} $$
A vector along the bisector of the angle $$BAC$$
$$\eqalign{ & = \frac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \frac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}} \cr & = \frac{{\overrightarrow i + 2\overrightarrow j + \overrightarrow k }}{{\sqrt {{1^2} + {2^2} + {1^2}} }} + \frac{{ - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k }}{{\sqrt {{{\left( { - 6} \right)}^2} + {3^2} + {{\left( { - 3} \right)}^2}} }} \cr & = \frac{1}{{\sqrt 6 }}\left( {\overrightarrow i + 2\overrightarrow j + \overrightarrow k } \right) + \frac{1}{{3\sqrt 6 }}\left( { - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k } \right) \cr & = \frac{1}{{3\sqrt 6 }}\left( { - 3\overrightarrow i + 9\overrightarrow j } \right) \cr & = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{\sqrt 6 }} \cr} $$
$$\therefore $$  the unit vector along $$AD = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{\sqrt {10} }}$$
$$\eqalign{ & \therefore \,\overrightarrow {AD} = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{10}}AD \cr & {\text{As }}D{\text{ is on }}BC,\,\overrightarrow {BD} = t\overrightarrow {BC} \cr & \therefore \,\overrightarrow {BA} + \overrightarrow {AD} = t\left( {\overrightarrow {BA} + \overrightarrow {AC} } \right) \cr & {\text{or }} - \overrightarrow i - 2\overrightarrow j - \overrightarrow k + \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{10}}AD = t\left( { - \overrightarrow i - 2\overrightarrow j - \overrightarrow k - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k } \right) \cr & = t\left( { - 7\overrightarrow i + \overrightarrow j - 4\overrightarrow k } \right) \cr & \Rightarrow \,\, - 1 - \frac{{AD}}{{10}} = - 7t,\,\,\, - 2 + \frac{3}{{10}}AD = t,\,\,\, - 1 = - 4t \cr & \therefore \,t = \frac{1}{4} \cr & \therefore \, - 1 - \frac{{AD}}{{10}} = - \frac{7}{4}{\text{ or }}\frac{{AD}}{{10}} = \frac{3}{4} \cr & \therefore \,AD = \frac{{15}}{2}. \cr} $$