If $$\vec { a }$$ and $$\vec { b }$$ are unit vectors such that $$\left( \overset { - }{ a } +\overset { - }{ b }  \right) .\left( \overset { - }{ 2a } +3\overset { - }{ b }  \right) \times \left( 3\overset { - }{ a } -2\overset { - }{ b }  \right) =\overset { - }{ 0 } $$ then angle between $$\vec { a }$$ and $$\vec { b }$$ is   
Velocity and acceleration of a particle at some instant are $$v=(3\hat{i}-4\hat{j}+2\hat{k})m/s$$ and $$a=(2\hat{i}+\hat{j}-2\hat{k})m/s^2$$.
Three vectors which are coplanar with respect to a certain coordinate system are given by  
$$\vec{a}=4\widehat{i}-\widehat{j}; \vec{b}=-3\widehat{i}+2\widehat{j}$$ and $$ \vec{c}=-3\widehat{j}$$
Let the two lines $$AB$$ and $$AC$$ meet at A whose position vector is $$\overline{a}$$ referred to origin $$o$$.Let $$\overline{b}$$ and $$\overline{c}$$ are vectors  to $$AC$$ & $$AD$$ respectively. Their equations of lines $$AB$$ & $$AC$$ are respectively given by
$$\displaystyle \overline{r}=\overline{a}+t\overline{c}$$ and $$\displaystyle \overline{r}=\overline{a}+s\overline{b}$$ where $$t$$ and $$s$$ are scalars Let P be any point on the
internal bisector of $$\displaystyle  \angle BAC $$ such that $$\displaystyle \overline{OP}=\overline{r}$$ Draw a line parallel to $$BA$$ which meet the line $$AC$$ at $$M$$
$$\displaystyle \therefore \angle PAM = \angle PAB = \angle APM $$ (Alternate angle)
$$\displaystyle \therefore AM=PM=\lambda$$ (say)
$$\displaystyle \therefore \overline{AM} $$ is collinear with $$\displaystyle  \overline{b} $$ $$\displaystyle \therefore \overline{AM}=AM \hat{b}=\frac{\lambda \overline{b} }{\left | \overline{b}  \right |}  $$
and $$\displaystyle \overline{MP} $$ is collinear with $$\displaystyle  \overline{c} $$ $$\displaystyle \therefore \overline{MP}=MP \hat{c}=\frac{\lambda \overline{c} }{\left | \overline{c}  \right |}  $$
Also $$\displaystyle \overline{AP}=\overline{AM}+\overline{MP}=\lambda \left ( \frac{\overline{b}}{\left | \overline{b} \right |} +\frac{\overline{c}}{\left | \overline{c} \right |}\right )=\lambda \left ( \hat{b} +\hat{c}\right )$$
$$\displaystyle \Rightarrow \overline{OP}-\overline{OA}=\lambda \left ( \hat{b} +\hat{c}\right )$$
$$\displaystyle \Rightarrow \overline{OP}= \overline{a}+\lambda \left ( \hat{b} +\hat{c}\right )$$ or $$\displaystyle  \overline{r}= \overline{a}+\lambda \left ( \hat{b} +\hat{c}\right )$$ ...(i)
called the vector equation of internal bisector of $$\displaystyle \angle BAC$$ Now by replacing $$\displaystyle \overline{c}$$ to $$\displaystyle - \overline{c}$$ Then the equation (i) becomes $$\displaystyle \overline{r}=\overline{a}+\lambda \left ( \hat{b} -\hat{c}\right )$$ called the vector equation of external bisector of $$\displaystyle \angle BAC$$
On the basis of above infonnations, answer the following
The maximum value of magnitude of ( A-B)