Vectors

goals
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
questions
The maximum value of magnitude of ( A-B)