Given a → = 3 i ^ - j ^ , b → = 2 i ^ + j ^ - 3 k ^ and b → = b 1 → + b 2 → ,…

Given:- $\overrightarrow{a}=3\hat{i}-\hat{j}$

and $\overrightarrow{b}=2\hat{i}+\hat{j}-3\hat{k}$

Also, $\overrightarrow{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$ where $\overrightarrow{b_1}\parallel \overrightarrow{a}$

$\because \overrightarrow{b_1}\parallel \overrightarrow{a}$

$\therefore \overrightarrow{b_1}=\lambda \left(3\hat{i}-\hat{j}\right)$

Let $\overrightarrow{b_2}=x\hat{i}+y\hat{j}+z\hat{k}$

$\because \overrightarrow{b_2}\perp \overrightarrow{a}$

$\therefore \overrightarrow{b_2}.\overrightarrow{a}=0$

$\Rightarrow 3x-y=0 ...\left(1\right)$

As, $\overrightarrow{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$

$2\hat{i}+\hat{j}-3\hat{k}=\left(3\lambda +x\right)\hat{i}+\left(y-\lambda \right)\hat{j}+z\hat{k}$

Comparing the corresponding components, we get

$2=3\lambda +x ...\left(2\right)$

$1=-\lambda +y$

$\therefore \lambda =y-1 ...\left(3\right)$

&  $-3=z ...\left(4\right)$

from equation $\left(2\right), 2=3\left(y-1\right)+x$

$\Rightarrow x+3y=5 ...\left(5\right)$

equation $\left(1\right)\times 3$& equation $\left(5\right)$, we get

$10x=5$

$\therefore x=\frac{1}{2}$

Substituting $x=\frac{1}{2}$in $\left(1\right)$, we get

$y=\frac{3}{2}$

$\because \overrightarrow{b_2}=x\hat{i}+y\hat{j}+z\hat{k}$

$\therefore \overrightarrow{b_2}=\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$