Question

If we have a vector as $\overrightarrow{a}=3\hat{i}-\hat{j}$ and $\overrightarrow{b}=2\hat{i}+\hat{j}-3\hat{k}$, then express $\overrightarrow{b}$ in the form $\overrightarrow{b}=\overrightarrow{{{b}_{1}}}+\overrightarrow{{{b}_{2}}}$ where $\overrightarrow{{{b}_{1}}}\parallel \overrightarrow{a}$ and $\overrightarrow{{{b}_{2}}}\bot \overrightarrow{a}$.

Answer 1

We start solving the problem by recalling the concept of parallel vectors. We use this to find the general form of the vector $\overrightarrow{{{b}_{1}}}$. We then assume a vector for $\overrightarrow{{{b}_{2}}}$ and recall the dot product of perpendicular vectors to find the relation between the coefficients in $\overrightarrow{{{b}_{2}}}$. We use this values in $\overrightarrow{b}=\overrightarrow{{{b}_{1}}}+\overrightarrow{{{b}_{2}}}$ and equate corresponding coefficients to get the required vectors to express $\overrightarrow{b}$ in the given form.

Complete step-by-step solution:
According to the problem, we have given vectors $\overrightarrow{a}=3\hat{i}-\hat{j}$ and $\overrightarrow{b}=2\hat{i}+\hat{j}-3\hat{k}$. We need to express $\overrightarrow{b}$ in the form $\overrightarrow{b}=\overrightarrow{{{b}_{1}}}+\overrightarrow{{{b}_{2}}}$, where $\overrightarrow{{{b}_{1}}}\parallel \overrightarrow{a}$ and $\overrightarrow{{{b}_{2}}}\bot \overrightarrow{a}$.
We know that if the vector $\overrightarrow{x}$ is parallel to another vector $\overrightarrow{y}$, then $\overrightarrow{x}=d\overrightarrow{y}$, when d is a scalar.
Since we have $\overrightarrow{{{b}_{1}}}$ parallel to $\overrightarrow{a}$, we get $\overrightarrow{{{b}_{1}}}=d\overrightarrow{a}$.
$\Rightarrow \overrightarrow{{{b}_{1}}}=d\left( 3\hat{i}-\hat{j} \right)$.
$\Rightarrow \overrightarrow{{{b}_{1}}}=3d\hat{i}-d\hat{j}$ ---(1).
Let us assume the vector $\overrightarrow{{{b}_{2}}}$ be $a\hat{i}+b\hat{j}+c\hat{k}$. According to the problem we have given that $\overrightarrow{{{b}_{2}}}$ is perpendicular to $\overrightarrow{a}$.
We know that if the vector $p\hat{i}+q\hat{j}+r\hat{k}$ is perpendicular to $s\hat{i}+t\hat{j}+u\hat{k}$, then $ps+qt+ru=0$.
So, we have $a\hat{i}+b\hat{j}+c\hat{k}$ perpendicular to $3\hat{i}-\hat{j}$.
We get $\left( a\times 3 \right)+\left( b\times -1 \right)+\left( c\times 0 \right)=0$.
$\Rightarrow \left( 3a \right)+\left( -b \right)+0=0$.
$\Rightarrow 3a-b=0$.
$\Rightarrow b=3a$ ---(2).
Now, we have $\overrightarrow{b}=\overrightarrow{{{b}_{1}}}+\overrightarrow{{{b}_{2}}}$.
From equation (1) we get,
$\Rightarrow 2\hat{i}+\hat{j}-3\hat{k}=\left( 3d\hat{i}-d\hat{j} \right)+\left( a\hat{i}+b\hat{j}+c\hat{k} \right)$.
From equation (2) we get,
$\Rightarrow 2\hat{i}+\hat{j}-3\hat{k}=\left( 3d\hat{i}-d\hat{j} \right)+\left( a\hat{i}+3a\hat{j}+c\hat{k} \right)$ ---(3).
$\Rightarrow 2\hat{i}+\hat{j}-3\hat{k}=\left( 3d+a \right)\hat{i}+\left( -d+3a \right)\hat{j}+c\hat{k}$.
Equating corresponding coefficients on both sides we get,
$\Rightarrow 3d+a=2$ ---(4).
$\Rightarrow -d+3a=1$ ---(5).
$\Rightarrow c=-3$ ---(6).
From equation (5), we have $-d+3a=1$.
$\Rightarrow d=3a-1$ ---(7).
We substitute equation (7) in equation (4).
$\Rightarrow 3\left( 3a-1 \right)+a=2$.
$\Rightarrow 9a-3+a=2$.
$\Rightarrow 10a=2+3$.
$\Rightarrow 10a=5$.
$\Rightarrow a=\dfrac{5}{10}$.
$\Rightarrow a=\dfrac{1}{2}$ ---(8). We substitute this in equation (7).
$\Rightarrow d=3\left( \dfrac{1}{2} \right)-1$.
$\Rightarrow d=\dfrac{3}{2}-1$.
$\Rightarrow d=\dfrac{1}{2}$ ---(9).
Substituting equation (6), (8), and (9) in equation (3), we get
$\Rightarrow 2\hat{i}+\hat{j}-3\hat{k}=\left( 3\left( \dfrac{1}{2} \right)\hat{i}-\left( \dfrac{1}{2} \right)\hat{j} \right)+\left( \left( \dfrac{1}{2} \right)\hat{i}+3\left( \dfrac{1}{2} \right)\hat{j}-3\hat{k} \right)$.
$\Rightarrow 2\hat{i}+\hat{j}-3\hat{k}=\left( \left( \dfrac{3}{2} \right)\hat{i}-\left( \dfrac{1}{2} \right)\hat{j} \right)+\left( \left( \dfrac{1}{2} \right)\hat{i}+\left( \dfrac{3}{2} \right)\hat{j}-3\hat{k} \right)$.
$\therefore$ We have got expressed $\overrightarrow{b}$ in the form $\overrightarrow{b}=\overrightarrow{{{b}_{1}}}+\overrightarrow{{{b}_{2}}}$ where $\overrightarrow{{{b}_{1}}}\parallel \overrightarrow{a}$ and $\overrightarrow{{{b}_{2}}}\bot \overrightarrow{a}$, where $\overrightarrow{{{b}_{1}}}=\left( \dfrac{3}{2} \right)\hat{i}-\left( \dfrac{1}{2} \right)\hat{j}$ and $\overrightarrow{{{b}_{2}}}=\left( \dfrac{1}{2} \right)\hat{i}+\left( \dfrac{3}{2} \right)\hat{j}-3\hat{k}$.

Note: We should not confuse with the concepts of parallel and perpendicular vectors. We should know that the direction ratios of the parallel vectors are the same (or) scalar multiples to each other. We can also solve this by converting the given vectors into the points in a 3-dimensional form and make necessary calculations to get the required answer. Similarly, we can also expect problems to express vectors in terms of projection vector.