Question

If $\vec a = 3\hat i - \hat j$ and $\vec b = 2\hat i + \hat j - 3\hat k$, then express $\vec b$ in the form of $\vec b = {\vec b_1} + {\vec b_2}$, where ${\vec b_1}\left\| {\vec a} \right.$ and ${\vec b_2}$ perpendicular to $\vec a$.

Answer 1

First we are to form the vectors from the given conditions. As given, ${\vec b_1}\left\| {\vec a} \right.$, therefore, we can say that ${\vec b_1} = \lambda \vec a$. And also, ${\vec b_2}$ perpendicular to $\vec a$, so, we know that, ${\vec b_2}.\vec a = 0$. These two conditions will give us two equations. On equating the two equations along with the third condition given, $\vec b = {\vec b_1} + {\vec b_2}$, we will get the required solution.

Complete step by step answer:
Here, given, $\vec a = 3\hat i - \hat j$ and $\vec b = 2\hat i + \hat j - 3\hat k$.
We are to express, $\vec b = {\vec b_1} + {\vec b_2} - - - \left( 1 \right)$
Given, ${\vec b_1}\left\| {\vec a} \right.$.
Therefore, ${\vec b_1}$ can be expressed as a scalar multiple of $\vec a$.
So, we can write, ${\vec b_1} = \lambda \vec a$
Now, substituting the values of $\vec a$, we get,
${\vec b_1} = \lambda \left( {3\hat i - \hat j} \right)$
$\Rightarrow {\vec b_1} = 3\lambda \hat i - \lambda \hat j$ ………….(A)
And, also given, ${\vec b_2}$ perpendicular to $\vec a$.
So, we know, if two vectors are perpendicular to each other, then their scalar product should be $0$.
Therefore, we can write, ${\vec b_2}.\vec a = 0$
Let us assume, ${\vec b_2} = x\hat i + y\hat j + z\hat k$ ……. (B)
Now, substituting the values of ${\vec b_2}$ and $\vec a$, we get,
${\vec b_2}.\vec a = 0$
$\Rightarrow (x\hat i + y\hat j + z\hat k).(3\hat i - \hat j) = 0$
$\Rightarrow 3x - y + 0 = 0$
$\Rightarrow 3x = y - - - \left( 2 \right)$
Now, substituting the corresponding values of $\vec b,{\vec b_1},{\vec b_2}$ from (A),(B) $\left( 1 \right)$, we get,
$\vec b = {\vec b_1} + {\vec b_2}$
$\Rightarrow \left( {2\hat i + \hat j - 3\hat k} \right) = \left( {3\lambda \hat i - \lambda \hat j} \right) + \left( {x\hat i + y\hat j + z\hat k} \right)$
$\Rightarrow 2\hat i + \hat j - 3\hat k = 3\lambda \hat i - \lambda \hat j + x\hat i + y\hat j + z\hat k$
Now, taking the coefficients of $\hat i,\hat j,\hat k$ together on right hand side, we get,
$\Rightarrow 2\hat i + \hat j - 3\hat k = \left( {3\lambda + x} \right)\hat i + \left( {y - \lambda } \right)\hat j + z\hat k$
Comparing the coefficients of $\hat i,\hat j,\hat k$ on both the sides, we get,
$3\lambda + x = 2 - - - \left( 3 \right)$
$y - \lambda = 1 - - - \left( 4 \right)$
$z = - 3 - - - \left( 5 \right)$
Substituting $3x=y$ which we got from $\left( 2 \right)$ in $\left( 4 \right)$, we get,
$3x - \lambda = 1 - - - \left( 6 \right)$
Now, multiplying $\left( 6 \right)$ by $3$ and adding it to $\left( 3 \right)$, we get,
$\left( {3\lambda + x} \right) + \left( {3\left( {3x - \lambda } \right)} \right) = 2 + 3\left( 1 \right)$
$\Rightarrow 3\lambda + x + 9x - 3\lambda = 2 + 3$
Cancelling, the $\lambda$ terms, we get,
$\Rightarrow 10x = 5$
$\Rightarrow x = \dfrac{1}{2}$
Substituting the value of $x$ in $\left( 2 \right)$, we get,
$y = 3.\dfrac{1}{2} = \dfrac{3}{2}$
Also, substituting the value of $x$ in $\left( 6 \right)$, we get,
$3.\dfrac{1}{2} - \lambda = 1$
$\Rightarrow \dfrac{3}{2} - \lambda = 1$
Subtracting $\dfrac{3}{2}$ on both sides, we get,
$\Rightarrow - \lambda = 1 - \dfrac{3}{2}$
$\Rightarrow - \lambda = - \dfrac{1}{2}$
Multiplying, both sides by $- 1$, we get,
$\lambda = \dfrac{1}{2}$
Therefore, $x = \dfrac{1}{2},y = \dfrac{3}{2},z = - 3,\lambda = \dfrac{1}{2}$.
Substituting these values in ${\vec b_1}$ and ${\vec b_2}$, we get,
${\vec b_1} = 3\lambda \hat i - \lambda \hat j$
$\Rightarrow {\vec b_1} = 3.\dfrac{1}{2}.\hat i - \dfrac{1}{2}.\hat j$
$\Rightarrow {\vec b_1} = \dfrac{3}{2}\hat i - \dfrac{1}{2}\hat j$
And, ${\vec b_2} = x\hat i + y\hat j + z\hat k$
$\Rightarrow {\vec b_2} = \dfrac{1}{2}.\hat i + \dfrac{3}{2}.\hat j + \left( { - 3} \right)\hat k$
$\Rightarrow {\vec b_2} = \dfrac{1}{2}\hat i + \dfrac{3}{2}\hat j - 3\hat k$
Therefore, $\vec b = \left( {\dfrac{3}{2}\hat i - \dfrac{1}{2}\hat j} \right) + \left( {\dfrac{1}{2}\hat i + \dfrac{3}{2}\hat j - 3\hat k} \right)$.

Note:
When two vectors are parallel to each other, all the points in one vector are at a certain distance from their corresponding points in the other vector. Hence, one vector can be expressed as a scalar multiple of the other vector. And, when two vectors are perpendicular to each other, their scalar product is $0$, because, in the scalar product there is a term of $\cos \theta$, where $\theta$ is the angle between the two vectors. So, $\cos 90^\circ = 0$, hence the whole product turns out to be $0$.