Question

If $\overrightarrow a = 2\widehat i - \widehat j - 2\widehat k$ and $\overrightarrow b = 7\widehat i + 2\widehat j - 3\widehat k$, then express $\overrightarrow b $ in the form of $\overrightarrow b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}} $, where $\overrightarrow {{b_1}} $ is parallel to $\overrightarrow {{a}} $ and $\overrightarrow{{b_2}} $ perpendicular to $\overrightarrow a $.

Answer 1

Hint:
In vector addition, you can directly add the two components. But, remember you can perform $\widehat i + \widehat i = 2\widehat i$ but you cannot add \[\widehat i + \widehat j\] as it is not possible. $\overrightarrow {{b_1}} $ parallel to vector a means $\overrightarrow {{b_1}} = \lambda \overrightarrow {{a_1}} $ and $\overrightarrow {{b_2}} $ is perpendicular to vector a means \[\overrightarrow {{b_2}} .\,\overrightarrow a = 0\].

Stepwise solution:
Given:
$\overrightarrow a = 2\widehat i - \widehat j - 2\widehat k$ and $\overrightarrow b = 7\widehat i + 2\widehat j - 3\widehat k$ .
 $\overrightarrow {{b_1}} $ is parallel to vector a and $\overrightarrow {{b_2}} $ is perpendicular to vector a.
Here,
$\overrightarrow a = 2\widehat i - \widehat j - 2\widehat k$
$\overrightarrow b = 7\widehat i + 2\widehat j - 3\widehat k$
To express $\overrightarrow b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}} $, we have to proceed according to the given condition ${b_1}||\,\overrightarrow a $.
Therefore, $\overrightarrow {{b_1}} = \lambda \overrightarrow {{a_1}} $
Therefore, we can now say
\[\overrightarrow {{b_1}} = \lambda (2\widehat i - \widehat j - 2\widehat k)\] eq. (1)
Now, let us assume $\overrightarrow {{b_2}} $ as
\[\overrightarrow {{b_2}} = (x\widehat i + y\widehat j + z\widehat k)\]
Now according to given condition $\overrightarrow {{b_2}} $ perpendicular to vector a.
Therefore, $\overrightarrow {{b_2}} .\overrightarrow a = 0$
Therefore, \[(x\widehat i + y\widehat j + z\widehat k)\,.\,(2\widehat i - \widehat j - 2\widehat k) = 0\]
$2x - y - 2z = 0$ eq. (2)
Now, equating $\overrightarrow b $ with the value of summation of $\overrightarrow {{b_1}} $ and $\overrightarrow {{b_2}} $, we will get result as
$\overrightarrow b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}} $
$ \Rightarrow 7\widehat i + 2\widehat j - 3\widehat k = \lambda (2\widehat i - \widehat j - 2\widehat k) + (x\widehat i + y\widehat j + x\widehat k)$
$ = (2\lambda + x)\widehat i - (\lambda - y)\widehat j - (2\lambda - z)\widehat k$
Completing the corresponding components,
$7 = (2\lambda + x)$ eq. (3)
$2 = - (\lambda - y)$ eq. (4)
$ - 3 = - (2\lambda - z)$ eq. (5)
And, from the perpendicular condition, we have
$2x - y - 2x = 0$ eq. (2)
Taking eq. (3), eq. (4) and eq. (5) and putting the values of x, y and z in eq. (2), we get
$x = 7 - 2\lambda $
$y = 2 + \lambda $
$x = 2\lambda - 3$
Therefore, the equation (2) becomes,
$2(7 - 2\lambda ) - (2 + \lambda ) - 2(2\lambda - 3) = 0$
$ \Rightarrow 14 - 4\lambda - 2 - \lambda - 4\lambda + 6 = 0$
$ \Rightarrow 20 - 2 - \lambda = 0$
$ \Rightarrow \, - 9\lambda = - 18$
Therefore, $\lambda = 2$
Now, putting the values of $\lambda $ in eq. (3) and eq. (4) and eq. (5) we will get the values of x, y and z.
Now,
$7(2 \times 2 + x)$
$ \Rightarrow \,x = 7 - 4$
$\therefore \,x = 3$
Again from eq. (4)
$ - 3z - (2\lambda - z)$
$ \Rightarrow \,z = 4 - 3$
Therefore, $z = 1$
Therefore, the value of x, y and z are 3, 4 and 1.
Therefore, $\overrightarrow b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}} $, is
\[\overrightarrow {{b_1}} = \lambda (2\widehat i - \widehat j - 2\widehat k)\]
\[\overrightarrow {{b_2}} = (x\widehat i + y\widehat j + z\widehat k)\]
Thus,
$\overrightarrow b = 2(2\widehat i - \widehat j - 2\widehat k)\, + \,3\widehat i + 4\widehat j + \widehat k$
Hence the form is represented.

Note: In this type of questions, students often get confused regarding the process of approach. Therefore, read the question carefully to get the desired answer. Also, the coordinates of co-linearity and parallel are both the same.