Question

Express $2\widehat i + 3\widehat j + \widehat k$ as a sum of two vectors out of which one vector is perpendicular to $2\widehat i - 4\widehat j + \widehat k$ and another is parallel to $2\widehat i - 4\widehat j + \widehat k$

Answer 1

In order to solve this question, we will find the vector which is parallel to a vector by using the Parallel condition of two vectors. Then we will find another vector which is perpendicular to the same vector by using the Perpendicularity condition of two vectors. After that we will write the given vector as the sum of the two vectors obtained in both the conditions to get the required result.

Formula used:
We will use the following formulas:

1. If $\overrightarrow {{a_1}}$ is parallel to $\overrightarrow b$ ,then the equation of line in vector form is given by $\overrightarrow {{a_1}} = \lambda \overrightarrow b$ where $\lambda$ is a scalar constant.
2. If $\overrightarrow {{a_2}}$ is perpendicular to $\overrightarrow b$ ,then $\overrightarrow {{a_2}} \cdot \overrightarrow b = 0$

Complete step by step answer:
Let the given vector be $\overrightarrow a$
${\text{ }}\overrightarrow a = 2\widehat i + 3\widehat j + \widehat k$
Also let $\overrightarrow b = 2\widehat i - 4\widehat j + \widehat k$
Now we will express the vector $\overrightarrow a$ as a sum of two vectors which is parallel to $\overrightarrow b = 2\widehat i - 4\widehat j + \widehat k$ and another is perpendicular to $\overrightarrow b = 2\widehat i - 4\widehat j + \widehat k$
Let $\overrightarrow {{a_1}}$ is a vector which is parallel to $\overrightarrow b$ and $\overrightarrow {{a_2}}$ is a vector which is perpendicular to $\overrightarrow b$
Therefore,
$\overrightarrow a = \overrightarrow {{a_1}} + \overrightarrow {{a_2}} {\text{ }} - - - \left( i \right)$
Now firstly we will find the vector $\overrightarrow {{a_1}}$ which is parallel to $\overrightarrow b$
Since $\overrightarrow {{a_1}}$ is parallel to $\overrightarrow b$
Therefore, by equation of line in vector form, $\overrightarrow {{a_1}} = \lambda \overrightarrow b$ where $\lambda$ is a scalar constant.
$\Rightarrow \overrightarrow {{a_1}} = \lambda \left( {2\widehat i - 4\widehat j + \widehat k} \right)$
Now by distributing the scalar quantity, we get
$\Rightarrow \overrightarrow {{a_1}} = 2\lambda \widehat i - 4\lambda \widehat j + \lambda \widehat k{\text{ }} - - - \left( {ii} \right)$
Now we will find the vector $\overrightarrow {{a_2}}$ which is perpendicular to $\overrightarrow b$ by using the equation $\left( i \right)$
So, by substituting equation $\left( {ii} \right)$ and value of $\overrightarrow a$ in the equation $\left( i \right)$ we get
$2\widehat i + 3\widehat j + \widehat k = 2\lambda \widehat i - 4\lambda \widehat j + \lambda \widehat k + \overrightarrow {{a_2}}$
By rewriting the equation, we get
$\Rightarrow \overrightarrow {{a_2}} = 2\widehat i + 3\widehat j + \widehat k - 2\lambda \widehat i + 4\lambda \widehat j - \lambda \widehat k{\text{ }}$
$\Rightarrow \overrightarrow {{a_2}} = \left( {2 - 2\lambda } \right)\widehat i + \left( {3 + 4\lambda } \right)\widehat j + \left( {1 - \lambda } \right)\widehat k{\text{ }} - - - \left( {iii} \right)$
Since $\overrightarrow {{a_2}}$ is perpendicular to $\overrightarrow b$
Therefore, by perpendicularity conditions we have $\overrightarrow {{a_2}} \cdot \overrightarrow b = 0$
$\Rightarrow \left( {\left( {2 - 2\lambda } \right)\widehat i + \left( {3 + 4\lambda } \right)\widehat j + \left( {1 - \lambda } \right)\widehat k} \right) \cdot \left( {2\widehat i - 4\widehat j + \widehat k} \right) = 0$
Now by multiplying the vectors, we get
$\Rightarrow 2\left( {2 - 2\lambda } \right) + \left( { - 4} \right)\left( {3 + 4\lambda } \right) + \left( 1 \right)\left( {1 - \lambda } \right) = 0$
By multiplying the terms, we get
$\Rightarrow 4 - 4\lambda - 12 - 16\lambda + 1 - \lambda = 0$
On simplifying, we get
$\Rightarrow - 21\lambda - 7 = 0$
$\Rightarrow \lambda = \dfrac{{ - 1}}{3}$
Now by substituting $\lambda = \dfrac{{ - 1}}{3}$ in equation $\left( {ii} \right)$ we get
$\Rightarrow \overrightarrow {{a_1}} = 2\left( {\dfrac{{ - 1}}{3}} \right)\widehat i - 4\left( {\dfrac{{ - 1}}{3}} \right)\widehat j + \left( {\dfrac{{ - 1}}{3}} \right)\widehat k$
$\Rightarrow \overrightarrow {{a_1}} = \dfrac{{ - 1}}{3}\left( {2\widehat i - 4\widehat j + \widehat k} \right)$
Now by substituting $\lambda = \dfrac{{ - 1}}{3}$ in equation $\left( {iii} \right)$ we get
$\Rightarrow \overrightarrow {{a_2}} = \left( {2 - 2\left( {\dfrac{{ - 1}}{3}} \right)} \right)\widehat i + \left( {3 + 4\left( {\dfrac{{ - 1}}{3}} \right)} \right)\widehat j + \left( {1 - \left( {\dfrac{{ - 1}}{3}} \right)} \right)\widehat k$
On simplifying the equation, we get
$\Rightarrow \overrightarrow {{a_2}} = \left( {2 + \dfrac{2}{3}} \right)\widehat i + \left( {3 - \dfrac{4}{3}} \right)\widehat j + \left( {1 + \dfrac{1}{3}} \right)\widehat k$
On simplifying the terms, we get
$\Rightarrow \overrightarrow {{a_2}} = \dfrac{8}{3}\widehat i + \dfrac{5}{3}\widehat j + \dfrac{4}{3}\widehat k$
Now by substituting $\overrightarrow {{a_1}}$ and $\overrightarrow {{a_2}}$ in equation $\left( i \right)$ we get
$\overrightarrow a = \dfrac{{ - 1}}{3}\left( {2\widehat i - 4\widehat j + \widehat k} \right) + \left( {\dfrac{8}{3}\widehat i + \dfrac{5}{3}\widehat j + \dfrac{4}{3}\widehat k} \right)$
Therefore, vector $\overrightarrow a$ can be expressed as sum of two vectors such as $\overrightarrow {{a_1}} = \dfrac{{ - 1}}{3}\left( {2\widehat i - 4\widehat j + \widehat k} \right)$ and $\overrightarrow {{a_2}} = \dfrac{8}{3}\widehat i + \dfrac{5}{3}\widehat j + \dfrac{4}{3}\widehat k$.

Note:
While solving these types of questions, we must know that two vectors are said to be parallel, if and only if both the vectors are scalar multiples of one another and two vectors are said to be perpendicular if and only if their scalar product is zero. Also remember when we are adding the two vectors which are parallel and perpendicular, we must get the given vector.