Question

How do you write the vector $a = 3i + 2j - 6k$ as a sum of two vectors, one parallel and one perpendicular to $d = 2i - 4j + k$?

Answer 1

Here, 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. We will then write the given vector as the sum of the two vectors obtained in both the condition to get the required answer.

Formula Used:
We will use the following formula:
1. If $\overrightarrow {{a_1}}$ is parallel to $\overrightarrow d$, then the equation of line in vector form is given by $\overrightarrow {{a_1}} = \lambda \overrightarrow d$ where $\lambda$ is a scalar constant.
2. If $\overrightarrow {{a_2}}$ is perpendicular to $\overrightarrow d$, then $\overrightarrow {{a_2}} \cdot \overrightarrow d = 0$

Complete step by step solution:
We are given with a vector $a = 3i + 2j - 6k$
Now, we will express the vector $a$ as a sum of two vectors which is parallel to $d = 2i - 4j + k$ and a vector $a = 3i + 2j - 6k$ which is perpendicular to $d = 2i - 4j + k$.
Let$\overrightarrow {{a_1}}$ is a vector which is parallel to $\overrightarrow d$ and $\overrightarrow {{a_2}}$ is a vector which is perpendicular to $\overrightarrow d$.
$\overrightarrow a = \overrightarrow {{a_1}} + \overrightarrow {{a_2}}$ …………………………………………$\left( 1 \right)$
Now, we will find the vector $\overrightarrow {{a_1}}$ which is parallel to $\overrightarrow d$.
Since $\overrightarrow {{a_1}}$ which is parallel to $\overrightarrow d$, Equation of line in vector form is given by $\overrightarrow {{a_1}} = \lambda \overrightarrow d$ where $\lambda$ is a scalar constant.
Now, by substituting $\overrightarrow d$ in $\overrightarrow {{a_1}} = \lambda \overrightarrow d$, we get
$\overrightarrow {{a_1}} = \lambda \left( {2i - 4j + k} \right)$
By distributing the scalar quantity, we get
$\Rightarrow \overrightarrow {{a_1}} = 2\lambda i - 4\lambda j + \lambda k$…………………………………$\left( 2 \right)$
Now, we will find the vector $\overrightarrow {{a_2}}$ which is perpendicular to $\overrightarrow d$ by using equation $\left( 1 \right)$.
By substituting equation $\left( 2 \right)$ and value of $\overrightarrow a$ in the equation $\left( 1 \right)$, we get
$3i + 2j - 6k = 2\lambda i - 4\lambda j + \lambda k + \overrightarrow {{a_2}}$
By rewriting the equation, we get
$\Rightarrow \overrightarrow {{a_2}} = 3i + 2j - 6k - 2\lambda i + 4\lambda j - \lambda k$
$\Rightarrow \overrightarrow {{a_2}} = \left( {3 - 2\lambda } \right)i + \left( {2 + 4\lambda } \right)j + \left( { - 6 - \lambda } \right)k$……………………………..$\left( 3 \right)$
Now, we will find the scalar constant $\lambda$ by using the perpendicularity condition of vectors.
Since$\overrightarrow {{a_2}}$ is perpendicular to $\overrightarrow d$, $\overrightarrow {{a_2}} \cdot \overrightarrow d = 0$
Now, by substituting $\overrightarrow d$ and $\overrightarrow {{a_2}}$ in $\overrightarrow {{a_2}} \cdot \overrightarrow d = 0$, we get
$\left( {\left( {3 - 2\lambda } \right)i + \left( {2 + 4\lambda } \right)j + \left( { - 6 - \lambda } \right)k} \right) \cdot \left( {2i - 4j + k} \right) = 0$
Now, by multiplying the vectors, we get
$\Rightarrow \left( {3 - 2\lambda } \right) \cdot 2 + \left( {2 + 4\lambda } \right) \cdot \left( { - 4} \right) + \left( { - 6 - \lambda } \right) \cdot 1 = 0$
Now, by multiplying the terms, we get
$\Rightarrow 6 - 4\lambda - 8 - 16\lambda - 6 - \lambda = 0$
Adding and subtracting like terms, we get
$\Rightarrow - 8 - 21\lambda = 0$
Adding 8 on both the sides, we get
$\Rightarrow - 21\lambda = 8$
Dividing both sides by $- 21$, we get
$\Rightarrow \lambda = - \dfrac{8}{{21}}$
Now, by substituting $\lambda = - \dfrac{8}{{21}}$in equation$\left( 2 \right)$, we get
$\overrightarrow {{a_1}} = 2\left( { - \dfrac{8}{{21}}} \right)i - 4\left( { - \dfrac{8}{{21}}} \right)j + \left( { - \dfrac{8}{{21}}} \right)k$
Now, by simplifying the equation, we get
$\Rightarrow \overrightarrow {{a_1}} = \dfrac{1}{{21}}\left( { - 16i + 32j - 8k} \right)$
Now, by substituting $\lambda = - \dfrac{8}{{21}}$in equation$\left( 3 \right)$, we get
$\overrightarrow {{a_2}} = \left( {3 - 2\left( { - \dfrac{8}{{21}}} \right)} \right)i + \left( {2 + 4\left( { - \dfrac{8}{{21}}} \right)} \right)j + \left( { - 6 - \left( { - \dfrac{8}{{21}}} \right)} \right)k$
Simplifying the equation, we get
$\Rightarrow \overrightarrow {{a_2}} = \left( {\dfrac{{63 + 16}}{{21}}} \right)i + \left( {\dfrac{{42 - 32}}{{21}}} \right)j + \left( {\dfrac{{ - 126 + 8}}{{21}}} \right)k$
Now, by simplifying the terms, we get
$\Rightarrow \overrightarrow {{a_2}} = \dfrac{1}{{21}}\left( {79i + 10j - 118k} \right)$
By substituting $\overrightarrow {{a_1}}$ and $\overrightarrow {{a_2}}$ in equation $\left( 1 \right)$, we get
$\Rightarrow \overrightarrow a = \left( {\dfrac{1}{{21}}\left( { - 16i + 32j - 8k} \right)} \right) + \left( {\dfrac{1}{{21}}\left( {79i + 10j - 118k} \right)} \right)$

Therefore, a vector $\overrightarrow a$ can be expressed as a sum of two vectors $\dfrac{1}{{21}}\left( { - 16i + 32j - 8k} \right)$ and $\dfrac{1}{{21}}\left( {79i + 10j - 118k} \right)$

Note:
We know that a vector is an object which has both a magnitude and a direction. We should 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. Scalar product is also called a dot product of two vectors. When we are adding the vectors which are parallel and perpendicular to a common vector, we should get the given vector if both the vectors are correct.