Question

The value of $\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$ in terms of vector $a$ .

Answer 1

Hint: In this problem we need to calculate the value of the given expression in terms of the vector $a$. So we will first assume the vector $a$ like $a={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k$ . now we will calculate the values of $a.i$ , $a.j$ and $a.k$ by using the dot product rules of the vectors. After having the values of $a.i$ , $a.j$ and $a.k$ we will substitute those values in the given expression and simplify the equation to get the required result.

Complete step by step solution:
Given expression is $\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$.
Let us assume the value of vector $a$ as $a={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k$.
Now the value of $a.i$will be written as
$\begin{align}
  & a.i=\left( {{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k \right)i \\
 & \Rightarrow a.i={{a}_{1}}\left( i.i \right)+{{a}_{2}}\left( j.i \right)+{{a}_{3}}\left( k.i \right) \\
\end{align}$
In the scalar product of vectors we have the values $i.i=1$ , $j.i=0$ , $k.i=0$ . Substituting these values in the above equation, then we will have
$\begin{align}
  & a.i={{a}_{1}}\left( 1 \right)+{{a}_{2}}\left( 0 \right)+{{a}_{3}}\left( 0 \right) \\
 & \Rightarrow a.i={{a}_{1}}....\left( \text{i} \right) \\
\end{align}$
Now the value of $a.j$will be written as
$\begin{align}
  & a.j=\left( {{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k \right)j \\
 & \Rightarrow a.j={{a}_{1}}\left( i.j \right)+{{a}_{2}}\left( j.j \right)+{{a}_{3}}\left( k.j \right) \\
\end{align}$
In the scalar product of vectors we have the values $i.j=0$ , $j.j=1$ , $k.j=0$ . Substituting these values in the above equation, then we will have
$\begin{align}
  & a.j={{a}_{1}}\left( 0 \right)+{{a}_{2}}\left( 1 \right)+{{a}_{3}}\left( 0 \right) \\
 & \Rightarrow a.j={{a}_{2}}....\left( \text{ii} \right) \\
\end{align}$
Now the value of $a.k$will be written as
$\begin{align}
  & a.k=\left( {{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k \right)k \\
 & \Rightarrow a.k={{a}_{1}}\left( i.k \right)+{{a}_{2}}\left( j.k \right)+{{a}_{3}}\left( k.k \right) \\
\end{align}$
In scalar product of vectors we have the values $i.k=0$ , $j.k=0$ , $k.k=1$ . Substituting these values in the above equation, then we will have
$\begin{align}
  & a.k={{a}_{1}}\left( 0 \right)+{{a}_{2}}\left( 0 \right)+{{a}_{3}}\left( 1 \right) \\
 & \Rightarrow a.k={{a}_{3}}....\left( \text{iii} \right) \\
\end{align}$
Substituting the values of $a.i$ , $a.j$ and $a.k$ in the given expression, then we will get
$\begin{align}
  & \left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k=\left( {{a}_{1}} \right)i+\left( {{a}_{2}} \right)j+\left( {{a}_{3}} \right)k \\
 & \Rightarrow \left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k \\
\end{align}$
We have assumed the value ${{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k$ as vector $a$ , so the above values will become as
$\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k=a$

Note: In this problem we have used the values of the dot product or scalar product of the unit vectors. In some cases we need to use the cross product or the vector product the unit vectors to solve the problem. The cross product of the unit vectors is given below
$i\times j=k$ , $j\times k=i$ , $k\times i=j$ .