Question
In vector combination, commutative and associative rules are followed. Explain.
✓
Answer
(i) Permutation rule of vector combination : If vectors $\vec{a}$ and $\vec{b}$ are two vectors then,
$\vec{a}+\vec{b}=\vec{b}+\vec{a}$
Proof : Let $\vec{a}$ and $\vec{b}$ are denoted by $\overrightarrow{ OA }$ and $\overrightarrow{ OB }$ respectively. Hence,
$\overrightarrow{ OA }=\vec{a}$ and $\overrightarrow{ AB }=\vec{b}$
By vector addition rule of triangle,
$\begin{array}{l}
\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB} \\
\overrightarrow{OB}=\vec{a}+\vec{b} \ldots(1)
\end{array}$
The parallelogram completes OABC whose two adjacent sides are $O A$ and $A B$ :$
\begin{array}{l}
\overrightarrow{CD}=\overrightarrow{OA}=\vec{a} \\
\overrightarrow{OC}=\overrightarrow{AB}=\vec{b}
\end{array}$
Again, by triangle law of vector addition
$\begin{array}{l}
\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}=\vec{b}+\vec{a} \\
\overrightarrow{OB}=\vec{b}+\vec{a} \ \ldots(2)
\end{array}$
From equations (1) and (2)
$\vec{a}+\vec{b}=\vec{b}+\vec{a}$So, this proves the rule.
(ii) Associative rule in vector addition : If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors.
Then, $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
Proof : Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are represented by $\overrightarrow{ OA }, \overrightarrow{ AB }$ and $\overrightarrow{ BC }$ respectively.
Hence, $\overrightarrow{ OA }=\vec{a}, \overrightarrow{ AB }=\vec{b}, \overrightarrow{ BC }=\vec{c}$
By applying triangle law of vector addition in $\triangle OAB$ and $\triangle OBC , \overrightarrow{ OB }=\overrightarrow{ OA }+\overrightarrow{ AB }$
$\begin{array}{l}
\overrightarrow{OB}=\vec{a}+\vec{b} \\
\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}
\end{array}$$=(\vec{a}+\vec{b})+\vec{c} \ \ldots(1)$
In $\triangle ABC$ and $\triangle OAC$, we get
$\begin{aligned}
\overrightarrow{AC} & =\overrightarrow{AB}+\overrightarrow{BC}=\vec{b}+\vec{c} \\
\overrightarrow{OC} & =\overrightarrow{OA}+\overrightarrow{AC} \\
& =\vec{a}+(\vec{b}+\vec{c}) \ \ldots(2)
\end{aligned}$
From equations (1) and (2)
$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
$\vec{a}+\vec{b}=\vec{b}+\vec{a}$
Proof : Let $\vec{a}$ and $\vec{b}$ are denoted by $\overrightarrow{ OA }$ and $\overrightarrow{ OB }$ respectively. Hence,
$\overrightarrow{ OA }=\vec{a}$ and $\overrightarrow{ AB }=\vec{b}$
By vector addition rule of triangle,
$\begin{array}{l}
\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB} \\
\overrightarrow{OB}=\vec{a}+\vec{b} \ldots(1)
\end{array}$
The parallelogram completes OABC whose two adjacent sides are $O A$ and $A B$ :$
\begin{array}{l}
\overrightarrow{CD}=\overrightarrow{OA}=\vec{a} \\
\overrightarrow{OC}=\overrightarrow{AB}=\vec{b}
\end{array}$
Again, by triangle law of vector addition
$\begin{array}{l}
\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}=\vec{b}+\vec{a} \\
\overrightarrow{OB}=\vec{b}+\vec{a} \ \ldots(2)
\end{array}$
From equations (1) and (2)
$\vec{a}+\vec{b}=\vec{b}+\vec{a}$So, this proves the rule.
(ii) Associative rule in vector addition : If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors.
Then, $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
Proof : Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are represented by $\overrightarrow{ OA }, \overrightarrow{ AB }$ and $\overrightarrow{ BC }$ respectively.
Hence, $\overrightarrow{ OA }=\vec{a}, \overrightarrow{ AB }=\vec{b}, \overrightarrow{ BC }=\vec{c}$
By applying triangle law of vector addition in $\triangle OAB$ and $\triangle OBC , \overrightarrow{ OB }=\overrightarrow{ OA }+\overrightarrow{ AB }$
$\begin{array}{l}
\overrightarrow{OB}=\vec{a}+\vec{b} \\
\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}
\end{array}$$=(\vec{a}+\vec{b})+\vec{c} \ \ldots(1)$
In $\triangle ABC$ and $\triangle OAC$, we get
$\begin{aligned}
\overrightarrow{AC} & =\overrightarrow{AB}+\overrightarrow{BC}=\vec{b}+\vec{c} \\
\overrightarrow{OC} & =\overrightarrow{OA}+\overrightarrow{AC} \\
& =\vec{a}+(\vec{b}+\vec{c}) \ \ldots(2)
\end{aligned}$
From equations (1) and (2)
$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
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