Question
Prove that: Vector addition is associative.
✓
Answer
Associative property of vector addition: According to associative property, for three vectors $\vec{P}, \vec{Q}$ and $\vec{R}$,
$(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })$
Proof:
i. Let $\overrightarrow{ OA }=\overrightarrow{ P }, \overrightarrow{ AB }=\overrightarrow{ Q }, \overrightarrow{ BC }=\overrightarrow{ R }$
ii. Join OB and AC
In $\triangle OAB$,
$\overrightarrow{ OA }+\overrightarrow{ AB }=\overrightarrow{ OB }$ (From triangle law of vector addition)
$\therefore \quad \overrightarrow{ P }+\overrightarrow{ Q }=\overrightarrow{ R }_1$ ....(1)
In $\triangle OBC$,
$\overrightarrow{ OB }+\overrightarrow{ BC }=\overrightarrow{ OC }$(From triangle law of vector addition)
$\therefore \quad \vec{R}_1+\vec{R}=\vec{S}$
From equation (1)
$(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ S }$ ....(2)
$
\begin{array}{ll}
\text { iii. } & \text { In } \Delta A B C, \\
& \overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\
\therefore \quad & \vec{Q}+\vec{R}=\vec{R}_2 .....(3)
\end{array}
$
$
\begin{array}{ll}
\text { iv. } \quad & \text { In } \triangle OAC \\
& \overrightarrow{ OA }+\overrightarrow{ AC }=\overrightarrow{ OC } \\
\therefore \quad & \overrightarrow{ P }+\overrightarrow{ R }_2=\overrightarrow{ S }
\end{array}
$
From equation (3)
$\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })=\overrightarrow{ S }$ .......(4)
On comparing, equation (2) and (4), we get,
$
(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })
$
Hence, associative law is proved.
$(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })$
Proof:
i. Let $\overrightarrow{ OA }=\overrightarrow{ P }, \overrightarrow{ AB }=\overrightarrow{ Q }, \overrightarrow{ BC }=\overrightarrow{ R }$
ii. Join OB and AC
In $\triangle OAB$,
$\overrightarrow{ OA }+\overrightarrow{ AB }=\overrightarrow{ OB }$ (From triangle law of vector addition)
$\therefore \quad \overrightarrow{ P }+\overrightarrow{ Q }=\overrightarrow{ R }_1$ ....(1)
In $\triangle OBC$,
$\overrightarrow{ OB }+\overrightarrow{ BC }=\overrightarrow{ OC }$(From triangle law of vector addition)
$\therefore \quad \vec{R}_1+\vec{R}=\vec{S}$
From equation (1)
$(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ S }$ ....(2)
$
\begin{array}{ll}
\text { iii. } & \text { In } \Delta A B C, \\
& \overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\
\therefore \quad & \vec{Q}+\vec{R}=\vec{R}_2 .....(3)
\end{array}
$
$
\begin{array}{ll}
\text { iv. } \quad & \text { In } \triangle OAC \\
& \overrightarrow{ OA }+\overrightarrow{ AC }=\overrightarrow{ OC } \\
\therefore \quad & \overrightarrow{ P }+\overrightarrow{ R }_2=\overrightarrow{ S }
\end{array}
$
From equation (3)
$\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })=\overrightarrow{ S }$ .......(4)
On comparing, equation (2) and (4), we get,
$
(\overrightarrow{ P }+\overrightarrow{ Q })+\overrightarrow{ R }=\overrightarrow{ P }+(\overrightarrow{ Q }+\overrightarrow{ R })
$
Hence, associative law is proved.
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