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In $\triangle O A C$, if $B$ is the mid-point of side $A C$ and $\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}$, then $\mathbf{O C}$ is equal to
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Verified Answer
The correct answer is:
$2 b-a$
Given, $\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}$
$$
\begin{aligned}
& \text { In } \triangle O A B, \mathbf{O A}+\mathbf{A B}+\mathbf{B O}=0 \\
& \begin{aligned}
\Rightarrow \quad \mathbf{A B} & =-\mathbf{a}-\mathbf{B O}=-\mathbf{a}-(\mathbf{b}) \\
& =-\mathbf{a}+\mathbf{b} \quad[\because \mathbf{O B}=\mathbf{b} \quad \therefore \mathbf{B O}=-\mathbf{b}] \\
\because \quad \mathbf{A C} & =2 \mathbf{A B}=2 \mathbf{b}-2 \mathbf{a}
\end{aligned}
\end{aligned}
$$
Now, in $\triangle O A C$,
$$
\begin{aligned}
& & \mathbf{O A}+\mathbf{A C}+\mathbf{C O} & =0 \\
\Rightarrow & & \mathbf{a}+2 \mathbf{b}-2 \mathbf{a}-\mathbf{O C} & =0 \\
\Rightarrow & & \mathbf{O C} & =2 \mathbf{b}-\mathbf{a}
\end{aligned}
$$
$$
\begin{aligned}
& \text { In } \triangle O A B, \mathbf{O A}+\mathbf{A B}+\mathbf{B O}=0 \\
& \begin{aligned}
\Rightarrow \quad \mathbf{A B} & =-\mathbf{a}-\mathbf{B O}=-\mathbf{a}-(\mathbf{b}) \\
& =-\mathbf{a}+\mathbf{b} \quad[\because \mathbf{O B}=\mathbf{b} \quad \therefore \mathbf{B O}=-\mathbf{b}] \\
\because \quad \mathbf{A C} & =2 \mathbf{A B}=2 \mathbf{b}-2 \mathbf{a}
\end{aligned}
\end{aligned}
$$
Now, in $\triangle O A C$,
$$
\begin{aligned}
& & \mathbf{O A}+\mathbf{A C}+\mathbf{C O} & =0 \\
\Rightarrow & & \mathbf{a}+2 \mathbf{b}-2 \mathbf{a}-\mathbf{O C} & =0 \\
\Rightarrow & & \mathbf{O C} & =2 \mathbf{b}-\mathbf{a}
\end{aligned}
$$
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