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If in the given figure $\overrightarrow{O A}=\mathbf{a}, \overrightarrow{O B}=\mathbf{b}$ and $A P: P B=m: n$, then $\overrightarrow{O P}=$
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The correct answer is:
$\frac{m \mathbf{a}+n \mathbf{b}}{m+n}$
$\begin{aligned} \therefore & \frac{A P}{P B}=\frac{n}{m} \\ \therefore & \frac{P B}{A P}=\frac{m}{n} \\ \therefore & \frac{P B+A P}{A P}=\frac{m+n}{n} \\ \therefore & A P=\frac{n}{m+n} \times(p B+A P)\end{aligned}$
$\begin{aligned} & =\frac{n}{m+n}(\overrightarrow{O B}-O \vec{A}) \\ & =\frac{n}{m+n}(b-a) \\ & =\frac{n b-n a}{m+n}\end{aligned}$
$\begin{aligned} \therefore O P & =O A+A P \\ & =a+\frac{n b-n a}{m+n} \\ & =\frac{m a+m a+n b-n a}{m+n} \\ & =\frac{m a+n b}{m+n} \end{aligned}$
$\begin{aligned} & =\frac{n}{m+n}(\overrightarrow{O B}-O \vec{A}) \\ & =\frac{n}{m+n}(b-a) \\ & =\frac{n b-n a}{m+n}\end{aligned}$
$\begin{aligned} \therefore O P & =O A+A P \\ & =a+\frac{n b-n a}{m+n} \\ & =\frac{m a+m a+n b-n a}{m+n} \\ & =\frac{m a+n b}{m+n} \end{aligned}$
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