Vectors 2 Question 24
24. If $A _1, A _2, \ldots, A _n$ are the vertices of a regular plane polygon with $n$ sides and $O$ is its centre. Then, show that $\sum _{i=1}^{n-1}\left(\overrightarrow{\mathbf{O A}} _i \times \overrightarrow{\mathbf{O A}} _{i+1}\right)=(1-n)\left(\overrightarrow{\mathbf{O A}} _2 \times \overrightarrow{\mathbf{O A}} _1\right)$.
(1982, 2M)
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Solution:
- Since, $\overrightarrow{\mathbf{O A}} _1, \overrightarrow{\mathbf{A A}} _2, \ldots, \overrightarrow{\mathbf{O A}} _n$ are all vectors of same magnitude and angle between any two consecutive vectors is same i.e. $(2 \pi / n)$.
$$ \therefore \quad \overrightarrow{\mathbf{O A}} _1 \times \overrightarrow{\mathbf{O A}} _2=a^{2} \cdot \sin \frac{2 \pi}{n} \cdot \hat{\mathbf{p}} $$
where, $\hat{\mathbf{p}}$ is perpendicular to plane of polygon.
Now, $\quad \sum _{i=1}^{n-1}\left(\overrightarrow{\mathbf{O A}}{ } _i \times \mathbf{O} \overrightarrow{\mathbf{A} _{i+1}}\right)=\sum _{i=1}^{n-1} a^{2} \cdot \sin \frac{2 \pi}{n} \cdot \hat{\mathbf{p}}$
$$ \begin{aligned} & =(n-1) \cdot a^{2} \cdot \sin \frac{2 \pi}{n} \cdot \hat{\mathbf{p}} \\ & =(n-1)\left[\overrightarrow{\mathbf{A A}} _1 \times \overrightarrow{\mathbf{O A}} _2\right] \\ & =(1-n)\left[\overrightarrow{\mathbf{O A}} _2 \times \overrightarrow{\mathbf{O A}} _1\right]=RHS \end{aligned} $$
