A1, A2, ..., An are the vertices of a...

MathematicsVector (Cross) ProductJEE Advanced 1982Moderate

$A_{1}, A_{2}, \dots, A_{n}$ are the vertices of a regular plane polygon with $n$ sides and $O$ is its centre. Show that $\sum_{i = 1}^{n - 1} (O A_{i} \times O A_{i + 1}) = (1 - n) (O A_{2} \times O A_{1})$ .

Visualized Solution (Hindi)

Visualizing the Regular Polygon

Consider a regular polygon with $n$ sides and center $O$ .
Let the vertices be $A_{1}, A_{2}, ..., A_{n}$ .
All vectors $O A_{i}$ have the same magnitude: $∣ O A_{i} ∣ = a$ for all $i$ .

The Angle Between Consecutive Vectors

The total angle around the center $O$ is $2 π$ .
Due to symmetry, the angle between any two consecutive vectors $O A_{i}$ and $O A_{i + 1}$ is $θ = \frac{2 π}{n}$ .

Defining the Cross Product

The cross product of two vectors is $A \times B = ∣ A ∣∣ B ∣ sin θ \overset{n}{^}$ .
For any $i$ , $O A_{i} \times O A_{i + 1} = a^{2} sin (\frac{2 π}{n}) \hat{P}$ , where $\hat{P}$ is the unit vector normal to the plane.

Summing the Terms

The summation is $\sum_{i = 1}^{n - 1} (O A_{i} \times O A_{i + 1})$ .
Since each term is equal to $a^{2} sin (\frac{2 π}{n}) \hat{P}$ , the sum becomes:
$(n - 1) a^{2} sin (\frac{2 π}{n}) \hat{P}$

The Anti-commutative Property

Recall the anti-commutative property: $A \times B = - (B \times A)$ .
Thus, $O A_{1} \times O A_{2} = - (O A_{2} \times O A_{1})$ .
Substituting this: $a^{2} sin (\frac{2 π}{n}) \hat{P} = - (O A_{2} \times O A_{1})$ .

Final Substitution and Conclusion

Substitute back into the sum:
Sum $= (n - 1) [- (O A_{2} \times O A_{1})]$
Sum $= - (n - 1) (O A_{2} \times O A_{1})$
Sum $= (1 - n) (O A_{2} \times O A_{1})$
Q.E.D. Hence Proved.

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Visualized Solution (Hindi)

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