Question

If ABCDEF is a regular hexagon and \(\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF} = p\overline{AD} = q\overline{AO}\), where O is the center of the hexagon, then the values of \(p\) and \(q\) respectively are

• A. \(2,3\)
• B. \(4,6\)
• C. \(3,6\)
• D. \(3,5\)

Hint:

Add the five vectors from A using position vectors from the centre; the six vertices sum to zero.

Answer 1

The Correct Option is C

Step 1: Set up coordinates:
Take the centre O as the origin and let the radius (also the side) of the regular hexagon be a. Put A = (a, 0), B = (a/2, \(\frac{\sqrt3}{2}a\)), C = (-a/2, \(\frac{\sqrt3}{2}a\)), D = (-a, 0), E = (-a/2, \(-\frac{\sqrt3}{2}a\)), F = (a/2, \(-\frac{\sqrt3}{2}a\)).

Step 2: Find each vector:
\(\vec{AB} = (-\frac a2, \frac{\sqrt3a}{2})\), \(\vec{AC} = (-\frac{3a}{2}, \frac{\sqrt3a}{2})\), \(\vec{AD} = (-2a, 0)\), \(\vec{AE} = (-\frac{3a}{2}, -\frac{\sqrt3a}{2})\), \(\vec{AF} = (-\frac a2, -\frac{\sqrt3a}{2})\).

Step 3: Add:
The y parts cancel in pairs. The x parts add to \(-\frac a2 - \frac{3a}{2} - 2a - \frac{3a}{2} - \frac a2 = -6a\). So the sum is \((-6a, 0)\).

Step 4: Compare:
\(\vec{AD} = (-2a,0)\), so the sum is \(3\vec{AD}\), giving \(p = 3\). \(\vec{AO} = (-a, 0)\), so the sum is \(6\vec{AO}\), giving \(q = 6\).

Step 5: Check the options:
Options (A) 2,3 and (B) 4,6 break the relation \(AD = 2AO\), because \(q = 2p\) must hold. Option (D) 3,5 also breaks \(q = 2p\). Only (C) 3,6 keeps it.

Final Answer:
\(p = 3\) and \(q = 6\), option (C). \[ \boxed{p = 3,\ q = 6} \]