Mathematics

Problem

OABCDE

is a regular hexagon of side 2 units in the

XY

-plane.

O

being the origin and

OA

taken along the

X

-axis. A point

P

is taken on a line parallel to

Z

-axis through the centre of the hexagon at a distance of 3 units from

O

. Then, the vector

\mathbf{AP}

is a.

-\hat{i} + 3\hat{j} + \sqrt{5}\hat{k}

\hat{i} - \sqrt{3}\hat{j} + 5\hat{k}

-\hat{i} + \sqrt{3}\hat{j} + \sqrt{5}\hat{k}

\hat{i} + \sqrt{3}\hat{j} + \sqrt{5}\hat{k}

Intuitive explanation

A regular hexagon has beautiful symmetry—its center lies at a predictable distance from each vertex. The key is to first locate the center of the hexagon using the geometry of regular hexagons, then find where point $P$ sits directly above (or below) that center. Once you have coordinates for both $A$ and $P$ , the vector $\mathbf{AP}$ is just their difference. The constraint that $P$ is 3 units from $O$ pins down its exact position in 3D space.

Intuitive explanation

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