OABCDE is a regular hexagon of side 2 units in the XY-plane in the first quadrant. O being the origin and OA taken along the x-axis. A point P is taken on a line parallel to the z-axis through the centre of the hexagon at a distance of 3 unit from O in the positive Z direction. Then find vector AP.

Options

A. `-hati+3hatj+sqrt(5)hatk`
B. `hati-sqrt(3)hatj+5hatk`
C. `-hati+sqrt(3)hatj+sqrt(5)hatk`
D. `hati+sqrt(3)hatj+sqrt(5)hatk`

Answer

C

Explanation

.
Here, coordinate of Q are `(2cos60^(@),2sin60^(@))`
`impliesQ(1,sqrt(3),0)`
`thereforeP(1,sqrt(3)z)`
`OP=3`
`implies sqrt(1+3+z^(2))=3` or `z^(2)=5`
`z=sqrt(5)`
`therefore P(1,sqrt(3),sqrt(5))implies OP=hati+sqrt(3)hatj+sqrt(5)hatk`
Now, `AP=OP-OA=hati+sqrt(3)hatj+sqrt(5)hatk-2hati`
`=-hati+sqrt(3)hatj+sqrt(5)hatk`.