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.

Explanation

`G-= (hati+sqrt(3)hatj)`
Let position vector of P be `vecP`
`because " "vec(GP)"||"hatk`
Then `" "vecp-(hati+sqrt(3)hatj)=lamdahatk`
`therefore " "vecp=hati +sqrt(3) hatj+lamdahatk`
Also`" "|vec(OP)|=3`
`rArr" "sqrt(1+3+lamda^(2))=3`
or `" "lamda^(2)=5`
or `" "lamda=pmsqrt(5)`
`rArr" "vecp=hati+sqrt(3)hatjpmsqrt(5) hatk`
For positive z-axis, `vecp=hati+sqrt(3)hatj+sqrt(5)hatk`
So `" "vec(AP)=vecp-2hati=-hati+sqrt(3)hatj+sqrt(5)hatk`