A pyramid with vertex at point P has a regular hexagonal base ABCDEF. Position vectors of points A and B are `hati and hati+ 2hatj`, respectively. The centre of the base has the position vector `hati+hatj+sqrt3hatk`.
Altitude drawn from P on the base meets the diagonal AD at point G. Find all possible vectors of G. It is given that the volume of the pyramid is `6sqrt3` cubic units and AP is 5 units.

Answer

`-(hati +2hatj+2sqrt3hatk), 3hati+2hatj +2sqrt3hatk`

Explanation

Let the centre of the base be (0). Therefore,
`" "|vec(AB)|=2`
`Delta OAB= (1)/(4)xx4xxsqrt3= sqrt3`
Base area `= 6sqrt3` sq. Units
Let height of the pyramid be h. Therefore,
`(1)/(3) xx 6sqrt3 h = 6 sqrt3 or h = 3units `

It is given that `|vec(AP)| = 5`. Therefore,
`" "AG= sqrt (25-9)= 4` units
`rArr |vec(AG)|= 4` units
Now `|vec(AG)| and |vec(AO)|` are collinear. Therefore
`" "vec(AG) = lamda vec(AO) rArr |vec(AG)| = |lamda||vec(AO)|`
or `" "2|lamda|= 4 or |lamda| = 2`
`rArr vec(AG)" "= pm 2(hati+hatj+ sqrt3hatk)`
`rArr vec(OG) = pm 2 (hati+hatj+ sqrt3hatk)+hati`
`" "= - (hatj+2hatj+ 2sqrt3hatk), 3hati+2hatj+2sqrt3hatk`