The position vectors of the vertices A, B and C of a tetrahedron ABCD are `hat i + hat j + hat k`, `hat k `, `hat i` and `hat 3i`,respectively. The altitude from vertex D to the opposite face ABC meets the median line through Aof triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is2√2/3, find the position vectors of the point E for all its possible positions

Answer

`-hati +_3hatj + 3hatk or 3hati-hatj-hatk`

Explanation

we are given AD=4
Volume of tetrahedron = `(2sqrt2)/3`
`Rightarrow 1/3 (" Area of "triangleABC)p=(2sqrt2)/3`
`1/2|vec(BA) xx vec(BC)|p=2sqrt2`
`1/2 |(hatj +hatk) xx2hati|p=2sqrt2`
`or |hatj - hatk| p= 2 sqrt2`
`or sqrt2 p =2 sqrt2or p=2`
we have to find the P.V of point E.Let it divide median AF in the ratio `lambda:1`
` P.V. "of" E = (lambda .2hati + (hati +hatj +hatk)/(lambda +1)`
`vec(AE)` P.V or E - P.V. of `A=(lambda(hati-hatj-hatk))/(lamda+1)`
`|vec(AE)|^(2)=3(lamda/(lambda+1))^(2)`
Now ` 4+3 (lamda/(lambda + 1))=16`
`(lamda/(lambda+_1))=+-2`
` lamda=-2 or -2//3`
Putting the value of `lamda` in (i), we get the P.V. of possible
Positions of E as `-hati+3hatj+3hatk or 3hati-hatj-hatk`