Question

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

Answer 1

ABCD is the tetrahedron.

Therefore, E falls outside of AF and lies on AF produced and such that AF = FE.

Therefore, the position vector of E is

2(2i) - (i + j + k) = 3i - j - k

This is one position of E (taken as E₁). 

Another position of E, taken as E₂, lies on FA produced, and for that position, position vector of E(E₂) which is an external division such that FA/FE = 2/3. Therefore,

This is a second possible position for E.