Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be i+2 j+ k, i+3 j-2 k and 2 i+ j- k respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is 1103 and the volume of the tetrahedron is 8056 2, then the position vector of E is
- 16 (7 i+12 j+ k )
- 112 (7 i+4 j+3 k )
- 16 (12 i+12 j+ k )
- 12 ( i+4 j+7 k )
Answer
(1) 16 (7 i+12 j+ k )
Solution
AB AC=-5 i-3 j- k area of ABC= 352
[figure]
Volume = 13(area) DE DE= 232, and DE plane, so in right ADE: AE2= 1109- 232= 1318
Mid-point of BC: M ( 32,2,- 32 ), AM= 12(1,0,-5), and E lies on AM
AE= 1318 (1,0,-5) 26= 16(1,0,-5)
E= 16 (7 i+12 j+ k )
Answer: (1)