JEE Main202523 Jan 2025Morning ShiftMathematicsVector AlgebraActual
Let the position vectors of the vertices A, B and C of a tetrahedron A B C D 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 A B C meets the median line segment through A of the triangle A B C at the point E . If the length of A D is 110 3 and the volume of the tetrahedron is 805 6 2 , then the position vector of E is
Options
- A1 12 (7 i +4 j +3 k )
- B1 2 ( i +4 j +7 k )
- C1 6 (12 i +12 j + k )
- D1 6 (7 i +12 j + k )
Correct answer
D. 1 6 (7 i +12 j + k )
Step-by-step solution
Area of ABC = 1 2 | AB AC |= 1 2 |5 i +3 j + k |= 1 2 35 volume of tetrahedron aligned & = 1 3 Base area h = 805 6 2 & 1 3 1 2 35 h = 805 6 2 & ~h = 23 2 aligned AE ^2= AD ^2- DE ^2= 13 18 AE = 13 18 aligned & AE =| AE | ( i -5 k 26 ) & = 13 18 ( i -5 k 26 ) & = 13 18 ( i -5 k 26 )= i -5 k 6 & P.V. of E = i -5 k 6 + i +2 j + k = 1 6 (7 i +12 j + k ) aligned