Question

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 positfons

Answer 1

As, F is the mid point of BC. So, `vecF=2hati`
As, AE is perpendicular to DE. Let E divides AF in the ratio `lambda:1`
`vecE=(2lambdahati+1(hati+ hatj +hatk))/(lambda+1)=(2lambda+1)/(lambda+1)hati+1/(lambda+1)hatj+1/(lambda+1)hatk`
Volume of tetrahedron=1/3*Area of base*Height=`2sqrt2/3`
Area of base=`1/2|vec(BC)Xvec(BA)|=sqrt2`
Hence, `1/3*sqrt2*DE=2sqrt2/3`
So, `DE=2`
As, `/_ADE` is right angled triangle.
So, `AD^2=AE^2 + DE^2`
`AE^2=12`
`vec(AE)=vecE - vecA=lambda/(lambda+1)hati - lambda/(lambda+1)hatj - lambda/(lambda+1)hatk`
`|vec(AE)|^2=12`
`3lambda^2/(lambda+1)^2=12`
On solving we get,
`lambda=-2 or -2/3`
Position vector correspond to `lambda=-2`is
`vecE=3hati-hatj-hatk`
Position vector correspond to `lambda=-2/3`is
`vecE=-hati+3hatj+3hatk`